The ProblemComplete the function printList that accepts a node (.pdf
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The Problem Complete the function printList that accepts a node * (the head of a linked list) as a parameter and print out the list in the following format: If the list is empty then the function should print \"Empty list\". You should complete the main function to test your printList in, at least, the following cases: An empty list A list with exactly one node A list with more than one node (eg: 100 -> 200 -> 300, as shown above) //main.cpp #include using namespace std; struct node { int val; node *next; }; void printList(node *head) { // printList function } int main() { // Test 1: An empty list node * head = NULL; printList(head); // Test 2: A list with exactly one node // Test 3: A list with more than one node return 0; } Solution #include using namespace std; struct node { int val; node *next; }; void printList(node *head) { int i = 0; if(head == NULL) // if head is NUll it is empty list { cout<<\"\ Empty list\"; } while(head != NULL) //while end of the list is not reached { cout<<\"\ Node \"<val; head = head->next; // traverse next node } } int main() { // Test 1: An empty list cout<<\"\ list with no node\"; node *head = NULL; printList(head); // Test 2: A list with exactly one node cout<<\"\ list with one node\"; node *node1 = new node; //make new node node1->val= 100; //assign its value to 100 node1->next = NULL; //assign its next pointer to null head= node1; // make it the head node as head is null printList(head); // call printList function // Test 3: A list with more than one node cout<<\"\ list with three nodes\"; node *node2 = new node; node2->val= 200; node2->next = NULL; node1->next = node2; node *node3 = new node; node3->val= 300; node3->next = NULL; node2->next = node3; printList(head); return 0; }.
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Complete the provided partial C++ Linked List program. Main.cpp is g.pdf
Complete the provided partial C++ Linked List program. Main.cpp is given and Link list header
file is also given. The given testfile listmain.cpp is given for demonstration of unsorted list
functionality. The functions header file is also given. Complete the functions of the header file
linked_list.h below.
=========================================================
// listmain.cpp
#include \"Linked_List.h\"
int main(int argc, char **argv)
{
float f;
Linked_List *theList;
cout << \"Simple List Demonstration\ \";
cout << \"(List implemented as an Array - Do not try this at home)\ \ \";
cout << \"Create a list and add a few tasks to the list\";
theList = new Linked_List(); // Instantiate a list object
theList->Insert(5, 3.1f); // Note: The argument to the funtion should be a float
theList->Insert(1, 5.6f); // A constant real number like 3.1 is interpreted as
theList->Insert(3, 8.3f); // a double unless it is explicitly defined as a float
theList->Insert(2, 7.4f); // by adding an \'f\' to the end of the number.
theList->Insert(4, 2.5f);
// Show what is in the list
theList->PrintList();
// Test the list length function
cout << \"\ List now contains \" << theList->ListLength() << \"items.\ \ \";
// Test delete function
cout << \"Testing delete of last item in list.\ \";
theList->Delete(4);
theList->PrintList();
// Test delete function
cout << \"Testing delete of first item in list.\ \";
theList->Delete(5);
theList->PrintList();
// Test delete function
cout << \"Testing delete of a middle item in list.\ \";
theList->Delete(3);
theList->PrintList();
// Test delete function with a known failure argument
cout << \"Testing failure in delete function.\ \";
if(theList->Delete(4))
cout << \"Oops! Should not have been able to delete.\ \";
else
cout << \"Unable to locate item to delete.\ \";
// Test search (known failure)
cout << \"Testing Search function. Search for key 3\ \";
if(theList->Search(3, &f))
cout << \"Search result: theData = %f\ \", f;
else
cout << \"Search result: Unable to locate item in list\ \";
// Test search (known success)
cout << \"Testing Search function. Search for key 2\ \";
if(theList->Search(2, &f))
cout << \"Search result: theData = \" << f << \"\ \";
else
cout << \"Search result: Unable to locate item in list\ \";
cout << \"\ \ End list demonstration...\";
return 0;
}
=====================================================================
===================
// linked_list.h functions
#include
using namespace std;
// Define a structure to use as the list item
struct ListItem
{
int key;
float theData;
ListItem *next;
};
class Linked_List
{
private:
ListItem *head; // Pointer to head of the list
public:
Linked_List(); // Class constructor
~Linked_List(); // Class destuctor
void ClearList(); // Remove all items from the list
bool Insert(int key, float f);// Add an item to the list
bool Delete(int key); // Delete an item from the list
bool Search(int key, float *retVal); // Search for an item in the list
int ListLength(); // Return numb.
This assignment and the next (#5) involve design and development of a.pdf
This assignment and the next (#5) involve design and development of a sequential
non contiguous and dynamic datastructure called LinkedList. A linked list object is
a container consisting of connected ListNode objects. As before, we are not going
to use pre-fabricated classes from the c++ library, but construct the LinkedList
ADT from scratch.
The first step is construction and testing of the ListNode class. A ListNode object
contains a data field and a pointer to the next ListNode object (note the recursive
definition).
#This assignment requires you to
1. Read the Assignment 4 Notes
2. Watch the Assignment 4 Support video
3. Implement the following methods of the ListNode class
-custom constructor
-setters for next pointer and data
4. Implement the insert and remove method in the main program
5. Scan the given template to find the above //TODO and implement the code
needed
//TODO in ListNodecpp.h file
public: ListNode(T idata, ListNode<T> * newNext);
public: void setNext(ListNode<T> * newNext);
public: void setData(T newData);
// TODO in main program
void remove(ListNode<int> * &front,int value)
void insert(ListNode<int> * &front,int value)
# The driver is given ListNodeMain.cpp is given to you that does the following
tests
1. Declares a pointer called front to point to a ListNode of datatype integer
2. Constructs four ListNodes with data 1,2,4 and adds them to form a linked
list.
3. Inserts ListNode with data 3 to the list
4. Removes node 1 and adds it back to test removing and adding the first
element
5. Removes node 3 to test removing a middle node
6. Removes node 4 to test removing the last node
7. Attempt to remove a non existent node
8. Remove all existing nodes to empty the list
9. Insert node 4 and then node 1 to test if insertions preserve order
10.Print the list
Main.cpp
#include <iostream>
#include "ListNodecpp.h"
// REMEMBER each ListNode has two parts : a data field
// and an address field. The address is either null or points to the next node
//in the chain
//Requires: integer value for searching, address of front
//Effects: traverses the list node chain starting from front until the end comparing search value
with listnode getData. Returns the original search value if found, if not adds +1 to indicate not
found
//Modifies: Nothing
int search(ListNode<int> * front, int value);
//Requires: integer value for inserting, address of front
//Effects: creates a new ListNode with value and inserts in proper position (increasing order)in
the chain. If chain is empty, adds to the beginning
//Modifies: front, if node is added at the beginning.
//Also changes the next pointer of the previous node to point to the
//newly inserted list node. the next pointer of the newly inserted pointer
//points to what was the next of the previous node.
//This way both previous and current links are adjusted
//******** NOTE the use of & in passing pointer to front as parameter -
// Why do you think this is needed ?**********
void insert(ListNode<int> * &fr.
in C++ , Design a linked list class named IntegerList to hold a seri.pdf
in C++ , Design a linked list class named IntegerList to hold a series of integers. The class must
have public member functions for appending , inserting, deleting and displaying nodes. The
append function must append a node to the end of the list. The insert function must insert node in
ascending order. The delete function must delete the node with the given value. The display
function should display the list from head to tail. The IntegerList class should also have a
destructor that destroys the list and release memory. The list node must be declared as a structure
using keyword struct. This structure is a private member of IntegerList. You should also store the
head of the list as a private member.
Write a test program to test the implementation your IntegerList class works properly.
Solution
Here is the code for the above scenario:
#include
using namespace std;
struct ListNode
{
float value;
ListNode *next;
};
ListNode *head;
class LinkedList
{
public:
int insertNode(float num);
int appendNode(float num);
void deleteNode(float num);
void destroyList();
void displayList();
LinkedList(void) {head = NULL;}
~LinkedList(void) {destroyList();}
};
int LinkedList::appendNode(float num)
{
ListNode *newNode, *nodePtr = head;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head == NULL)
{
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while(nodePtr->next != NULL)
nodePtr = nodePtr->next;
nodePtr->next = newNode;
}
return 0;
}
int LinkedList::insertNode(float num)
{
struct ListNode *newNode, *nodePtr = head, *prevNodePtr = NULL;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head==NULL) {
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while((nodePtr != NULL) && (nodePtr->value < num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(prevNodePtr==NULL)
newNode->next = head;
else
newNode->next = nodePtr; prevNodePtr->next = newNode;
}
return 0;
}
void LinkedList::deleteNode(float num)
{
struct ListNode *nodePtr = head, *prevNodePtr = NULL;
if(head==NULL) {
cout << \"The list was empty!\ \";
return;
}
if(head->value == num) {
head = nodePtr->next;
delete [] nodePtr;
}
else
{
while((nodePtr!= NULL)&&(nodePtr->value != num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(nodePtr==NULL)
cout << \"The value \" << num << \" is not in this list!\ \";
else {
prevNodePtr->next = nodePtr->next;
delete [] nodePtr;
}
}
}
void LinkedList::destroyList()
{
struct ListNode *nodePtr = head, *nextNode = nodePtr;
if(head==NULL) {
cout << \"The list is empty!\ \";
return;
}
while (nodePtr != NULL) {
nextNode = nodePtr->next;
delete [] nodePtr;
nodePtr = nextNode;
}
}
void LinkedList.
PROBLEM STATEMENTIn this assignment, you will complete DoubleEnde.pdf
PROBLEM STATEMENT:
In this assignment, you will complete DoubleEndedList.java that implements the ListInterface as
well as an interface called DoubleEndedInterface which represents the list's entries by using a
chain of nodes that has both a head reference and a tail reference. Be sure to read through the
code and understand the implementation.
WHAT IS PROVIDED:
- A driver class to test your code. You should not modify this file!
- A list interface (ListInterface.java)
- A double ended interface (DoubleEndedInterface.java)
- An incomplete DoubleEndedList class (DoubleEndedList.java)
WHAT YOU NEED TO DO:
4. Complete the DoubleEndedList class
4. Run the driver and make sure your output is exactly the same as mine (at the bottom of
Driver.java)
\} // end else numberofEntries--; else throw new IndexOut0fBoundsException("Illegal position
given to remove operation."); return result; // Return removed entry }//endreturnreve public T
replace(int givenPosition, T newEntry) \{ T replace(int givenPosition, T newEntry) \{ if
((givenPosition >=1)&& (givenPosition <= numberOfEntries)) \{ // Assertion: The list is not
empty Node desiredNode = getNodeAt (givenPosition); ToriginalEntry = desiredNode.getData();
desiredNode.setData(newEntry); return originalEntry; f // end if else throw new
IndexOut0fBoundsException("Illegal position given to replace operation."); replace if ((
givenPosition >=1)&& (givenPosition <= number0fEntries)) \{ // Assertion: The list is not
empty Node desiredNode = getNodeAt ( givenPosition); T originalEntry = desiredNode.
getData( ); desiredNode.setData(newEntry); return originalentry; \} // end if throw new
Index0ut0fBoundsException("Illegal position given to replace operation."); \} // end replace
public T getEntry(int givenPosition) \{ if ((givenPosition >=1) \&\& (givenPosition < =
number0fEntries ) ) \{ // Assertion: The list is not empty return getNodeAt (givenPosition).
getData(); else // end if throw new IndexOut0fBoundsException("Illegal position given to
getEntry operation."); \} // end getEntry public boolean contains ( T anEntry) \{ boolean found =
false; Node currentNode = firstNode; while (!found && (currentNode != null)) \{ if
(anEntry.equals (currentNode.getData())) else found = true; \} // end while currentNode =
currentNode. getNextNode () ; return found; \} // end contains public int getLength() \{ return
numberofEntries; \} // end getLength public boolean isEmpty() \{ return number0fEntries ==0;
\} // end isEmpty public T[] toArray() \{ // The cast is safe because the new array contains null
entries aSuppressWarnings ("unchecked") T[] result =(T[]) new 0bject [numberofEntries]; //
Unchecked cast int index =0; Node currentNode = firstNode; while ((index < numberOfEntries)
\&\& (currentNode != null)) \& result [ index ]= currentNode. getData () ; currentNode =
currentNode.getNextNode ( ); index++; 3 // end while return result; 3 // end toArray // Returns a
reference to the node at a given position. // Precondition: L.
Homework 05 - Linked Lists (C++)(1) Implement the concepts of a un.pdf
Homework 05 - Linked Lists (C++)
(1) Implement the concepts of a union, and difference as defined noted in Chapter 1 questions 6,
and 8 using a linked list. You should not use arrays, dynamic arrays or the STL vector or STL
list classes.
You will be linking a series of nodes together to implement a single linked list class. Define a
Node as a struct that holds a character (like they did in the book) and a pointer to another Node:
struct Node {
char data;
Node* next;
};
Then define a LinkedList class which has (at a minimum) these member functions:
LinkedList();
~LinkedList();
bool insertAtFront();
bool insertBeforePosition(int index);
bool insertAtBack();
bool deleteAtFront();
bool deleteBeforePosition(int index);
bool deleteAtBack();
Implement each of these functions to provide the correct functionality. These functions return
true if successful in inserting or deleting, otherwise they return false (indicating the operation
was not successful).
Finally, create an overloaded + operator to handle the Union of two linked lists and an
overloaded - operator to handle the Difference of two linked lists.
Because we are dealing with pointers you should have both a LinkedList Constructor and
Destructor. Remember that you do not directly call a Constructor or Destructor Function. The
Destructor is automatically called when the variable loses scope or the program ends.
Remember, that we are dealing with not just one dynamically allocated Node (with the new
operator), but many, so you will have to start at the head of the list and go until the Node points
to nullptr. Then keep deleting the previous Node pointer until there are no Nodes left to delete.
(2) To verify your set class, write a main function that takes two lines of characters from the
input file input.txt and store each line of characters into two separate linked lists. Then using
these two lists, perform the Union and set Difference of the characters from the file, and print the
results to the console.
(3) Please also complete an asymptotic (Big O) analysis of your insertAtFront() and
insertAtBack() member functions. Place this in a file called analysis.txt.
Solution
Here is the complete code for your question. Please do rate the aswer if you are happy with
program. Please read comments in the program. Also pay special attention to comments in + and
- operators. You WILL NEED to comment out if condition in + if you want all elements in both
lists irrespective of if they duplicates or no. For now implemented to have true behaviour of no
duplicates in union.
check the functions by creating some lists and playing around different fucntions like its done in
main and commented out.
As far as complexity:
for insertAtFront() its O(1) since it does not depend on the size of the list. It just replaces the start
of the list. Hence it is O(1) and independent of the size of list.
for insertAtBack(), its O(n) since it depends on the number of elements in the list. It should
traverse till end of list and .
Recommended
C Homework Help
I am Andrew O. I am a C Homework Expert at programminghomeworkhelp.com. I hold a master’s in Programming from, the University of Southampton, UK. I have been helping students with their homework for the past 10 years. I solve homework related to C.
Visit programminghomeworkhelp.com or email support@programminghomeworkhelp.com.
You can also call on +1 678 648 4277 for any assistance with C Homework.
C Exam Help
I am Gabriel C. I am a C Exam Expert at programmingexamhelp.com. I hold a PhD. in Business analyst of Information Technology, Montreal College of Information Technology, Canada. I have been helping students with their exams for the past 8 years. You can hire me to take your exam in C.
Visit programmingexamhelp.com or email support@programmingexamhelp.com. You can also call on +1 678 648 4277 for any assistance with the C Exam.
Complete the provided partial C++ Linked List program. Main.cpp is g.pdf
Complete the provided partial C++ Linked List program. Main.cpp is given and Link list header
file is also given. The given testfile listmain.cpp is given for demonstration of unsorted list
functionality. The functions header file is also given. Complete the functions of the header file
linked_list.h below.
=========================================================
// listmain.cpp
#include \"Linked_List.h\"
int main(int argc, char **argv)
{
float f;
Linked_List *theList;
cout << \"Simple List Demonstration\ \";
cout << \"(List implemented as an Array - Do not try this at home)\ \ \";
cout << \"Create a list and add a few tasks to the list\";
theList = new Linked_List(); // Instantiate a list object
theList->Insert(5, 3.1f); // Note: The argument to the funtion should be a float
theList->Insert(1, 5.6f); // A constant real number like 3.1 is interpreted as
theList->Insert(3, 8.3f); // a double unless it is explicitly defined as a float
theList->Insert(2, 7.4f); // by adding an \'f\' to the end of the number.
theList->Insert(4, 2.5f);
// Show what is in the list
theList->PrintList();
// Test the list length function
cout << \"\ List now contains \" << theList->ListLength() << \"items.\ \ \";
// Test delete function
cout << \"Testing delete of last item in list.\ \";
theList->Delete(4);
theList->PrintList();
// Test delete function
cout << \"Testing delete of first item in list.\ \";
theList->Delete(5);
theList->PrintList();
// Test delete function
cout << \"Testing delete of a middle item in list.\ \";
theList->Delete(3);
theList->PrintList();
// Test delete function with a known failure argument
cout << \"Testing failure in delete function.\ \";
if(theList->Delete(4))
cout << \"Oops! Should not have been able to delete.\ \";
else
cout << \"Unable to locate item to delete.\ \";
// Test search (known failure)
cout << \"Testing Search function. Search for key 3\ \";
if(theList->Search(3, &f))
cout << \"Search result: theData = %f\ \", f;
else
cout << \"Search result: Unable to locate item in list\ \";
// Test search (known success)
cout << \"Testing Search function. Search for key 2\ \";
if(theList->Search(2, &f))
cout << \"Search result: theData = \" << f << \"\ \";
else
cout << \"Search result: Unable to locate item in list\ \";
cout << \"\ \ End list demonstration...\";
return 0;
}
=====================================================================
===================
// linked_list.h functions
#include
using namespace std;
// Define a structure to use as the list item
struct ListItem
{
int key;
float theData;
ListItem *next;
};
class Linked_List
{
private:
ListItem *head; // Pointer to head of the list
public:
Linked_List(); // Class constructor
~Linked_List(); // Class destuctor
void ClearList(); // Remove all items from the list
bool Insert(int key, float f);// Add an item to the list
bool Delete(int key); // Delete an item from the list
bool Search(int key, float *retVal); // Search for an item in the list
int ListLength(); // Return numb.
This assignment and the next (#5) involve design and development of a.pdf
This assignment and the next (#5) involve design and development of a sequential
non contiguous and dynamic datastructure called LinkedList. A linked list object is
a container consisting of connected ListNode objects. As before, we are not going
to use pre-fabricated classes from the c++ library, but construct the LinkedList
ADT from scratch.
The first step is construction and testing of the ListNode class. A ListNode object
contains a data field and a pointer to the next ListNode object (note the recursive
definition).
#This assignment requires you to
1. Read the Assignment 4 Notes
2. Watch the Assignment 4 Support video
3. Implement the following methods of the ListNode class
-custom constructor
-setters for next pointer and data
4. Implement the insert and remove method in the main program
5. Scan the given template to find the above //TODO and implement the code
needed
//TODO in ListNodecpp.h file
public: ListNode(T idata, ListNode<T> * newNext);
public: void setNext(ListNode<T> * newNext);
public: void setData(T newData);
// TODO in main program
void remove(ListNode<int> * &front,int value)
void insert(ListNode<int> * &front,int value)
# The driver is given ListNodeMain.cpp is given to you that does the following
tests
1. Declares a pointer called front to point to a ListNode of datatype integer
2. Constructs four ListNodes with data 1,2,4 and adds them to form a linked
list.
3. Inserts ListNode with data 3 to the list
4. Removes node 1 and adds it back to test removing and adding the first
element
5. Removes node 3 to test removing a middle node
6. Removes node 4 to test removing the last node
7. Attempt to remove a non existent node
8. Remove all existing nodes to empty the list
9. Insert node 4 and then node 1 to test if insertions preserve order
10.Print the list
Main.cpp
#include <iostream>
#include "ListNodecpp.h"
// REMEMBER each ListNode has two parts : a data field
// and an address field. The address is either null or points to the next node
//in the chain
//Requires: integer value for searching, address of front
//Effects: traverses the list node chain starting from front until the end comparing search value
with listnode getData. Returns the original search value if found, if not adds +1 to indicate not
found
//Modifies: Nothing
int search(ListNode<int> * front, int value);
//Requires: integer value for inserting, address of front
//Effects: creates a new ListNode with value and inserts in proper position (increasing order)in
the chain. If chain is empty, adds to the beginning
//Modifies: front, if node is added at the beginning.
//Also changes the next pointer of the previous node to point to the
//newly inserted list node. the next pointer of the newly inserted pointer
//points to what was the next of the previous node.
//This way both previous and current links are adjusted
//******** NOTE the use of & in passing pointer to front as parameter -
// Why do you think this is needed ?**********
void insert(ListNode<int> * &fr.
in C++ , Design a linked list class named IntegerList to hold a seri.pdf
in C++ , Design a linked list class named IntegerList to hold a series of integers. The class must
have public member functions for appending , inserting, deleting and displaying nodes. The
append function must append a node to the end of the list. The insert function must insert node in
ascending order. The delete function must delete the node with the given value. The display
function should display the list from head to tail. The IntegerList class should also have a
destructor that destroys the list and release memory. The list node must be declared as a structure
using keyword struct. This structure is a private member of IntegerList. You should also store the
head of the list as a private member.
Write a test program to test the implementation your IntegerList class works properly.
Solution
Here is the code for the above scenario:
#include
using namespace std;
struct ListNode
{
float value;
ListNode *next;
};
ListNode *head;
class LinkedList
{
public:
int insertNode(float num);
int appendNode(float num);
void deleteNode(float num);
void destroyList();
void displayList();
LinkedList(void) {head = NULL;}
~LinkedList(void) {destroyList();}
};
int LinkedList::appendNode(float num)
{
ListNode *newNode, *nodePtr = head;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head == NULL)
{
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while(nodePtr->next != NULL)
nodePtr = nodePtr->next;
nodePtr->next = newNode;
}
return 0;
}
int LinkedList::insertNode(float num)
{
struct ListNode *newNode, *nodePtr = head, *prevNodePtr = NULL;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head==NULL) {
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while((nodePtr != NULL) && (nodePtr->value < num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(prevNodePtr==NULL)
newNode->next = head;
else
newNode->next = nodePtr; prevNodePtr->next = newNode;
}
return 0;
}
void LinkedList::deleteNode(float num)
{
struct ListNode *nodePtr = head, *prevNodePtr = NULL;
if(head==NULL) {
cout << \"The list was empty!\ \";
return;
}
if(head->value == num) {
head = nodePtr->next;
delete [] nodePtr;
}
else
{
while((nodePtr!= NULL)&&(nodePtr->value != num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(nodePtr==NULL)
cout << \"The value \" << num << \" is not in this list!\ \";
else {
prevNodePtr->next = nodePtr->next;
delete [] nodePtr;
}
}
}
void LinkedList::destroyList()
{
struct ListNode *nodePtr = head, *nextNode = nodePtr;
if(head==NULL) {
cout << \"The list is empty!\ \";
return;
}
while (nodePtr != NULL) {
nextNode = nodePtr->next;
delete [] nodePtr;
nodePtr = nextNode;
}
}
void LinkedList.
PROBLEM STATEMENTIn this assignment, you will complete DoubleEnde.pdf
PROBLEM STATEMENT:
In this assignment, you will complete DoubleEndedList.java that implements the ListInterface as
well as an interface called DoubleEndedInterface which represents the list's entries by using a
chain of nodes that has both a head reference and a tail reference. Be sure to read through the
code and understand the implementation.
WHAT IS PROVIDED:
- A driver class to test your code. You should not modify this file!
- A list interface (ListInterface.java)
- A double ended interface (DoubleEndedInterface.java)
- An incomplete DoubleEndedList class (DoubleEndedList.java)
WHAT YOU NEED TO DO:
4. Complete the DoubleEndedList class
4. Run the driver and make sure your output is exactly the same as mine (at the bottom of
Driver.java)
\} // end else numberofEntries--; else throw new IndexOut0fBoundsException("Illegal position
given to remove operation."); return result; // Return removed entry }//endreturnreve public T
replace(int givenPosition, T newEntry) \{ T replace(int givenPosition, T newEntry) \{ if
((givenPosition >=1)&& (givenPosition <= numberOfEntries)) \{ // Assertion: The list is not
empty Node desiredNode = getNodeAt (givenPosition); ToriginalEntry = desiredNode.getData();
desiredNode.setData(newEntry); return originalEntry; f // end if else throw new
IndexOut0fBoundsException("Illegal position given to replace operation."); replace if ((
givenPosition >=1)&& (givenPosition <= number0fEntries)) \{ // Assertion: The list is not
empty Node desiredNode = getNodeAt ( givenPosition); T originalEntry = desiredNode.
getData( ); desiredNode.setData(newEntry); return originalentry; \} // end if throw new
Index0ut0fBoundsException("Illegal position given to replace operation."); \} // end replace
public T getEntry(int givenPosition) \{ if ((givenPosition >=1) \&\& (givenPosition < =
number0fEntries ) ) \{ // Assertion: The list is not empty return getNodeAt (givenPosition).
getData(); else // end if throw new IndexOut0fBoundsException("Illegal position given to
getEntry operation."); \} // end getEntry public boolean contains ( T anEntry) \{ boolean found =
false; Node currentNode = firstNode; while (!found && (currentNode != null)) \{ if
(anEntry.equals (currentNode.getData())) else found = true; \} // end while currentNode =
currentNode. getNextNode () ; return found; \} // end contains public int getLength() \{ return
numberofEntries; \} // end getLength public boolean isEmpty() \{ return number0fEntries ==0;
\} // end isEmpty public T[] toArray() \{ // The cast is safe because the new array contains null
entries aSuppressWarnings ("unchecked") T[] result =(T[]) new 0bject [numberofEntries]; //
Unchecked cast int index =0; Node currentNode = firstNode; while ((index < numberOfEntries)
\&\& (currentNode != null)) \& result [ index ]= currentNode. getData () ; currentNode =
currentNode.getNextNode ( ); index++; 3 // end while return result; 3 // end toArray // Returns a
reference to the node at a given position. // Precondition: L.
Homework 05 - Linked Lists (C++)(1) Implement the concepts of a un.pdf
Homework 05 - Linked Lists (C++)
(1) Implement the concepts of a union, and difference as defined noted in Chapter 1 questions 6,
and 8 using a linked list. You should not use arrays, dynamic arrays or the STL vector or STL
list classes.
You will be linking a series of nodes together to implement a single linked list class. Define a
Node as a struct that holds a character (like they did in the book) and a pointer to another Node:
struct Node {
char data;
Node* next;
};
Then define a LinkedList class which has (at a minimum) these member functions:
LinkedList();
~LinkedList();
bool insertAtFront();
bool insertBeforePosition(int index);
bool insertAtBack();
bool deleteAtFront();
bool deleteBeforePosition(int index);
bool deleteAtBack();
Implement each of these functions to provide the correct functionality. These functions return
true if successful in inserting or deleting, otherwise they return false (indicating the operation
was not successful).
Finally, create an overloaded + operator to handle the Union of two linked lists and an
overloaded - operator to handle the Difference of two linked lists.
Because we are dealing with pointers you should have both a LinkedList Constructor and
Destructor. Remember that you do not directly call a Constructor or Destructor Function. The
Destructor is automatically called when the variable loses scope or the program ends.
Remember, that we are dealing with not just one dynamically allocated Node (with the new
operator), but many, so you will have to start at the head of the list and go until the Node points
to nullptr. Then keep deleting the previous Node pointer until there are no Nodes left to delete.
(2) To verify your set class, write a main function that takes two lines of characters from the
input file input.txt and store each line of characters into two separate linked lists. Then using
these two lists, perform the Union and set Difference of the characters from the file, and print the
results to the console.
(3) Please also complete an asymptotic (Big O) analysis of your insertAtFront() and
insertAtBack() member functions. Place this in a file called analysis.txt.
Solution
Here is the complete code for your question. Please do rate the aswer if you are happy with
program. Please read comments in the program. Also pay special attention to comments in + and
- operators. You WILL NEED to comment out if condition in + if you want all elements in both
lists irrespective of if they duplicates or no. For now implemented to have true behaviour of no
duplicates in union.
check the functions by creating some lists and playing around different fucntions like its done in
main and commented out.
As far as complexity:
for insertAtFront() its O(1) since it does not depend on the size of the list. It just replaces the start
of the list. Hence it is O(1) and independent of the size of list.
for insertAtBack(), its O(n) since it depends on the number of elements in the list. It should
traverse till end of list and .
Help please, I have attached LinkedList.cpp and LinkedList.hPlease.pdf
Help please, I have attached LinkedList.cpp and LinkedList.h
Please help with the implementation of the Stack.h and Stack.cpp as well
LinkedList.cpp
~~~~~~~~~~~~~~
#include \"LinkedList.h\"
LinkedList::LinkedList()
{
first = last = NULL;
}
LinkedList::~LinkedList()
{
while (first != NULL)
{
Node *temp;
temp = first;
first = first->next;
free(temp);
}
last = NULL;
}
void LinkedList::insertAtBack(int valueToInsert)
{
Node *temp = (Node *)malloc(sizeof(Node));
temp->val = valueToInsert;
temp->next = NULL;
if (last == NULL)
first = last = temp;
else
{
last->next = temp;
last = temp;
}
}
bool LinkedList::removeFromBack()
{
if (first == NULL)
return false;
else if (first == last)
{
free(first);
first = last = NULL;
}
else
{
Node *temp;
temp = first;
while (temp->next != last)
temp = temp->next;
last = temp;
temp = temp->next;
free(temp);
last->next = NULL;
}
return true;
}
void LinkedList::print()
{
Node *temp = first;
while (temp != last)
{
cout << temp->val << \" \";
temp = temp->next;
}
if (temp != NULL)
cout << temp->val;
}
bool LinkedList::isEmpty()
{
if (first == last)
return true;
return false;
}
int LinkedList::size()
{
int count = 0;
Node *temp = first;
if (temp == NULL)
return count;
while (temp != last)
{
count++;
temp = temp->next;
}
return count + 1;
}
void LinkedList::clear()
{
while (first != NULL)
{
Node *temp;
temp = first;
first = first->next;
free(temp);
}
last = NULL;
}
/*Exercise2 */
void LinkedList::insertAtFront(int valueToInsert)
{
Node *temp = (Node *)malloc(sizeof(Node));
temp->val = valueToInsert;
if (last == NULL)
first = last = temp;
else
{
temp->next = first;
first = temp;
}
}
bool LinkedList::removeFromFront()
{
if (first == NULL)
return false;
Node *temp = first;
first = first->next;
free(temp);
if (first == NULL)
last = NULL;
return true;
}
LinkedList.h
~~~~~~~~~~~
#include \"LinkedList.h\"
LinkedList::LinkedList()
{
first = last = NULL;
}
LinkedList::~LinkedList()
{
while (first != NULL)
{
Node *temp;
temp = first;
first = first->next;
free(temp);
}
last = NULL;
}
void LinkedList::insertAtBack(int valueToInsert)
{
Node *temp = (Node *)malloc(sizeof(Node));
temp->val = valueToInsert;
temp->next = NULL;
if (last == NULL)
first = last = temp;
else
{
last->next = temp;
last = temp;
}
}
bool LinkedList::removeFromBack()
{
if (first == NULL)
return false;
else if (first == last)
{
free(first);
first = last = NULL;
}
else
{
Node *temp;
temp = first;
while (temp->next != last)
temp = temp->next;
last = temp;
temp = temp->next;
free(temp);
last->next = NULL;
}
return true;
}
void LinkedList::print()
{
Node *temp = first;
while (temp != last)
{
cout << temp->val << \" \";
temp = temp->next;
}
if (temp != NULL)
cout << temp->val;
}
bool LinkedList::isEmpty()
{
if (first == last)
return true;
return false;
}
int LinkedList::size()
{
int count = 0;
Node *temp = first;
if (temp == NULL)
return count;
while (temp != last)
{
count++;
temp = temp->next;
}
return count + 1;
}
void LinkedList::clear()
{
while (first != NULL)
{
Node *.
When we first examined the array based and node based implementations.docx
When we first examined the array based and node based implementations
C++ problemPart 1 Recursive Print (40 pts)Please write the recu.pdf
C++ problem
Part 1: Recursive Print (40 pts)
Please write the recursive List reverse print function, whose iterative version we wrote in class.
Below are the function signatures for the functions you are going to need:
public:
/**Additional Operations*/
void print_reverse();
//Wrapper function that calls the reverse helper function to print a list in reverse
//prints nothing if the List is empty
private:
void reverse(Nodeptr node);
//Helper function for the public printReverse() function.
//Recursively prints the data in a List in reverse.
Why do we need the private helper function here?
Since we are going to be reversing our list node by node, in a recursive fashion, we want to pass
a one node at a time to our reverse function.
However, since our nodes are private, we cannot access them if we call the function inside of
main.
Add these function signatures to your List.h file along with your other function prototypes inside
the class definition.
Make sure that you place the reverse function inside the private portion of your List class
definition and the print_reverse function prototype to the public portion of your List class
definition.
Now, implement these two functions inside of List.h, under your section for additional
operations.
Important: Test each function carefully inside of your ListTest.cpp to make sure that it is
working properly.
Part 2: Adding an Index to Your List Nodes (20 pts)
Next, you will add the following functions to your List.h
/**Accessor Functions*/
int get_index();
//Indicates the index of the Node where the iterator is currently pointing
//Nodes are numbered from 1 to length of the list
//Pre: length != 0
//Pre: !off_end()
...
int List::get_index()
{
//Implement the function here
}
/**Manipulation Procedures*/
void scroll_to_index(int index);
//Moves the iterator to the node whose index is specified by the user
//Pre: length != 0
...
void scroll_to_index(int index)
{
//Implement function here
}
Part 3: Implementing Search as Part of Your List (40 pts)
Now, we are going to add two search functions to our List so that we can search for elements in
our List.
The first of these functions is going to be a simple linear search function.
You will need to add the following function prototype and function definition to your List.h:
/**Additional Operations*/
int linear_search(listitem item);
//Searchs the list, element by element, from the start of the List to the end of the List
//Returns the index of the element, if it is found in the List
//Returns -1 if the element is not in the List
//Pre: length != 0
...
int List::linear_search(listitem item)
{
//Implement the function here
}
You are also going to add a function to perform recursive binary search on your List.
You will need to add the following function prototype and function definition to your List.h:
int binary_search(int low, int high, listitem item);
//Recursively searchs the list by dividing the search space in half
//Returns the index of the element, if it is fo.
Please find the answer to the above problem as follows- Program.pdf
Please find the answer to the above problem as follows:-
/* Program to insert in a sorted list */
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
void sortedDelete(struct node** head_ref, int number)
{
struct node *temp = *head_ref;
struct node *prev = NULL;
if(*head_ref != NULL && (*head_ref)->data == number){
*head_ref = (*head_ref)->next;
return;
}
while(temp != NULL)
{
if(temp->data == number){
prev->next = temp->next;
free(temp);
return;
}
prev = temp;
temp = temp->next;
}
}
/* function to insert a new_node in a list. Note that this
* function expects a pointer to head_ref as this can modify the
* head of the input linked list (similar to push())*/
void sortedInsert(struct node** head_ref, struct node* new_node)
{
struct node* current;
/* Special case for the head end */
if (*head_ref == NULL || (*head_ref)->data >= new_node->data)
{
if(*head_ref != NULL&&(*head_ref)->data==new_node->data){
return;
}
new_node->next = *head_ref;
*head_ref = new_node;
}
else
{
/* Locate the node before the point of insertion */
current = *head_ref;
while (current->next!=NULL &&
current->next->data < new_node->data)
{
current = current->next;
}
if(current->next!=NULL && current->next->data == new_node->data){
return;
}
new_node->next = current->next;
current->next = new_node;
}
}
/* BELOW FUNCTIONS ARE JUST UTILITY TO TEST sortedInsert */
/* A utility function to create a new node */
struct node *newNode(int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
new_node->next = NULL;
return new_node;
}
/* Function to print linked list */
void printList(struct node *head)
{
struct node *temp = head;
if(temp == NULL)
printf(\"\ \");
while(temp != NULL)
{
printf(\"%d \", temp->data);
temp = temp->next;
}
}
/* Drier program to test count function*/
int main(int argc, char *argv[])
{
/* Start with the empty list */
struct node* head = NULL;
struct node *new_node;
FILE *fp;
fp = fopen(argv[1], \"r+\");
if(!fp){
printf(\"error\");
return 1;
}
char op = \' \';
int number = 0;
while(fscanf(fp, \"%c\\t%d\ \", &op, &number)!=EOF){
if(op==\'i\'){
new_node = newNode(number);
sortedInsert(&head, new_node);
}else if(op==\'d\'){
sortedDelete(&head, number);
}else{
printf(\"error\");
return;
}
}
printList(head);
printf(\"\ \");
return 0;
}
Solution
Please find the answer to the above problem as follows:-
/* Program to insert in a sorted list */
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
void sortedDelete(struct node** head_ref, int number)
{
struct node *temp = *head_ref;
struct node *prev = NULL;
if(*head_ref != NULL && (*head_ref)->data == number){
*head_ref = (*head_ref)->next;
return;
}
while(temp != NULL)
{
if(temp->data == number){
prev->next = temp->next;
free(temp);
return;
}
prev = temp;
temp = temp->next;
}
}
/* function to insert a new_node in a list. Note that this
* function expects a point.
Data Structures in C++I am really new to C++, so links are really .pdf
Data Structures in C++
I am really new to C++, so links are really hard topic for me. It would be nice if you can provide
explanations of what doubly linked lists are and of some of you steps... Thank you
In this assignment, you will implement a doubly-linked list class, together with some list
operations. To make things easier, you’ll implement a list of int, rather than a template class.
Solution
A variable helps us to identify the data. For ex: int a = 5; Here 5 is identified through variable a.
Now, if we have collection of integers, we need some representation to identify them. We call it
array. For ex: int arr[5]
This array is nothing but a Data Structure.
So, a Data Structure is a way to group the data.
There are many Data Structures available like Arrays, Linked List, Doubly-Linked list, Stack,
Queue, etc.
Doubly-Linked list are the ones where you can traverse from the current node both in left and
right directions.
Why so many different types of Data Structures are required ?
Answer is very simple, grouping of data, storage of data and accessing the data is different.
For example, in case of Arrays we store all the data in contiguous locations.
What if we are not able to store the data in contiguous locations because we have huge data.
Answer is go for Linked List/Doubly-Linked list.
Here we can store the data anywhere and link the data through pointers.
I will try to provide comments for the code you have given. May be this can help you.
#pragma once
/*
dlist.h
Doubly-linked lists of ints
*/
#include
class dlist {
public:
dlist() { }
// Here we are creating a NODE, it has a integer value and two pointers.
// One pointer is to move to next node and other to go back to previous node.
struct node {
int value;
node* next;
node* prev;
};
// To return head pointer, i.e. start of the Doubly-Linked list.
node* head() const { return _head; }
// To return Tail pointer, i.e. end of the Doubly-Linked list.
node* tail() const { return _tail; }
// **** Implement ALL the following methods ****
// Returns the node at a particular index (0 is the head).
node* at(int index){
int cnt = 0;
struct node* tmp = head();
while(tmp!=NULL)
{
if (cnt+1 == index)
return tmp;
tmp = tmp->next;
}
}
// Insert a new value, after an existing one
void insert(node *previous, int value){
// check if the given previous is NULL
if (previous == NULL)
{
printf(\"the given previous node cannot be NULL\");
return;
}
// allocate new node
struct node* new_node =(struct node*) malloc(sizeof(struct node));
// put in the data
new_node->data = new_data;
// Make next of new node as next of previous
new_node->next = previous->next;
// Make the next of previous as new_node
previous->next = new_node;
// Make previous as previous of new_node
new_node->prev = previous;
// Change previous of new_node\'s next node
if (new_node->next != NULL)
new_node->next->prev = new_node;
}
// Delete the given node
void del(node* which){
struct node* head_ref = head();
/* base case */
if(*head_ref == NUL.
finish the following code based on green line requirement us.pdf
finish the following code based on green line requirement use c++
void addIntToEndOfList(LinkedList list, int value) { assert(list != NULL); // if list is NULL, we can do
nothing. Node *p; // temporary pointer // TOD0: // (1) Allocate a new node, p will point to it. // (2)
Set p's data field to the value passed in // (3) Set p's next field to NULL if (list->head == NULL) { //
(4) Make both head and tail of this list point to p } else { // Add p at the end of the list. // (5) The
current node at the tail? Make it point to p instead of NULL }void addIntToStartOfList(LinkedList
list, int value) assert(list != NULL); // if list is NULL, we can do nothing, // Add code for this. // HINT:
// consider all edge cases such as when list->head is or is not null AND // consider all edge cases
such as when list->tail is or is not null. // Visualizing the problem with a box and pointer diagram
can help. / // list: ptr to a linked list of Node (each with int data, and Node next) // Return a pointer
to node with the largest value. // You may assume list has at least one element // If more than one
element has largest value, // break tie by returning a pointer to the one the occurs // earlier in the
list, i.e. closer to the head.
#include sstream #include linkylist.h #include iostream.pdf
#include
#include \"linkylist.h\"
#include
#include
#include
#include
using namespace std;
int main(int argc, char *argv[])
{
//node* head = new node;
LinkedList list(1);
ifstream infile1 (argv[1]);
//make sure the file opens properly and read the contents of the file in to a list
if(!infile1)
{
cout<<\"Your Testfile could not be opened!\"<>users;
if (users.find(\"I\")!=string::npos)
{
//int spot = atoi((users[3]).c_str());
string pot = string(users.substr(2));
stringstream spot(pot);
string data;
cin>>data;
insertAfter(list.getNode(spot),data);
}
/*else if (users.find(\"D\")!=string::npos)
{
int spoty = atoi((users[3]).c_str());
deleteNode(spoty);
}*/
else if (users==\"L\")
{
list.traverse_and_print();
}
}
return 0;
#include
using namespace std;
class Node
{
friend class LinkedList;
private:
int value; /* data, can be any data type, but use integer for easiness */
Node *Next; /* pointer to the next node */
public:
/* Constructors with No Arguments */
Node(void)
: Next(NULL)
{ }
/* Constructors with a given value */
Node(int val)
: value(val), Next(NULL)
{ }
/* Constructors with a given value and a link of the next node */
Node(int val, Node* next)
: value(val), Next(next)
{}
/* Getters */
int getValue(void)
{ return value; }
Node* getNext(void)
{ return Next; }
};
/* definition of the linked list class */
class LinkedList
{
private:
/* pointer of head node */
Node *Head;
/* pointer of tail node */
Node *Tail;
public:
/* Constructors with a given value of a list node */
LinkedList(int val);
/* Destructor */
~LinkedList(void);
/* Function to append a node to the end of a linked list */
void tailAppend(string val);
/* Remove a node with a specific value if it exists */
void remove(int val);
void insertAfter(Node* afterMe, string val);
Node* getNode(int pos);
/* Traversing the list and printing the value of each node */
void traverse_and_print();
};
#include
#include \"linkylist.h\"
#include
using namespace std;
LinkedList::LinkedList(int val)
{
/* Create a new node, acting as both the head and tail node */
Head = new Node(val);
Tail = Head;
}
Node* LinkedList::getNode(int pos)
{
Node* p = Head;
/* Counter */
while (pos--) {
if (p != NULL)
p = p->Next;
else
return NULL;
}
return p;
}
void LinkedList::insertAfter(Node* afterMe, string val)
{
if(afterme != NULL)
{
afterMe->Next = new Node(val, afterMe->Next);
}
}
LinkedList::~LinkedList()
{
}
void LinkedList::tailAppend(string val)
{
/* The list is empty? */
if (Head == NULL) {
/* the same to create a new list with a given value */
Tail = Head = new Node(val);
}
else
{
/* Append the new node to the tail */
Tail->Next = new Node(val);
/* Update the tail pointer */
Tail = Tail->Next;
}
}
void LinkedList::remove(int val)
{
Node *pPre = NULL, *pDel = NULL;
/* Check whether it is the head node?
if it is, delete and update the head node */
if (Head->value == val) {
/* point to the node to be deleted */
pDel = Head;
/* update */
Head = pDel->Next;
delete pDel;
return;
}
pPre = Head;
pDel = Head->Next;
/*.
C++ Doubly-Linked ListsThe goal of the exercise is to implement a.pdf
C++: Doubly-Linked Lists
The goal of the exercise is to implement a class List of doubly- linked lists.
My goal is to implement the class in a file doubly-linked.cpp.
The class List implements lists using the following structures for list ele- ments:
struct Node { int val ;
Node next;
Node prev; };
Each Node contains a value (an integer) and two pointers: one meant to point to the next element
of the list and one meant to point to the previous element in the list. The class provides the
following methods that you need to implement:
• A constructor and a destructor, with no arguments;
• void insert(int n): insert an integer at the end of the list;
• void reverse(): reverse the list;
• void print(): print the list to cout in the same format as the input (i.e. integers separated by
spaces);
doubly-linked.h
#ifndef __dll__
#define __dll__
#include
using namespace std;
// Basic structure to store elements of a list
struct Node {
int val; // contains the value
Node * next; // pointer to the next element in the list
Node * prev; // pointer to the previous element in the list
};
// Class List
class List {
public:
List(void); // Constructor
~List(void); // Destructor
void insert(int n); // This should insert n in the list
void reverse(void); // This should reverse the list
void print(void); // This shoiuld print the list
private:
Node * first; // Pointer to the first (if any) element in the list
};
#endif
main.cpp
#include
#include \"doubly-linked.h\"
using namespace std;
int main(void){
List l;
int n;
while(cin >> n){
l.insert(n);
}
// Print list as read from cin
l.print();
// Reverse the list and print it
l.reverse();
l.print();
// Reverse again and print it
l.reverse();
l.print();
return 0;
}
Solution
#include
using namespace std;
/* Linked list structure */
struct list {
struct list *prev;
int data;
struct list *next;
} *node = NULL, *first = NULL, *last = NULL, *node1 = NULL, *node2 = NULL;
class linkedlist {
public:
/* Function for create/insert node at the beginning of Linked list */
void insrt_frnt() {
list *addBeg = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addBeg->data;
if(first == NULL) {
addBeg->prev = NULL;
addBeg->next = NULL;
first = addBeg;
last = addBeg;
cout << \"Linked list Created!\" << endl;
}
else {
addBeg->prev = NULL;
first->prev = addBeg;
addBeg->next = first;
first = addBeg;
cout << \"Data Inserted at the beginning of the Linked list!\" << endl;
}
}
/* Function for create/insert node at the end of Linked list */
void insrt_end() {
list *addEnd = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addEnd->data;
if(first == NULL) {
addEnd->prev = NULL;
addEnd->next = NULL;
first = addEnd;
last = addEnd;
cout << \"Linked list Created!\" << endl;
}
else {
addEnd->next = NULL;
last->next = addEnd;
addEnd->prev = last;
last = addEnd;
cout << \"Data Inserted at the end of the Linked list!\" << endl;
}
}
/* Function for Display Linked list */
void display() {
node = first;
cout << \"List of data in L.
In C++Write a recursive function to determine whether or not a Lin.pdf
In C++
Write a recursive function to determine whether or not a Linked List is in sorted order (smallest
value to largest value).
Add the following function prototypes to your List class under the section for access functions,
then implement the functions below the class definition:
public:
/**Access Functions*/
bool isSorted();
//Wrapper function that calls the isSorted helper function to determine whether
//a list is sorted in ascending order.
//We will consider that a list is trivially sorted if it is empty.
//Therefore, no precondition is needed for this function
private:
bool isSorted(Nodeptr node);
//Helper function for the public isSorted() function.
//Recursively determines whether a list is sorted in ascending order.
#include //for NULL
#include
#include
using namespace std;
template //list stores generic list data, not any specific C++ type
class List
{
private:
struct Node
{
listdata data;
Node* next;
Node* previous;
Node(listdata data): data(data), next(NULL), previous(NULL){}
};
typedef struct Node* Nodeptr;
Nodeptr first;
Nodeptr last;
Nodeptr iterator;
int size;
public:
/**Constructors and Destructors*/
List();
//Default constructor; initializes and empty list
//Postcondition: numeric values equated to zero, or strings should be empty.
List(const List &list);
~List();
//Destructor. Frees memory allocated to the list
//Postcondition: position NodePtr at next Node
/**Accessors*/
listdata getFirst();
//Returns the first element in the list
//Precondition:NodePtr points to the first node on list
listdata getLast();
//Returns the last element in the list
//Precondition: make new node first on the list
listdata getIterator();
bool isEmpty();
//Determines whether a list is empty.
int getSize();
//Returns the size of the list
/**Manipulation Procedures*/
void startIterator();
void advanceIterator();
void removeLast();
//Removes the value of the last element in the list
//Precondition: list is not empty, not the first element.
//Postcondition: one remaining node
void removeIterator();
void removeFirst();
//Removes the value of the first element in the list
//Precondition: list is not empty
//Postcondition: no nodes left
void insertLast(listdata data);
//Inserts a new element at the end of the list
//If the list is empty, the new element becomes both first and last
//Postcondition: next equal to null
void insertFirst(listdata data);
//Inserts a new element at the start of the list
//If the list is empty, the new element becomes both first and last
//Postcondition:point nodePtr next node on list
/**Additional List Operations*/
bool offEnd();
void printList();
//Prints to the console the value of each element in the list sequentially
//and separated by a blank space
//Prints nothing if the list is empty
void insertIterator(listdata data);
bool operator==(const List &list);
};
// constructor definition
template
List::List(): first(NULL), last(NULL), iterator(NULL), size(0) {}
template
List::List(const List &list): size(list.size)
{
if(list..
I need to implment a function that can reverse a single linked list..pdf
I need to implment a function that can reverse a single linked list. Not just print it, but actually
change the order of the list.
THIS MUST BE DONE USING RECURSION
EX: a, b, c, d -> d, c, b, a
You have access to the start of the list. Code must be in C++
Solution
#include
using namespace std;
struct node
{
char data;
struct node* next;
};
void reverse(struct node* head)
{
if (head == NULL) //base case
return;
reverse(head->next); //calling function recursively
cout<data<<\",\";
}
//add data to the front of the linked list
void add(struct node** head_ref, char data)
{
//create new node
struct node* newnode=new node();
//adding data
newnode->data = data;
//adding node to the front
newnode->next = (*head_ref);
//making new node head
(*head_ref) = newnode;
}
int main()
{
// create linked list a, b, c, d
struct node* head = NULL;
add(&head, \'d\');
add(&head, \'c\');
add(&head, \'b\');
add(&head, \'a\');
reverse(head);
return 0;
}
//OUTPUT***********
d,c,b,a,
//OUTPUT***********
//This code has been tested on g++ compiler,Please ask in case of any doubt,Thanks..
Using the C++ programming language1. Implement the UnsortedList cl.pdf
Using the C++ programming language
1. Implement the UnsortedList class to store a list of numbers that are input into the list from
data.txt.
- create a main.cpp file that gets the numbers from the file
- insert the number 7 into the list
- insert another number 300 into the list
- delete the number 6 from the list
- print out the following:
--the entire list
- the greatest
- the least
2. Attach the main.cpp, UnsortedList.cpp, the ItemType.h, and the output file two called
outfile2.txt
Use the files below:
// listDriver.cpp
// Test driver
#include
#include
#include
#include
#include
#include \"unsorted.h\"
using namespace std;
void PrintList(ofstream& outFile, UnsortedType& list);
int main()
{
ifstream inFile; // file containing operations
ofstream outFile; // file containing output
string inFileName; // input file external name
string outFileName; // output file external name
string outputLabel;
string command; // operation to be executed
int number;
ItemType item;
UnsortedType list;
bool found;
int numCommands;
// Prompt for file names, read file names, and prepare files
cout << \"Enter name of input command file; press return.\" << endl;
cin >> inFileName;
inFile.open(inFileName.c_str());
cout << \"Enter name of output file; press return.\" << endl;
cin >> outFileName;
outFile.open(outFileName.c_str());
cout << \"Enter name of test run; press return.\" << endl;
cin >> outputLabel;
outFile << outputLabel << endl;
if (!inFile)
{
cout << \"file not found\" << endl;
exit(2)
}
inFile >> command;
numCommands = 0;
while (command != \"Quit\")
{
if (command == \"PutItem\")
{
inFile >> number;
item.Initialize(number);
list.PutItem(item);
item.Print(outFile);
outFile << \" is in list\" << endl;
}
else if (command == \"DeleteItem\")
{
inFile >> number;
item.Initialize(number);
list.DeleteItem(item);
item.Print(outFile);
outFile << \" is deleted\" << endl;
}
else if (command == \"GetItem\")
{
inFile >> number;
item.Initialize(number);
item = list.GetItem(item, found);
item.Print(outFile);
if (found)
outFile << \" found in list.\" << endl;
else outFile << \" not in list.\" << endl;
}
else if (command == \"GetLength\")
outFile << \"Length is \" << list.GetLength() << endl;
else if (command == \"IsFull\")
if (list.IsFull())
outFile << \"List is full.\" << endl;
else outFile << \"List is not full.\" << endl;
else if (command == \"MakeEmpty\")
list.MakeEmpty();
else if (command == \"PrintList\")
PrintList(outFile, list);
else
cout << command << \" is not a valid command.\" << endl;
numCommands++;
cout << \" Command number \" << numCommands << \" completed.\"
<< endl;
inFile >> command;
};
cout << \"Testing completed.\" << endl;
inFile.close();
outFile.close();
return 0;
}
void PrintList(ofstream& dataFile, UnsortedType& list)
// Pre: list has been initialized.
// dataFile is open for writing.
// Post: Each component in list has been written to dataFile.
// dataFile is still open.
{
int length;
ItemType item;
list.ResetList();
length = list.GetLength();
for .
please help me in C++Objective Create a singly linked list of num.pdf
please help me in C++
Objective: Create a singly linked list of numbers based upon user input.
Program logic: Ask for a number, add that number to the front of the list, print the list. Repeat
until they enter -1 for the number. .
Sample Input: 10, 15, 5, 2, 4, -1 Output: 4, 2, 5, 15, 10
Solution
#include
#include
using namespace std;
// It contains the value and the pointer to the next element in the list.
struct element {
int value;
struct element *next;
};
struct element *head = NULL;
// The list is stored in the object of this class.
class linkedlist {
public:
// To display list, we first see if head is NULL,
// If head is NULL it means the list is empty.
// Then we traverse along the list and display each item, until the next pointer is NULL.
void display() {
if(head == NULL){
cout << \"The list is empty\ \";
} else {
struct element *temp;
temp = head;
while(temp != NULL){
cout << temp->value << \" \" ;
temp = temp->next;
}
cout << \"\ \";
}
}
// To add an element to the list,create a newElement and store the value.
// If head is NULL, add the newElement to the head.
// To add an element at the starting of the list.
// The next pointer of the newElement must point to the old head.
// The new head must be pointed to the newElement.
void add(int a){
struct element *newElement;
newElement = (struct element *)malloc(sizeof(struct element));
newElement->value = a;
newElement->next = NULL;
if(head == NULL) {
head = newElement;
} else {
newElement->next = head;
head = newElement;
}
}
};
int main()
{
// Create an object of the class linkedlist.
linkedlist l = linkedlist();
int a;
do {
cout << \"Enter a number : \";
cin >> a;
if(a!=-1) { // If the number is -1 do not add it to the list.
l.add(a); // Add the number to the list.
}
l.display(); // Display the list.
} while(a!=-1); // Keep asking for a number until -1 is pressed.
return 0;
}.
#include iostream #includestdlib.h using namespace std;str.pdf
#include
#include
using namespace std;
struct node { //Task 1
char val;
node *next;
};
void printList(node *head) { //Task 2
if (head == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head;
int count = 0;
while(temp != NULL) {
cout << \"Node \" << count << \": \" << temp ->val << endl;
count++;
temp = temp->next;
}
}
struct node *globalhead = NULL;
void deepCopy(node *head1, node *head2) { //Task 3
if (head1 == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head1;
node *curr = NULL;
node *prev = NULL;
while(temp != NULL) {
curr = (struct node *)malloc(sizeof(struct node));
curr->val = temp->val;
if(prev==NULL){
head2 = curr;
globalhead = curr;
}else{
prev->next = curr;
}
prev = curr;
temp = temp->next;
}
}
int main() { //Task 4
// Example #1
node p00, p01, p02, p03, p10, p11, p12, p13;
// List 1: A B B B
p00.val = \'A\'; p00.next = &p01;
p01.val = \'B\'; p01.next = &p02;
p02.val = \'B\'; p02.next = &p03;
p03.val = \'B\'; p03.next = NULL;
p10.next = NULL;
cout<<\"First List:\"<
Solution
//Tested on Windos OS Dev c++
/*********************LinkedList.cpp********************/
#include
#include
using namespace std;
struct node { //Task 1
char val;
node *next;
};
void printList(node *head) { //Task 2
if (head == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head;
int count = 0;
while(temp != NULL) {
cout << \"Node \" << count << \": \" << temp ->val << endl;
count++;
temp = temp->next;
}
}
struct node *globalhead = NULL;
void deepCopy(node *head1, node **head2) { //Task 3
if (head1 == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head1;
node *curr = NULL;
node *prev = NULL;
while(temp != NULL) {
curr = (struct node *)malloc(sizeof(struct node));
curr->val = temp->val;
curr->next = NULL;
if(*head2==NULL){
*head2=curr;
}else{
prev=*head2;
while(prev->next!=NULL){
prev=prev->next;
}
prev->next=curr;
}
temp = temp->next;
}
}
int main() { //Task 4
// Example #1
node p00, p01, p02, p03, p10, p11, p12, p13;
// List 1: A B B B
p00.val = \'A\'; p00.next = &p01;
p01.val = \'B\'; p01.next = &p02;
p02.val = \'B\'; p02.next = &p03;
p03.val = \'B\'; p03.next = NULL;
p10.next = NULL;
cout<<\"First List:\"<.
c++ Computational Complexity filling in the following three .pdf
c++ Computational Complexity
/* filling in the
* following three functions:
*
* 1) void disp(Node* head)
* Given a head pointer to a list, display the list using cout.
* If the list is empty, you should output \"HEAD->NULL\" (without the enclosing \").
* If the list contains one node whose data value is 3, you should output \"HEAD->3->NULL\".
* If the list contains three nodes and the data values are 1, 2, and 3, respectively, then you
should output
* \"HEAD->1->2->3->NULL\"
*
* 2) Node* add(Node* head, int data)
* Given the old head pointer to a list, create a new node using the value of data, and add the new
node to the front of
* the list. Return the new head pointer.
* Example: If the original list is \"HEAD->1->2->NULL\", then after adding 5, it should
become \"HEAD->5->1->2->NULL\". The
* return value should be the address of node 5.
*
* 3) void magic(Node* p1, Node* p2)
* p1 and p2 each points to a node from a same list. p1 may point to a later node than p2. Your
task is to calculate how
* many nodes are between p1 and p2. If there are k nodes in between (excluding nodes pointed
to by p1 and p2), then the
* result is k (k could be 0). Replace the data fields of all nodes between p1 and p2 by k.
* Example: The original list is \"HEAD->1->7->5->0->3->NULL\". p1 points to node 1 (the
node with value 1), p2 points to
* node 0. There are two nodes in between p1 and p2. These two nodes\' data fields need to be
replaced by k=2. The
* resulting list should be \"HEAD->1->2->2->0->3->NULL\".
*
* You should modify sections labelled as \"fill in here\". Please leave the other sections (e.g., the
main
* function and all function protocols) intact.
*/
#include
using namespace std;
struct Node {
int data;
Node* link;
};
void disp(Node* head) {
// fill in here
}
Node* add(Node* head, int data) {
// fill in here
}
void magic(Node* p1, Node* p2) {
// fill in here
}
main() {
Node* head = NULL;
Node *p4, *p7;
for(int i = 0; i < 10; i++) {
head = add(head, i);
if (i==4) p4 = head;
if (i==7) p7 = head;
}
// at this point, we have created the following list: HEAD->9->8->7->6->5->4->3->2->1->0-
>NULL
// p4 now points to node 4 (the node containing 4); p7 now points to node 7 (the node
containing 7)
magic(p4, p7);
disp(head);
// between p4 and p7, there are k=2 nodes, and these two nodes\' values will be replaced by
k=2
// the output should be: HEAD->9->8->7->2->2->4->3->2->1->0->NULL
// housekeeping (free all nodes)
Node* tmp;
while (head) {
tmp = head;
head = head->link;
delete tmp;
}
}
Comments
Expert Answer
c++ Computational Complexity
/* Your task is to fill in the
* following three functions:
*
* 1) void disp(Node* head)
* Given a head pointer to a list, display the list using cout.
* If the list is empty, you should output \"HEAD->NULL\" (without the enclosing \").
* If the list contains one node whose data value is 3, you should output \"HEAD->3->NULL\".
* If the list contains three nodes and the data values are 1, 2, and 3, respectively, then you
shou.
in Java (ignore the last line thats hidden) Create a doubly linked l.pdf
in Java (ignore the last line thats hidden) Create a doubly linked list whose nodes contain Strings.
Your list should include the following methods: Insert a node in the list in alphabetical order
Find a node that matches a String Traverse the list forwards and print Traverse the list backwards
and print Delete a node from the list Delete/destroy the list In addition to creating the class(es)
for your linked list, you\'ll need to create a main method that thoroughly tests each function you
build to show that it works. You should do these things in
Solution
/ Java program to create , insert , traverse and delete a node from doubly linked list
class LinkedList {
static Node head = null;
class Node {
int data;
Node next, prev;
Node(int d) {
data = d;
next = prev = null;
}
}
/* Let us create the doubly linked list 10<->8<->4<->2 */
list.push(head, 2);
list.push(head, 4);
list.push(head, 8);
list.push(head, 10);
System.out.println(\"Original Linked list \");
list.printList(head);
/*Function to delete a node in a Doubly Linked List.
head_ref --> pointer to head node pointer.
del --> pointer to node to be deleted. */
void deleteNode(Node head_ref, Node del) {
/* base case */
if (head == null || del == null) {
return;
}
/* If node to be deleted is head node */
if (head == del) {
head = del.next;
}
/* Change next only if node to be deleted is NOT the last node */
if (del.next != null) {
del.next.prev = del.prev;
}
/* Change prev only if node to be deleted is NOT the first node */
if (del.prev != null) {
del.prev.next = del.next;
}
/* Finally, free the memory occupied by del*/
return;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning of the Doubly Linked List */
void push(Node head_ref, int new_data) {
/* allocate node */
Node new_node = new Node(new_data);
/* since we are adding at the begining,
prev is always NULL */
new_node.prev = null;
/* link the old list off the new node */
new_node.next = (head);
/* change prev of head node to new node */
if ((head) != null) {
(head).prev = new_node;
}
/* move the head to point to the new node */
(head) = new_node;
}
/*Function to print nodes in a given doubly linked list
This function is same as printList() of singly linked lsit */
void printList(Node node) {
while (node != null) {
System.out.print(node.data + \" \");
node = node.next;
}
}
public static void main(String[] args) {
LinkedList list = new LinkedList();
/* delete nodes from the doubly linked list */
list.deleteNode(head, head); /*delete first node*/
list.deleteNode(head, head.next); /*delete middle node*/
list.deleteNode(head, head.next); /*delete last node*/
System.out.println(\"\");
/* Modified linked list will be NULL<-8->NULL */
System.out.println(\"Modified Linked List\");
list.printList(head);
}
}.
Java AssignmentUsing the ListNode.java file below Write method.pdf
Java Assignment
Using the ListNode.java file below
Write methods called min and max that return the smallest and largest values in the linked list.
These methods will be added to your ListNode class. For example if a variable called list stores
{11, -7, 3, 42, 0, 14], the call of list.min() should return -7 and the call of list.max() should return
42. If the list is empty, return -1. Print the returned value.
Write a method called insertNode that inserts a new node anywhere in your linked list. Display
your linked list with new value.
Write a program to test these methods to ensure accuracy and successful implementation. Upload
your modified version of the ListNode java file only.
Note: Original code should still be included. You will make additions and slight modifications
where necessary. Throw exceptions where you think is necessary. // Listlode is a class for storing
a single node of a linked // list. This node class is for a list of integer values. public class Listlode
\{ public int data; //data stored in this node public Listliode front;//points to head node public
Listliode next; /f link to next node in the list // post: constructs a node with data and null link
public Listlode() \{ 3. // post: constructs a node with given data and null link: public Listlode(int
data) \{ this(data, null); 3 // post: constructs a node with given data and given link. public
Listlode(int data, Listliode next) \{ this.data = data; this. next = next; 3 public String toString() \{
if (front = null) : \} else \{ String result = " ["+ front data; Listliode current = front. next; while
(current != null) \{ result +=m,m+ current data, current = current. next; 3 result += " ]n - return
result; 3. }// post: appends the given value to the end of the list public void add(int value) \&
puolic void aod if (front = nult) \{. front = new Listlode (value); 3 else \{ Listlode current =
front; whistiode current = front; while (current next I= null) \{ current = current. next; 3 current.
next = new Listliode (value ); 3. 3 J// post : returns the position of the first occurrence of the
given /l value (-1 if not found ) public void remove (int index) \{ if ( index =){ \} else \{ Listiode
current = nodeAt ( index - 1); current. next = current. next. next, 3 3 3// pre : <=i< size() // post:
returns a reference to the node at the given index public Listlode nodeAt (int index) : List Node
current = front; for (int i=0;i< index; i++) \{ current = current. next. \}. return current; 3.
--INSTRUCTION- --It helps to first create if-then-else structure to fi.pdf
//INSTRUCTION:
//It helps to first create if-then-else structure to fill in later.
//- Make sure PURPOSE and PARAMETER comments are given in detail
// Add sufficient comments to your code body to describe what it does.
// - Make sure all if-then-else are commented describing which case it is
// - Make sure all local variables are described fully with their purposes
//Complete the TODO parts.
// ====================================================
//Complier: g++
//File type: linkedList.cpp implementation file
//=====================================================
#include<iostream>
#include"linkedlist.h"
using namespace std;
//cunstructor that initializes an empty list
LinkedList::LinkedList() {
front = nullptr ;
rear = nullptr ;
count = 0;
}
/**
* @brief Destructor to destroy all nodes and release memory
*/
LinkedList::~LinkedList() {
Node *p, *q;
p = front;
while (p != nullptr ) {
q = p->next;
delete p;
p = q;
}
}
/**
* @brief Purpose: Checks if the list is empty
* @return true if the list is empty, false otherwise
*/
bool LinkedList::isEmpty() const {
// TODO: Add code here
return count == 0;
}
/**
* @brief Get the number of nodes in the list
* @return int The number of nodes in the list
*/
int LinkedList::length() const {
return count;
}
/**
* @brief Convert the list to a string
*
*/
string LinkedList::toString() {
string str = "[";
Node *ptr = front;
if (ptr != nullptr ) {
// Head node is not preceded by separator
str += to_string(ptr->val);
ptr = ptr->next;
}
while (ptr != nullptr ) {
str += ", " + to_string(ptr->val);
ptr = ptr->next;
}
str += "]";
return str;
}
/**
* @brief Displays the contents of the list
*/
void LinkedList::displayAll() {
cout << toString() << endl;
}
//Add a node to the end of the list
void LinkedList::addRear(T val) {
Node *p = new Node;
p->val = val;
p->next = nullptr ;
if (isEmpty()) {
front = p;
}
else {
rear->next = p;
}
rear = p;
count++;
// consider the two cases of whether the list was empty
}
//Add a node to the front of the list
void LinkedList::addFront(T val) {
Node *p = new Node;
p->val = val;
p->next = front;
if (isEmpty()) {
rear = p;
}
front = p;
count++;
// consider the two cases of whether the list was empty
}
//Delete the first node of the list
bool LinkedList::deleteFront(T &OldNum) {
if (isEmpty()) {
return false ;
}
Node *p = front;
OldNum = p->val;
front = front->next;
if (front == nullptr ) {
rear = nullptr ;
}
delete p;
count--;
return true ;
// consider if the list was empty and return false if the list is empty
// consider the special case of deleting the only node in the list
}
//Delete the last node of the list
bool LinkedList::deleteRear(T &OldNum) {
if (isEmpty()) {
return false ;
}
Node *p = rear;
OldNum = p->val;
if (front == rear) {
front = rear = nullptr ;
}
else {
Node *q = front;
while (q->next != rear) {
q = q->next;
}
rear = q;
q->next = nullptr ;
}
delete p;
count--;
return true ;
// consider if the list was empty and return false if the list is empty
// consider the specia.
H&M The Challenges of Global Expansion and the Move to adopt Intern.pdf
H&M: The Challenges of Global Expansion and the Move to adopt International Financial
Reporting standards Hennes & Mauritz AB (also known as H&M) , the Swedish MNE that is a
trendsetter in the latest fashion trends, has a stated goal “to give customers unbeatable value by
offering fashion and quality at the best price.” It doesn’t own any factories, but rather outsources
production to independent suppliers, primarily in Asia and Europe. H&M also rents space from
international and local landlords rather than owning its own stores. H&M is a major firm in the
apparel retail market where fashion trends are critical and where goods move quickly. In this
industry, the key buyers are consumers, and the key suppliers are clothing manufacturers and
wholesalers, and designers are king and a fast, well-organized supply chain is essential.
Depending on the individual firm strategy, the apparel retail market doesn’t have to be capital
intensive, but the largest players in the industry are very international, both in retail footprint and
suppliers. The biggest companies in the industry are U.S.-based The Gap, H&M, and Spain-
based Industria de Diseno Textil, S.A. (Inditex), better known by its flagship brand, Zara. All
three companies have different store brands: Gap, Banana Republic, Old Navy, and Athleta for
The Gap; H&M, COS (collection of style), Monki, Weekday, and Cheap Monday for H&M; and
Zara, Bershka, Pull and bear, Massimo Dutti, Stradivarius, Oysho, Zara Home and Uterque for
Inditex. Both of H&M’s competitors are very international. H&M operates about 2,800 stores in
49 markets, whereas The Gap operates 3,100 company stores and over 300 franchise stores
worldwide, and Inditex operates in a network of over 5,000 stores in 77 countries. Zara has the
largest geographic spread within Inditex with stores in 74 countries. Hennes & Mauritz AB
started as a single women’s wear store in Sweden in 1947. Today, H&M’s business is much
broader and currently includes the sales of clothing, accessories, footwear, cosmetics, and home
textiles. Although H&M is known as one of Sweden’s premier MNEs, it generates 21.5% of its
sales in Germany, 8.9% in the United States, and 7.4% in the United Kingdom, compared to only
5.8% of total sales in Sweden. H&M and Zara have very different strategies. Zara delivers new
products to its stores twice a week. Because of its highly organized supply chain, Zara only takes
10–15 days to go from design to the stores. Although it sources its apparel from around the
world, it has adopted just-in-time manufacturing from the auto industry and established 14 highly
automated Spanish factories where robots cut and dye fabrics creating the unfinished “gray
goods” which are the foundation for their final products. It then takes the gray goods and
outsources them to a network of small shops in Portugal and Spain to do the finish work. Store
managers are constantly sending updated information on consumer demand so that they can
move to the n.
C++ ProgrammingYou are to develop a program to read Baseball playe.pdf
C++ Programming
You are to develop a program to read Baseball player statistics from an input file. Each player
should bestored in a Player object. Therefore, you need to reuse the Player class from program 1.
Each playerwill have a first name, last name, a position (strings) and a batting average (floating
point number).
This program adds on to the requirements for Program #1. This will read int he same type of
player data from an input file and write the results to an output file.
New Requirements
You are to reuse your Player class from Program1. You are to implement a PlayerList data
structure/data type that stores players in a Linked List structure (internal to the class). Store the
players in the list in alphabetical order (By lastname, use firstname to further distinguish in the
event there is more than 1 player with the same last name).
You will need to add the class PlayerList and the class Node to facilitate the linked list
operations.
You must implement the following operations on your PlayerList along with any other utility
functionsyou might need. You may also need to add operations to the class Player to facilitate
comparisons sincePlayer data is private.
Operations
Default Constructor
Destructor
Add a Player to the List
Iterate through the List so that you can get each player
Clear out the list to make it empty
Test if a list isEmpty
Get the size of the list
Find a Player in the List (get) - Put a couple of hard-coded calls in your main program to make
sure this feature works.
Summary of Operation
Prompt the user for the input and output file names. DO NOT hardcode file names into your
program.
Open input file
Read each player and add them to your PlayerList Keep track of the number of players in the list
Store the players in the list in alphabetical order (lastname then firstname in the event of
twopeople with the same last name). You should be adding them in between other nodes
asappropriate
Test your find operation a couple of times. This can be hardcoded into your program. You donot
have to prompt the user for players, unless you wish to. Just make sure you test for playersthat
exist and that do not
Open an output file Write each player from the list into the output file, along with any other
output required by theassignment
Split your program code files into the appropriate .cpp and .h files for each of your
classes.(Exception - your NODE class can be defined in the same file as the PlayerList).
Detailed comments describing what is happening
Here is the input file:
Chipper Jones 3B 0.303
Rafael Furcal SS 0.281
Hank Aaron RF 0.305
Here is my player.h file
// Class Definition: Player
// Version 1
// This class defines an object to hold information about
// baseball player for use in a player database.
//--------------------------------------------------------
#include
using namespace std;
class Player {
string m_firstName; // player\'s first name
string m_lastName; // player\'s last name
string m_position; // player\'s primar.
More Related Content
Similar to The ProblemComplete the function printList that accepts a node (.pdf
Help please, I have attached LinkedList.cpp and LinkedList.hPlease.pdf
Help please, I have attached LinkedList.cpp and LinkedList.h
Please help with the implementation of the Stack.h and Stack.cpp as well
LinkedList.cpp
~~~~~~~~~~~~~~
#include \"LinkedList.h\"
LinkedList::LinkedList()
{
first = last = NULL;
}
LinkedList::~LinkedList()
{
while (first != NULL)
{
Node *temp;
temp = first;
first = first->next;
free(temp);
}
last = NULL;
}
void LinkedList::insertAtBack(int valueToInsert)
{
Node *temp = (Node *)malloc(sizeof(Node));
temp->val = valueToInsert;
temp->next = NULL;
if (last == NULL)
first = last = temp;
else
{
last->next = temp;
last = temp;
}
}
bool LinkedList::removeFromBack()
{
if (first == NULL)
return false;
else if (first == last)
{
free(first);
first = last = NULL;
}
else
{
Node *temp;
temp = first;
while (temp->next != last)
temp = temp->next;
last = temp;
temp = temp->next;
free(temp);
last->next = NULL;
}
return true;
}
void LinkedList::print()
{
Node *temp = first;
while (temp != last)
{
cout << temp->val << \" \";
temp = temp->next;
}
if (temp != NULL)
cout << temp->val;
}
bool LinkedList::isEmpty()
{
if (first == last)
return true;
return false;
}
int LinkedList::size()
{
int count = 0;
Node *temp = first;
if (temp == NULL)
return count;
while (temp != last)
{
count++;
temp = temp->next;
}
return count + 1;
}
void LinkedList::clear()
{
while (first != NULL)
{
Node *temp;
temp = first;
first = first->next;
free(temp);
}
last = NULL;
}
/*Exercise2 */
void LinkedList::insertAtFront(int valueToInsert)
{
Node *temp = (Node *)malloc(sizeof(Node));
temp->val = valueToInsert;
if (last == NULL)
first = last = temp;
else
{
temp->next = first;
first = temp;
}
}
bool LinkedList::removeFromFront()
{
if (first == NULL)
return false;
Node *temp = first;
first = first->next;
free(temp);
if (first == NULL)
last = NULL;
return true;
}
LinkedList.h
~~~~~~~~~~~
#include \"LinkedList.h\"
LinkedList::LinkedList()
{
first = last = NULL;
}
LinkedList::~LinkedList()
{
while (first != NULL)
{
Node *temp;
temp = first;
first = first->next;
free(temp);
}
last = NULL;
}
void LinkedList::insertAtBack(int valueToInsert)
{
Node *temp = (Node *)malloc(sizeof(Node));
temp->val = valueToInsert;
temp->next = NULL;
if (last == NULL)
first = last = temp;
else
{
last->next = temp;
last = temp;
}
}
bool LinkedList::removeFromBack()
{
if (first == NULL)
return false;
else if (first == last)
{
free(first);
first = last = NULL;
}
else
{
Node *temp;
temp = first;
while (temp->next != last)
temp = temp->next;
last = temp;
temp = temp->next;
free(temp);
last->next = NULL;
}
return true;
}
void LinkedList::print()
{
Node *temp = first;
while (temp != last)
{
cout << temp->val << \" \";
temp = temp->next;
}
if (temp != NULL)
cout << temp->val;
}
bool LinkedList::isEmpty()
{
if (first == last)
return true;
return false;
}
int LinkedList::size()
{
int count = 0;
Node *temp = first;
if (temp == NULL)
return count;
while (temp != last)
{
count++;
temp = temp->next;
}
return count + 1;
}
void LinkedList::clear()
{
while (first != NULL)
{
Node *.
When we first examined the array based and node based implementations.docx
When we first examined the array based and node based implementations
C++ problemPart 1 Recursive Print (40 pts)Please write the recu.pdf
C++ problem
Part 1: Recursive Print (40 pts)
Please write the recursive List reverse print function, whose iterative version we wrote in class.
Below are the function signatures for the functions you are going to need:
public:
/**Additional Operations*/
void print_reverse();
//Wrapper function that calls the reverse helper function to print a list in reverse
//prints nothing if the List is empty
private:
void reverse(Nodeptr node);
//Helper function for the public printReverse() function.
//Recursively prints the data in a List in reverse.
Why do we need the private helper function here?
Since we are going to be reversing our list node by node, in a recursive fashion, we want to pass
a one node at a time to our reverse function.
However, since our nodes are private, we cannot access them if we call the function inside of
main.
Add these function signatures to your List.h file along with your other function prototypes inside
the class definition.
Make sure that you place the reverse function inside the private portion of your List class
definition and the print_reverse function prototype to the public portion of your List class
definition.
Now, implement these two functions inside of List.h, under your section for additional
operations.
Important: Test each function carefully inside of your ListTest.cpp to make sure that it is
working properly.
Part 2: Adding an Index to Your List Nodes (20 pts)
Next, you will add the following functions to your List.h
/**Accessor Functions*/
int get_index();
//Indicates the index of the Node where the iterator is currently pointing
//Nodes are numbered from 1 to length of the list
//Pre: length != 0
//Pre: !off_end()
...
int List::get_index()
{
//Implement the function here
}
/**Manipulation Procedures*/
void scroll_to_index(int index);
//Moves the iterator to the node whose index is specified by the user
//Pre: length != 0
...
void scroll_to_index(int index)
{
//Implement function here
}
Part 3: Implementing Search as Part of Your List (40 pts)
Now, we are going to add two search functions to our List so that we can search for elements in
our List.
The first of these functions is going to be a simple linear search function.
You will need to add the following function prototype and function definition to your List.h:
/**Additional Operations*/
int linear_search(listitem item);
//Searchs the list, element by element, from the start of the List to the end of the List
//Returns the index of the element, if it is found in the List
//Returns -1 if the element is not in the List
//Pre: length != 0
...
int List::linear_search(listitem item)
{
//Implement the function here
}
You are also going to add a function to perform recursive binary search on your List.
You will need to add the following function prototype and function definition to your List.h:
int binary_search(int low, int high, listitem item);
//Recursively searchs the list by dividing the search space in half
//Returns the index of the element, if it is fo.
Please find the answer to the above problem as follows- Program.pdf
Please find the answer to the above problem as follows:-
/* Program to insert in a sorted list */
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
void sortedDelete(struct node** head_ref, int number)
{
struct node *temp = *head_ref;
struct node *prev = NULL;
if(*head_ref != NULL && (*head_ref)->data == number){
*head_ref = (*head_ref)->next;
return;
}
while(temp != NULL)
{
if(temp->data == number){
prev->next = temp->next;
free(temp);
return;
}
prev = temp;
temp = temp->next;
}
}
/* function to insert a new_node in a list. Note that this
* function expects a pointer to head_ref as this can modify the
* head of the input linked list (similar to push())*/
void sortedInsert(struct node** head_ref, struct node* new_node)
{
struct node* current;
/* Special case for the head end */
if (*head_ref == NULL || (*head_ref)->data >= new_node->data)
{
if(*head_ref != NULL&&(*head_ref)->data==new_node->data){
return;
}
new_node->next = *head_ref;
*head_ref = new_node;
}
else
{
/* Locate the node before the point of insertion */
current = *head_ref;
while (current->next!=NULL &&
current->next->data < new_node->data)
{
current = current->next;
}
if(current->next!=NULL && current->next->data == new_node->data){
return;
}
new_node->next = current->next;
current->next = new_node;
}
}
/* BELOW FUNCTIONS ARE JUST UTILITY TO TEST sortedInsert */
/* A utility function to create a new node */
struct node *newNode(int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
new_node->next = NULL;
return new_node;
}
/* Function to print linked list */
void printList(struct node *head)
{
struct node *temp = head;
if(temp == NULL)
printf(\"\ \");
while(temp != NULL)
{
printf(\"%d \", temp->data);
temp = temp->next;
}
}
/* Drier program to test count function*/
int main(int argc, char *argv[])
{
/* Start with the empty list */
struct node* head = NULL;
struct node *new_node;
FILE *fp;
fp = fopen(argv[1], \"r+\");
if(!fp){
printf(\"error\");
return 1;
}
char op = \' \';
int number = 0;
while(fscanf(fp, \"%c\\t%d\ \", &op, &number)!=EOF){
if(op==\'i\'){
new_node = newNode(number);
sortedInsert(&head, new_node);
}else if(op==\'d\'){
sortedDelete(&head, number);
}else{
printf(\"error\");
return;
}
}
printList(head);
printf(\"\ \");
return 0;
}
Solution
Please find the answer to the above problem as follows:-
/* Program to insert in a sorted list */
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
void sortedDelete(struct node** head_ref, int number)
{
struct node *temp = *head_ref;
struct node *prev = NULL;
if(*head_ref != NULL && (*head_ref)->data == number){
*head_ref = (*head_ref)->next;
return;
}
while(temp != NULL)
{
if(temp->data == number){
prev->next = temp->next;
free(temp);
return;
}
prev = temp;
temp = temp->next;
}
}
/* function to insert a new_node in a list. Note that this
* function expects a point.
Data Structures in C++I am really new to C++, so links are really .pdf
Data Structures in C++
I am really new to C++, so links are really hard topic for me. It would be nice if you can provide
explanations of what doubly linked lists are and of some of you steps... Thank you
In this assignment, you will implement a doubly-linked list class, together with some list
operations. To make things easier, you’ll implement a list of int, rather than a template class.
Solution
A variable helps us to identify the data. For ex: int a = 5; Here 5 is identified through variable a.
Now, if we have collection of integers, we need some representation to identify them. We call it
array. For ex: int arr[5]
This array is nothing but a Data Structure.
So, a Data Structure is a way to group the data.
There are many Data Structures available like Arrays, Linked List, Doubly-Linked list, Stack,
Queue, etc.
Doubly-Linked list are the ones where you can traverse from the current node both in left and
right directions.
Why so many different types of Data Structures are required ?
Answer is very simple, grouping of data, storage of data and accessing the data is different.
For example, in case of Arrays we store all the data in contiguous locations.
What if we are not able to store the data in contiguous locations because we have huge data.
Answer is go for Linked List/Doubly-Linked list.
Here we can store the data anywhere and link the data through pointers.
I will try to provide comments for the code you have given. May be this can help you.
#pragma once
/*
dlist.h
Doubly-linked lists of ints
*/
#include
class dlist {
public:
dlist() { }
// Here we are creating a NODE, it has a integer value and two pointers.
// One pointer is to move to next node and other to go back to previous node.
struct node {
int value;
node* next;
node* prev;
};
// To return head pointer, i.e. start of the Doubly-Linked list.
node* head() const { return _head; }
// To return Tail pointer, i.e. end of the Doubly-Linked list.
node* tail() const { return _tail; }
// **** Implement ALL the following methods ****
// Returns the node at a particular index (0 is the head).
node* at(int index){
int cnt = 0;
struct node* tmp = head();
while(tmp!=NULL)
{
if (cnt+1 == index)
return tmp;
tmp = tmp->next;
}
}
// Insert a new value, after an existing one
void insert(node *previous, int value){
// check if the given previous is NULL
if (previous == NULL)
{
printf(\"the given previous node cannot be NULL\");
return;
}
// allocate new node
struct node* new_node =(struct node*) malloc(sizeof(struct node));
// put in the data
new_node->data = new_data;
// Make next of new node as next of previous
new_node->next = previous->next;
// Make the next of previous as new_node
previous->next = new_node;
// Make previous as previous of new_node
new_node->prev = previous;
// Change previous of new_node\'s next node
if (new_node->next != NULL)
new_node->next->prev = new_node;
}
// Delete the given node
void del(node* which){
struct node* head_ref = head();
/* base case */
if(*head_ref == NUL.
finish the following code based on green line requirement us.pdf
finish the following code based on green line requirement use c++
void addIntToEndOfList(LinkedList list, int value) { assert(list != NULL); // if list is NULL, we can do
nothing. Node *p; // temporary pointer // TOD0: // (1) Allocate a new node, p will point to it. // (2)
Set p's data field to the value passed in // (3) Set p's next field to NULL if (list->head == NULL) { //
(4) Make both head and tail of this list point to p } else { // Add p at the end of the list. // (5) The
current node at the tail? Make it point to p instead of NULL }void addIntToStartOfList(LinkedList
list, int value) assert(list != NULL); // if list is NULL, we can do nothing, // Add code for this. // HINT:
// consider all edge cases such as when list->head is or is not null AND // consider all edge cases
such as when list->tail is or is not null. // Visualizing the problem with a box and pointer diagram
can help. / // list: ptr to a linked list of Node (each with int data, and Node next) // Return a pointer
to node with the largest value. // You may assume list has at least one element // If more than one
element has largest value, // break tie by returning a pointer to the one the occurs // earlier in the
list, i.e. closer to the head.
#include sstream #include linkylist.h #include iostream.pdf
#include
#include \"linkylist.h\"
#include
#include
#include
#include
using namespace std;
int main(int argc, char *argv[])
{
//node* head = new node;
LinkedList list(1);
ifstream infile1 (argv[1]);
//make sure the file opens properly and read the contents of the file in to a list
if(!infile1)
{
cout<<\"Your Testfile could not be opened!\"<>users;
if (users.find(\"I\")!=string::npos)
{
//int spot = atoi((users[3]).c_str());
string pot = string(users.substr(2));
stringstream spot(pot);
string data;
cin>>data;
insertAfter(list.getNode(spot),data);
}
/*else if (users.find(\"D\")!=string::npos)
{
int spoty = atoi((users[3]).c_str());
deleteNode(spoty);
}*/
else if (users==\"L\")
{
list.traverse_and_print();
}
}
return 0;
#include
using namespace std;
class Node
{
friend class LinkedList;
private:
int value; /* data, can be any data type, but use integer for easiness */
Node *Next; /* pointer to the next node */
public:
/* Constructors with No Arguments */
Node(void)
: Next(NULL)
{ }
/* Constructors with a given value */
Node(int val)
: value(val), Next(NULL)
{ }
/* Constructors with a given value and a link of the next node */
Node(int val, Node* next)
: value(val), Next(next)
{}
/* Getters */
int getValue(void)
{ return value; }
Node* getNext(void)
{ return Next; }
};
/* definition of the linked list class */
class LinkedList
{
private:
/* pointer of head node */
Node *Head;
/* pointer of tail node */
Node *Tail;
public:
/* Constructors with a given value of a list node */
LinkedList(int val);
/* Destructor */
~LinkedList(void);
/* Function to append a node to the end of a linked list */
void tailAppend(string val);
/* Remove a node with a specific value if it exists */
void remove(int val);
void insertAfter(Node* afterMe, string val);
Node* getNode(int pos);
/* Traversing the list and printing the value of each node */
void traverse_and_print();
};
#include
#include \"linkylist.h\"
#include
using namespace std;
LinkedList::LinkedList(int val)
{
/* Create a new node, acting as both the head and tail node */
Head = new Node(val);
Tail = Head;
}
Node* LinkedList::getNode(int pos)
{
Node* p = Head;
/* Counter */
while (pos--) {
if (p != NULL)
p = p->Next;
else
return NULL;
}
return p;
}
void LinkedList::insertAfter(Node* afterMe, string val)
{
if(afterme != NULL)
{
afterMe->Next = new Node(val, afterMe->Next);
}
}
LinkedList::~LinkedList()
{
}
void LinkedList::tailAppend(string val)
{
/* The list is empty? */
if (Head == NULL) {
/* the same to create a new list with a given value */
Tail = Head = new Node(val);
}
else
{
/* Append the new node to the tail */
Tail->Next = new Node(val);
/* Update the tail pointer */
Tail = Tail->Next;
}
}
void LinkedList::remove(int val)
{
Node *pPre = NULL, *pDel = NULL;
/* Check whether it is the head node?
if it is, delete and update the head node */
if (Head->value == val) {
/* point to the node to be deleted */
pDel = Head;
/* update */
Head = pDel->Next;
delete pDel;
return;
}
pPre = Head;
pDel = Head->Next;
/*.
C++ Doubly-Linked ListsThe goal of the exercise is to implement a.pdf
C++: Doubly-Linked Lists
The goal of the exercise is to implement a class List of doubly- linked lists.
My goal is to implement the class in a file doubly-linked.cpp.
The class List implements lists using the following structures for list ele- ments:
struct Node { int val ;
Node next;
Node prev; };
Each Node contains a value (an integer) and two pointers: one meant to point to the next element
of the list and one meant to point to the previous element in the list. The class provides the
following methods that you need to implement:
• A constructor and a destructor, with no arguments;
• void insert(int n): insert an integer at the end of the list;
• void reverse(): reverse the list;
• void print(): print the list to cout in the same format as the input (i.e. integers separated by
spaces);
doubly-linked.h
#ifndef __dll__
#define __dll__
#include
using namespace std;
// Basic structure to store elements of a list
struct Node {
int val; // contains the value
Node * next; // pointer to the next element in the list
Node * prev; // pointer to the previous element in the list
};
// Class List
class List {
public:
List(void); // Constructor
~List(void); // Destructor
void insert(int n); // This should insert n in the list
void reverse(void); // This should reverse the list
void print(void); // This shoiuld print the list
private:
Node * first; // Pointer to the first (if any) element in the list
};
#endif
main.cpp
#include
#include \"doubly-linked.h\"
using namespace std;
int main(void){
List l;
int n;
while(cin >> n){
l.insert(n);
}
// Print list as read from cin
l.print();
// Reverse the list and print it
l.reverse();
l.print();
// Reverse again and print it
l.reverse();
l.print();
return 0;
}
Solution
#include
using namespace std;
/* Linked list structure */
struct list {
struct list *prev;
int data;
struct list *next;
} *node = NULL, *first = NULL, *last = NULL, *node1 = NULL, *node2 = NULL;
class linkedlist {
public:
/* Function for create/insert node at the beginning of Linked list */
void insrt_frnt() {
list *addBeg = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addBeg->data;
if(first == NULL) {
addBeg->prev = NULL;
addBeg->next = NULL;
first = addBeg;
last = addBeg;
cout << \"Linked list Created!\" << endl;
}
else {
addBeg->prev = NULL;
first->prev = addBeg;
addBeg->next = first;
first = addBeg;
cout << \"Data Inserted at the beginning of the Linked list!\" << endl;
}
}
/* Function for create/insert node at the end of Linked list */
void insrt_end() {
list *addEnd = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addEnd->data;
if(first == NULL) {
addEnd->prev = NULL;
addEnd->next = NULL;
first = addEnd;
last = addEnd;
cout << \"Linked list Created!\" << endl;
}
else {
addEnd->next = NULL;
last->next = addEnd;
addEnd->prev = last;
last = addEnd;
cout << \"Data Inserted at the end of the Linked list!\" << endl;
}
}
/* Function for Display Linked list */
void display() {
node = first;
cout << \"List of data in L.
In C++Write a recursive function to determine whether or not a Lin.pdf
In C++
Write a recursive function to determine whether or not a Linked List is in sorted order (smallest
value to largest value).
Add the following function prototypes to your List class under the section for access functions,
then implement the functions below the class definition:
public:
/**Access Functions*/
bool isSorted();
//Wrapper function that calls the isSorted helper function to determine whether
//a list is sorted in ascending order.
//We will consider that a list is trivially sorted if it is empty.
//Therefore, no precondition is needed for this function
private:
bool isSorted(Nodeptr node);
//Helper function for the public isSorted() function.
//Recursively determines whether a list is sorted in ascending order.
#include //for NULL
#include
#include
using namespace std;
template //list stores generic list data, not any specific C++ type
class List
{
private:
struct Node
{
listdata data;
Node* next;
Node* previous;
Node(listdata data): data(data), next(NULL), previous(NULL){}
};
typedef struct Node* Nodeptr;
Nodeptr first;
Nodeptr last;
Nodeptr iterator;
int size;
public:
/**Constructors and Destructors*/
List();
//Default constructor; initializes and empty list
//Postcondition: numeric values equated to zero, or strings should be empty.
List(const List &list);
~List();
//Destructor. Frees memory allocated to the list
//Postcondition: position NodePtr at next Node
/**Accessors*/
listdata getFirst();
//Returns the first element in the list
//Precondition:NodePtr points to the first node on list
listdata getLast();
//Returns the last element in the list
//Precondition: make new node first on the list
listdata getIterator();
bool isEmpty();
//Determines whether a list is empty.
int getSize();
//Returns the size of the list
/**Manipulation Procedures*/
void startIterator();
void advanceIterator();
void removeLast();
//Removes the value of the last element in the list
//Precondition: list is not empty, not the first element.
//Postcondition: one remaining node
void removeIterator();
void removeFirst();
//Removes the value of the first element in the list
//Precondition: list is not empty
//Postcondition: no nodes left
void insertLast(listdata data);
//Inserts a new element at the end of the list
//If the list is empty, the new element becomes both first and last
//Postcondition: next equal to null
void insertFirst(listdata data);
//Inserts a new element at the start of the list
//If the list is empty, the new element becomes both first and last
//Postcondition:point nodePtr next node on list
/**Additional List Operations*/
bool offEnd();
void printList();
//Prints to the console the value of each element in the list sequentially
//and separated by a blank space
//Prints nothing if the list is empty
void insertIterator(listdata data);
bool operator==(const List &list);
};
// constructor definition
template
List::List(): first(NULL), last(NULL), iterator(NULL), size(0) {}
template
List::List(const List &list): size(list.size)
{
if(list..
I need to implment a function that can reverse a single linked list..pdf
I need to implment a function that can reverse a single linked list. Not just print it, but actually
change the order of the list.
THIS MUST BE DONE USING RECURSION
EX: a, b, c, d -> d, c, b, a
You have access to the start of the list. Code must be in C++
Solution
#include
using namespace std;
struct node
{
char data;
struct node* next;
};
void reverse(struct node* head)
{
if (head == NULL) //base case
return;
reverse(head->next); //calling function recursively
cout<data<<\",\";
}
//add data to the front of the linked list
void add(struct node** head_ref, char data)
{
//create new node
struct node* newnode=new node();
//adding data
newnode->data = data;
//adding node to the front
newnode->next = (*head_ref);
//making new node head
(*head_ref) = newnode;
}
int main()
{
// create linked list a, b, c, d
struct node* head = NULL;
add(&head, \'d\');
add(&head, \'c\');
add(&head, \'b\');
add(&head, \'a\');
reverse(head);
return 0;
}
//OUTPUT***********
d,c,b,a,
//OUTPUT***********
//This code has been tested on g++ compiler,Please ask in case of any doubt,Thanks..
Using the C++ programming language1. Implement the UnsortedList cl.pdf
Using the C++ programming language
1. Implement the UnsortedList class to store a list of numbers that are input into the list from
data.txt.
- create a main.cpp file that gets the numbers from the file
- insert the number 7 into the list
- insert another number 300 into the list
- delete the number 6 from the list
- print out the following:
--the entire list
- the greatest
- the least
2. Attach the main.cpp, UnsortedList.cpp, the ItemType.h, and the output file two called
outfile2.txt
Use the files below:
// listDriver.cpp
// Test driver
#include
#include
#include
#include
#include
#include \"unsorted.h\"
using namespace std;
void PrintList(ofstream& outFile, UnsortedType& list);
int main()
{
ifstream inFile; // file containing operations
ofstream outFile; // file containing output
string inFileName; // input file external name
string outFileName; // output file external name
string outputLabel;
string command; // operation to be executed
int number;
ItemType item;
UnsortedType list;
bool found;
int numCommands;
// Prompt for file names, read file names, and prepare files
cout << \"Enter name of input command file; press return.\" << endl;
cin >> inFileName;
inFile.open(inFileName.c_str());
cout << \"Enter name of output file; press return.\" << endl;
cin >> outFileName;
outFile.open(outFileName.c_str());
cout << \"Enter name of test run; press return.\" << endl;
cin >> outputLabel;
outFile << outputLabel << endl;
if (!inFile)
{
cout << \"file not found\" << endl;
exit(2)
}
inFile >> command;
numCommands = 0;
while (command != \"Quit\")
{
if (command == \"PutItem\")
{
inFile >> number;
item.Initialize(number);
list.PutItem(item);
item.Print(outFile);
outFile << \" is in list\" << endl;
}
else if (command == \"DeleteItem\")
{
inFile >> number;
item.Initialize(number);
list.DeleteItem(item);
item.Print(outFile);
outFile << \" is deleted\" << endl;
}
else if (command == \"GetItem\")
{
inFile >> number;
item.Initialize(number);
item = list.GetItem(item, found);
item.Print(outFile);
if (found)
outFile << \" found in list.\" << endl;
else outFile << \" not in list.\" << endl;
}
else if (command == \"GetLength\")
outFile << \"Length is \" << list.GetLength() << endl;
else if (command == \"IsFull\")
if (list.IsFull())
outFile << \"List is full.\" << endl;
else outFile << \"List is not full.\" << endl;
else if (command == \"MakeEmpty\")
list.MakeEmpty();
else if (command == \"PrintList\")
PrintList(outFile, list);
else
cout << command << \" is not a valid command.\" << endl;
numCommands++;
cout << \" Command number \" << numCommands << \" completed.\"
<< endl;
inFile >> command;
};
cout << \"Testing completed.\" << endl;
inFile.close();
outFile.close();
return 0;
}
void PrintList(ofstream& dataFile, UnsortedType& list)
// Pre: list has been initialized.
// dataFile is open for writing.
// Post: Each component in list has been written to dataFile.
// dataFile is still open.
{
int length;
ItemType item;
list.ResetList();
length = list.GetLength();
for .
please help me in C++Objective Create a singly linked list of num.pdf
please help me in C++
Objective: Create a singly linked list of numbers based upon user input.
Program logic: Ask for a number, add that number to the front of the list, print the list. Repeat
until they enter -1 for the number. .
Sample Input: 10, 15, 5, 2, 4, -1 Output: 4, 2, 5, 15, 10
Solution
#include
#include
using namespace std;
// It contains the value and the pointer to the next element in the list.
struct element {
int value;
struct element *next;
};
struct element *head = NULL;
// The list is stored in the object of this class.
class linkedlist {
public:
// To display list, we first see if head is NULL,
// If head is NULL it means the list is empty.
// Then we traverse along the list and display each item, until the next pointer is NULL.
void display() {
if(head == NULL){
cout << \"The list is empty\ \";
} else {
struct element *temp;
temp = head;
while(temp != NULL){
cout << temp->value << \" \" ;
temp = temp->next;
}
cout << \"\ \";
}
}
// To add an element to the list,create a newElement and store the value.
// If head is NULL, add the newElement to the head.
// To add an element at the starting of the list.
// The next pointer of the newElement must point to the old head.
// The new head must be pointed to the newElement.
void add(int a){
struct element *newElement;
newElement = (struct element *)malloc(sizeof(struct element));
newElement->value = a;
newElement->next = NULL;
if(head == NULL) {
head = newElement;
} else {
newElement->next = head;
head = newElement;
}
}
};
int main()
{
// Create an object of the class linkedlist.
linkedlist l = linkedlist();
int a;
do {
cout << \"Enter a number : \";
cin >> a;
if(a!=-1) { // If the number is -1 do not add it to the list.
l.add(a); // Add the number to the list.
}
l.display(); // Display the list.
} while(a!=-1); // Keep asking for a number until -1 is pressed.
return 0;
}.
#include iostream #includestdlib.h using namespace std;str.pdf
#include
#include
using namespace std;
struct node { //Task 1
char val;
node *next;
};
void printList(node *head) { //Task 2
if (head == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head;
int count = 0;
while(temp != NULL) {
cout << \"Node \" << count << \": \" << temp ->val << endl;
count++;
temp = temp->next;
}
}
struct node *globalhead = NULL;
void deepCopy(node *head1, node *head2) { //Task 3
if (head1 == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head1;
node *curr = NULL;
node *prev = NULL;
while(temp != NULL) {
curr = (struct node *)malloc(sizeof(struct node));
curr->val = temp->val;
if(prev==NULL){
head2 = curr;
globalhead = curr;
}else{
prev->next = curr;
}
prev = curr;
temp = temp->next;
}
}
int main() { //Task 4
// Example #1
node p00, p01, p02, p03, p10, p11, p12, p13;
// List 1: A B B B
p00.val = \'A\'; p00.next = &p01;
p01.val = \'B\'; p01.next = &p02;
p02.val = \'B\'; p02.next = &p03;
p03.val = \'B\'; p03.next = NULL;
p10.next = NULL;
cout<<\"First List:\"<
Solution
//Tested on Windos OS Dev c++
/*********************LinkedList.cpp********************/
#include
#include
using namespace std;
struct node { //Task 1
char val;
node *next;
};
void printList(node *head) { //Task 2
if (head == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head;
int count = 0;
while(temp != NULL) {
cout << \"Node \" << count << \": \" << temp ->val << endl;
count++;
temp = temp->next;
}
}
struct node *globalhead = NULL;
void deepCopy(node *head1, node **head2) { //Task 3
if (head1 == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head1;
node *curr = NULL;
node *prev = NULL;
while(temp != NULL) {
curr = (struct node *)malloc(sizeof(struct node));
curr->val = temp->val;
curr->next = NULL;
if(*head2==NULL){
*head2=curr;
}else{
prev=*head2;
while(prev->next!=NULL){
prev=prev->next;
}
prev->next=curr;
}
temp = temp->next;
}
}
int main() { //Task 4
// Example #1
node p00, p01, p02, p03, p10, p11, p12, p13;
// List 1: A B B B
p00.val = \'A\'; p00.next = &p01;
p01.val = \'B\'; p01.next = &p02;
p02.val = \'B\'; p02.next = &p03;
p03.val = \'B\'; p03.next = NULL;
p10.next = NULL;
cout<<\"First List:\"<.
c++ Computational Complexity filling in the following three .pdf
c++ Computational Complexity
/* filling in the
* following three functions:
*
* 1) void disp(Node* head)
* Given a head pointer to a list, display the list using cout.
* If the list is empty, you should output \"HEAD->NULL\" (without the enclosing \").
* If the list contains one node whose data value is 3, you should output \"HEAD->3->NULL\".
* If the list contains three nodes and the data values are 1, 2, and 3, respectively, then you
should output
* \"HEAD->1->2->3->NULL\"
*
* 2) Node* add(Node* head, int data)
* Given the old head pointer to a list, create a new node using the value of data, and add the new
node to the front of
* the list. Return the new head pointer.
* Example: If the original list is \"HEAD->1->2->NULL\", then after adding 5, it should
become \"HEAD->5->1->2->NULL\". The
* return value should be the address of node 5.
*
* 3) void magic(Node* p1, Node* p2)
* p1 and p2 each points to a node from a same list. p1 may point to a later node than p2. Your
task is to calculate how
* many nodes are between p1 and p2. If there are k nodes in between (excluding nodes pointed
to by p1 and p2), then the
* result is k (k could be 0). Replace the data fields of all nodes between p1 and p2 by k.
* Example: The original list is \"HEAD->1->7->5->0->3->NULL\". p1 points to node 1 (the
node with value 1), p2 points to
* node 0. There are two nodes in between p1 and p2. These two nodes\' data fields need to be
replaced by k=2. The
* resulting list should be \"HEAD->1->2->2->0->3->NULL\".
*
* You should modify sections labelled as \"fill in here\". Please leave the other sections (e.g., the
main
* function and all function protocols) intact.
*/
#include
using namespace std;
struct Node {
int data;
Node* link;
};
void disp(Node* head) {
// fill in here
}
Node* add(Node* head, int data) {
// fill in here
}
void magic(Node* p1, Node* p2) {
// fill in here
}
main() {
Node* head = NULL;
Node *p4, *p7;
for(int i = 0; i < 10; i++) {
head = add(head, i);
if (i==4) p4 = head;
if (i==7) p7 = head;
}
// at this point, we have created the following list: HEAD->9->8->7->6->5->4->3->2->1->0-
>NULL
// p4 now points to node 4 (the node containing 4); p7 now points to node 7 (the node
containing 7)
magic(p4, p7);
disp(head);
// between p4 and p7, there are k=2 nodes, and these two nodes\' values will be replaced by
k=2
// the output should be: HEAD->9->8->7->2->2->4->3->2->1->0->NULL
// housekeeping (free all nodes)
Node* tmp;
while (head) {
tmp = head;
head = head->link;
delete tmp;
}
}
Comments
Expert Answer
c++ Computational Complexity
/* Your task is to fill in the
* following three functions:
*
* 1) void disp(Node* head)
* Given a head pointer to a list, display the list using cout.
* If the list is empty, you should output \"HEAD->NULL\" (without the enclosing \").
* If the list contains one node whose data value is 3, you should output \"HEAD->3->NULL\".
* If the list contains three nodes and the data values are 1, 2, and 3, respectively, then you
shou.
in Java (ignore the last line thats hidden) Create a doubly linked l.pdf
in Java (ignore the last line thats hidden) Create a doubly linked list whose nodes contain Strings.
Your list should include the following methods: Insert a node in the list in alphabetical order
Find a node that matches a String Traverse the list forwards and print Traverse the list backwards
and print Delete a node from the list Delete/destroy the list In addition to creating the class(es)
for your linked list, you\'ll need to create a main method that thoroughly tests each function you
build to show that it works. You should do these things in
Solution
/ Java program to create , insert , traverse and delete a node from doubly linked list
class LinkedList {
static Node head = null;
class Node {
int data;
Node next, prev;
Node(int d) {
data = d;
next = prev = null;
}
}
/* Let us create the doubly linked list 10<->8<->4<->2 */
list.push(head, 2);
list.push(head, 4);
list.push(head, 8);
list.push(head, 10);
System.out.println(\"Original Linked list \");
list.printList(head);
/*Function to delete a node in a Doubly Linked List.
head_ref --> pointer to head node pointer.
del --> pointer to node to be deleted. */
void deleteNode(Node head_ref, Node del) {
/* base case */
if (head == null || del == null) {
return;
}
/* If node to be deleted is head node */
if (head == del) {
head = del.next;
}
/* Change next only if node to be deleted is NOT the last node */
if (del.next != null) {
del.next.prev = del.prev;
}
/* Change prev only if node to be deleted is NOT the first node */
if (del.prev != null) {
del.prev.next = del.next;
}
/* Finally, free the memory occupied by del*/
return;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning of the Doubly Linked List */
void push(Node head_ref, int new_data) {
/* allocate node */
Node new_node = new Node(new_data);
/* since we are adding at the begining,
prev is always NULL */
new_node.prev = null;
/* link the old list off the new node */
new_node.next = (head);
/* change prev of head node to new node */
if ((head) != null) {
(head).prev = new_node;
}
/* move the head to point to the new node */
(head) = new_node;
}
/*Function to print nodes in a given doubly linked list
This function is same as printList() of singly linked lsit */
void printList(Node node) {
while (node != null) {
System.out.print(node.data + \" \");
node = node.next;
}
}
public static void main(String[] args) {
LinkedList list = new LinkedList();
/* delete nodes from the doubly linked list */
list.deleteNode(head, head); /*delete first node*/
list.deleteNode(head, head.next); /*delete middle node*/
list.deleteNode(head, head.next); /*delete last node*/
System.out.println(\"\");
/* Modified linked list will be NULL<-8->NULL */
System.out.println(\"Modified Linked List\");
list.printList(head);
}
}.
Java AssignmentUsing the ListNode.java file below Write method.pdf
Java Assignment
Using the ListNode.java file below
Write methods called min and max that return the smallest and largest values in the linked list.
These methods will be added to your ListNode class. For example if a variable called list stores
{11, -7, 3, 42, 0, 14], the call of list.min() should return -7 and the call of list.max() should return
42. If the list is empty, return -1. Print the returned value.
Write a method called insertNode that inserts a new node anywhere in your linked list. Display
your linked list with new value.
Write a program to test these methods to ensure accuracy and successful implementation. Upload
your modified version of the ListNode java file only.
Note: Original code should still be included. You will make additions and slight modifications
where necessary. Throw exceptions where you think is necessary. // Listlode is a class for storing
a single node of a linked // list. This node class is for a list of integer values. public class Listlode
\{ public int data; //data stored in this node public Listliode front;//points to head node public
Listliode next; /f link to next node in the list // post: constructs a node with data and null link
public Listlode() \{ 3. // post: constructs a node with given data and null link: public Listlode(int
data) \{ this(data, null); 3 // post: constructs a node with given data and given link. public
Listlode(int data, Listliode next) \{ this.data = data; this. next = next; 3 public String toString() \{
if (front = null) : \} else \{ String result = " ["+ front data; Listliode current = front. next; while
(current != null) \{ result +=m,m+ current data, current = current. next; 3 result += " ]n - return
result; 3. }// post: appends the given value to the end of the list public void add(int value) \&
puolic void aod if (front = nult) \{. front = new Listlode (value); 3 else \{ Listlode current =
front; whistiode current = front; while (current next I= null) \{ current = current. next; 3 current.
next = new Listliode (value ); 3. 3 J// post : returns the position of the first occurrence of the
given /l value (-1 if not found ) public void remove (int index) \{ if ( index =){ \} else \{ Listiode
current = nodeAt ( index - 1); current. next = current. next. next, 3 3 3// pre : <=i< size() // post:
returns a reference to the node at the given index public Listlode nodeAt (int index) : List Node
current = front; for (int i=0;i< index; i++) \{ current = current. next. \}. return current; 3.
--INSTRUCTION- --It helps to first create if-then-else structure to fi.pdf
//INSTRUCTION:
//It helps to first create if-then-else structure to fill in later.
//- Make sure PURPOSE and PARAMETER comments are given in detail
// Add sufficient comments to your code body to describe what it does.
// - Make sure all if-then-else are commented describing which case it is
// - Make sure all local variables are described fully with their purposes
//Complete the TODO parts.
// ====================================================
//Complier: g++
//File type: linkedList.cpp implementation file
//=====================================================
#include<iostream>
#include"linkedlist.h"
using namespace std;
//cunstructor that initializes an empty list
LinkedList::LinkedList() {
front = nullptr ;
rear = nullptr ;
count = 0;
}
/**
* @brief Destructor to destroy all nodes and release memory
*/
LinkedList::~LinkedList() {
Node *p, *q;
p = front;
while (p != nullptr ) {
q = p->next;
delete p;
p = q;
}
}
/**
* @brief Purpose: Checks if the list is empty
* @return true if the list is empty, false otherwise
*/
bool LinkedList::isEmpty() const {
// TODO: Add code here
return count == 0;
}
/**
* @brief Get the number of nodes in the list
* @return int The number of nodes in the list
*/
int LinkedList::length() const {
return count;
}
/**
* @brief Convert the list to a string
*
*/
string LinkedList::toString() {
string str = "[";
Node *ptr = front;
if (ptr != nullptr ) {
// Head node is not preceded by separator
str += to_string(ptr->val);
ptr = ptr->next;
}
while (ptr != nullptr ) {
str += ", " + to_string(ptr->val);
ptr = ptr->next;
}
str += "]";
return str;
}
/**
* @brief Displays the contents of the list
*/
void LinkedList::displayAll() {
cout << toString() << endl;
}
//Add a node to the end of the list
void LinkedList::addRear(T val) {
Node *p = new Node;
p->val = val;
p->next = nullptr ;
if (isEmpty()) {
front = p;
}
else {
rear->next = p;
}
rear = p;
count++;
// consider the two cases of whether the list was empty
}
//Add a node to the front of the list
void LinkedList::addFront(T val) {
Node *p = new Node;
p->val = val;
p->next = front;
if (isEmpty()) {
rear = p;
}
front = p;
count++;
// consider the two cases of whether the list was empty
}
//Delete the first node of the list
bool LinkedList::deleteFront(T &OldNum) {
if (isEmpty()) {
return false ;
}
Node *p = front;
OldNum = p->val;
front = front->next;
if (front == nullptr ) {
rear = nullptr ;
}
delete p;
count--;
return true ;
// consider if the list was empty and return false if the list is empty
// consider the special case of deleting the only node in the list
}
//Delete the last node of the list
bool LinkedList::deleteRear(T &OldNum) {
if (isEmpty()) {
return false ;
}
Node *p = rear;
OldNum = p->val;
if (front == rear) {
front = rear = nullptr ;
}
else {
Node *q = front;
while (q->next != rear) {
q = q->next;
}
rear = q;
q->next = nullptr ;
}
delete p;
count--;
return true ;
// consider if the list was empty and return false if the list is empty
// consider the specia.
Similar to The ProblemComplete the function printList that accepts a node (.pdf (20)
Help please, I have attached LinkedList.cpp and LinkedList.hPlease.pdf
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I need to implment a function that can reverse a single linked list..pdf
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More from fazanmobiles
H&M The Challenges of Global Expansion and the Move to adopt Intern.pdf
H&M: The Challenges of Global Expansion and the Move to adopt International Financial
Reporting standards Hennes & Mauritz AB (also known as H&M) , the Swedish MNE that is a
trendsetter in the latest fashion trends, has a stated goal “to give customers unbeatable value by
offering fashion and quality at the best price.” It doesn’t own any factories, but rather outsources
production to independent suppliers, primarily in Asia and Europe. H&M also rents space from
international and local landlords rather than owning its own stores. H&M is a major firm in the
apparel retail market where fashion trends are critical and where goods move quickly. In this
industry, the key buyers are consumers, and the key suppliers are clothing manufacturers and
wholesalers, and designers are king and a fast, well-organized supply chain is essential.
Depending on the individual firm strategy, the apparel retail market doesn’t have to be capital
intensive, but the largest players in the industry are very international, both in retail footprint and
suppliers. The biggest companies in the industry are U.S.-based The Gap, H&M, and Spain-
based Industria de Diseno Textil, S.A. (Inditex), better known by its flagship brand, Zara. All
three companies have different store brands: Gap, Banana Republic, Old Navy, and Athleta for
The Gap; H&M, COS (collection of style), Monki, Weekday, and Cheap Monday for H&M; and
Zara, Bershka, Pull and bear, Massimo Dutti, Stradivarius, Oysho, Zara Home and Uterque for
Inditex. Both of H&M’s competitors are very international. H&M operates about 2,800 stores in
49 markets, whereas The Gap operates 3,100 company stores and over 300 franchise stores
worldwide, and Inditex operates in a network of over 5,000 stores in 77 countries. Zara has the
largest geographic spread within Inditex with stores in 74 countries. Hennes & Mauritz AB
started as a single women’s wear store in Sweden in 1947. Today, H&M’s business is much
broader and currently includes the sales of clothing, accessories, footwear, cosmetics, and home
textiles. Although H&M is known as one of Sweden’s premier MNEs, it generates 21.5% of its
sales in Germany, 8.9% in the United States, and 7.4% in the United Kingdom, compared to only
5.8% of total sales in Sweden. H&M and Zara have very different strategies. Zara delivers new
products to its stores twice a week. Because of its highly organized supply chain, Zara only takes
10–15 days to go from design to the stores. Although it sources its apparel from around the
world, it has adopted just-in-time manufacturing from the auto industry and established 14 highly
automated Spanish factories where robots cut and dye fabrics creating the unfinished “gray
goods” which are the foundation for their final products. It then takes the gray goods and
outsources them to a network of small shops in Portugal and Spain to do the finish work. Store
managers are constantly sending updated information on consumer demand so that they can
move to the n.
C++ ProgrammingYou are to develop a program to read Baseball playe.pdf
C++ Programming
You are to develop a program to read Baseball player statistics from an input file. Each player
should bestored in a Player object. Therefore, you need to reuse the Player class from program 1.
Each playerwill have a first name, last name, a position (strings) and a batting average (floating
point number).
This program adds on to the requirements for Program #1. This will read int he same type of
player data from an input file and write the results to an output file.
New Requirements
You are to reuse your Player class from Program1. You are to implement a PlayerList data
structure/data type that stores players in a Linked List structure (internal to the class). Store the
players in the list in alphabetical order (By lastname, use firstname to further distinguish in the
event there is more than 1 player with the same last name).
You will need to add the class PlayerList and the class Node to facilitate the linked list
operations.
You must implement the following operations on your PlayerList along with any other utility
functionsyou might need. You may also need to add operations to the class Player to facilitate
comparisons sincePlayer data is private.
Operations
Default Constructor
Destructor
Add a Player to the List
Iterate through the List so that you can get each player
Clear out the list to make it empty
Test if a list isEmpty
Get the size of the list
Find a Player in the List (get) - Put a couple of hard-coded calls in your main program to make
sure this feature works.
Summary of Operation
Prompt the user for the input and output file names. DO NOT hardcode file names into your
program.
Open input file
Read each player and add them to your PlayerList Keep track of the number of players in the list
Store the players in the list in alphabetical order (lastname then firstname in the event of
twopeople with the same last name). You should be adding them in between other nodes
asappropriate
Test your find operation a couple of times. This can be hardcoded into your program. You donot
have to prompt the user for players, unless you wish to. Just make sure you test for playersthat
exist and that do not
Open an output file Write each player from the list into the output file, along with any other
output required by theassignment
Split your program code files into the appropriate .cpp and .h files for each of your
classes.(Exception - your NODE class can be defined in the same file as the PlayerList).
Detailed comments describing what is happening
Here is the input file:
Chipper Jones 3B 0.303
Rafael Furcal SS 0.281
Hank Aaron RF 0.305
Here is my player.h file
// Class Definition: Player
// Version 1
// This class defines an object to hold information about
// baseball player for use in a player database.
//--------------------------------------------------------
#include
using namespace std;
class Player {
string m_firstName; // player\'s first name
string m_lastName; // player\'s last name
string m_position; // player\'s primar.
Discuss the differences between a lists regular iterator, which su.pdf
Discuss the differences between a list\'s regular iterator, which supports the use of a \"for\" loop,
and its other list iterator that supports position-based operations. Discuss the advantages and
disadvantages of each.
Solution
Both iterator or list iterator is same for purpose ..i.e is to iterate through the list we have.
but the major difference between list iterator and iterator is quite simple
list iterator can iterate in both direction forward and backward direction ..it can move previous
and even next.
but in regular iterator..it can iterate only in one direction.that is only unidirection.it can move
only forward .
and one thing we should keep in mind that ..list iterator itarates only for list..but not for set map
etc.
where as regular iterator works beyond than that.it uses form list ,map.set etc.
listiterator comes up with other benifit is..it can modify the data inside list during iteration like
adding removing etc.
but in regualar iterator we cannot add element at a time of itearation..we can remove element ..
and another advantage of listiterator is..listiterator can get curent index position when traversing
but ..normal iterator cannot give that ..as it get the index positon by using prev and next but
normal iterator doesn not
iterator is not a class it is an interface and even listiterator is also an interface.
both are comes in java 1.2 and part of collection framework.
Feasibility is an important issue in security system formulation. Wh.pdf
Feasibility is an important issue in security system formulation. What are the two different
dimensions of feasibility that might affect the development of a real-world solution?
Solution
Feasibility in security system formulation is always going to help the security industry to grow in
a better manner possible. Because feasibility means convenience and whenever convenience is
there any organization or industry can work better on there technology level, management level
as well as their monetization model which is most important of them all to sustain. One
dimension of feasibility is convenience and another one is the ability to uncover the strengths
and weaknesses of a problem..
CopyPaste the sentence into the text box. Then, insert punctuat.pdf
Copy/Paste the sentence into the text box.
Then, insert punctuation or correct spelling where needed. All sentences contain one or more
errors.
1.) The three girls teacher arrived to school late.
2.) My associates clients do not seem satisfied with the service he provides.
Solution
1.
Correct Sentence:
The three girl teachers arrived at school, late.
2.
Correct answer:
My associate’s clients do not seem to be satisfied with the services, he provides..
C programming Create a system managing a mini library system. Eve.pdf
// C programming Create a system managing a mini library system. Every book corresponds to a
record (line) in a text file named \"mylibrary.txt\". Each record consists of 6 fields (Book ID,
Title, Author, Possession, checked out Date, Due Date) separated by comma: No comma \" \"in
title or author name. This mini library keeps the record for each book in the library. Different
books can share the book \"Title\". But the \"Book ID\" for each book is unique. One author may
have multiple books. Eg. Both book2 and book3 are written by author Possession field is
\"Library\" means this book is kept in library, such as book #001. Otherwise, this book is
checked out by the corresponding user. E.g. book #3 is checked out by User. A user is allowed to
check out up to 3 books. The \"Checked out Date\" field specifies the time of the latest checked
out date. The loan period for each book is 30 days. \"The Due Data\" filed specifies the time the
book should be returned. This system should provide a menu with following functionalities that
the bookkeeper can choose from at the top-level menu: Ask for title (read). Ask for author.
Generate a book ID by adding one to the current largest book ID. Add the record of the new
book into library with generated book ID, title, author, possession=\"library\", checked out
date=\"null\" and due date=\"null\". After that, print message \"book*** added successfully!\"
and go to step 5). Provide options \"t\" for \"\" and \"b\" for \"Back to main menu\". (If \"\", go to
step 1))
Solution
#include
#include
#include //contains delay(),getch()etc.
#include
#include //contains strcmp(),strcpy(),strlen(),etc
#include //contains toupper(), tolower(),etc
#include //contains _dos_getdate
#include
#define RETURNTIME 15
FILE *fp,*ft,*fs;
struct l_Date
{
int mm,dd,yy;
};
struct record
{
int bookid;
char title[20];
char Author[20];
char possession[20];
struct l_Date checkedout_date;
struct l_Date due_date;
};
struct record r;
void main()
{
clrscr();
int i;
printf(\" a. Add Books \");
printf(\"b. Delete books\");
printf(\"c. Search Books\");
printf(\"d. Issue Books\");
printf(\"e. View Book list\");
printf(\"f. Edit Book\'s Record\");
printf(\"g. Close Application\");
printf(\"Enter your choice:\");
switch(getch())
{
case \'a\':
addbooks();
break;
case \'b\':
deletebooks();
break;
case \'c\':
searchbooks();
break;
case \'d\':
issuebooks();
break;
case \'e\':
viewbooks();
break;
case \'f\':
editbooks();
break;
case \'g\':
{
clrscr();
printf(\"application is closing..\ \");
exit(0);
}
default:
{
printf(\"\ Please re-entered correct option\");
}
}
}
void addbooks(void) //funtion to add books
{
int i;
fp=fopen(\"Bibek.dat\",\"ab+\");
fseek(fp,0,SEEK_END);
fwrite(&r,sizeof(r),1,fp);
fclose(fp);
printf(\"The record is sucessfully saved\");
printf(\"add any more?(t / b):\");//t to try again and b to back
if(getch()==\'b\')
else
addbooks();
}
}
void deletebooks() //function delete record from file fp
{
int d;
char another=\'y\';
while(another=.
estion 47 (1 point) Raul Prebisch, based at the Economic Commission f.pdf
estion 47 (1 point) Raul Prebisch, based at the Economic Commission for Latin America and
Hans Singer, based t the United Nations Industrial Development Organization, both provided the
intellectual ationale for O 1) import substituting industrialization O 2) the rapid development of
the service sector industry. O 3) free trade. 4) liberation theology
Solution
Raul Prebisch and Hans Singer gave Prebisch-Singer hypothesis which formed the basis of
dependency theory and supported theories like import substitution.
The correct answer to this question is \"1\" import-substituting industrialization..
Dr. Muehls mother was a nurse and her father was a dairy farmer. S.pdf
Dr. Muehl\'s mother was a nurse and her father was a dairy farmer. She has gone to college and
has earned her PhD, which allowed her to become a professor at SUNY Canton. This is an
example of:
resiliency
ascribed status
social mobility
social stratificationA.
resiliencyB.
ascribed statusC.
social mobilityD.
social stratification
Solution
When a person moves from one societal status to another which the case mentioned in the
scenario of the question, it can be termed as social mobility
Therefore (C) social mobility is the answer.
CHAPTER 19 Reproductive system Disorders 24. List the STD and the org.pdf
CHAPTER 19 Reproductive system Disorders 24. List the STD and the organism that a.
Chancre causes a b. Vesicle C. Gumma d. Purulent exudate e. Pharyngitis f. Wart
Solution
A. chancre
The exposure of Treponema pallidum, the gram-negative spirochaete bacterium yielding
syphilis. Chancres transmit the sexually transmissible disease of syphilis through direct physical
contact. These ulcers usually form on or around the anus, mouth, penis, and vagina..
can Fluorescence Resonance Emission Transfer (FRET) Microscopy be us.pdf
can Fluorescence Resonance Emission Transfer (FRET) Microscopy be used to image static
structures??
Solution
The technique of fluorescence resonance energy transfer , when applied to optical microscopy,
permits determination of the approach between two molecules within several nanometers.FRET
is a powerful technique for studying molecular interactions inside living cells with improved
spatial (angstrom) and temporal (nanosecond) resolution. FRET microscopy relies on the ability
to capture fluorescent signals from the interactions of labeled molecules in single living or fixed
cells. Hence FRET Microscopy is used to image static structures..
An analyst has decided to capitalize the operating leases of Company.pdf
An analyst has decided to capitalize the operating leases of Company A. Using information in
the notes to the company’s 2015 financial statements, she has determined that the present value
of future minimum lease payments, at a discount rate of 10 percent, on December 31, 2015
equals €500 million. All lease contracts last another 5 years on December 31, 2015. As expected
at the beginning of the year, the company reports an operating lease expense in its income
statement for 2016 of €80 million. The company’s tax rate equals 30 percent. The company does
not engage in any new operating leases in 2016. The following information is also available from
Company A’s financial statements (all ratios use beginning-of-the-year balance sheet values)
Debt to capital (at beginning of 2016) = 0.55
Return on beginning equity in 2016 = 0.10
Assets / Capital (at beginning of 2016) = €3,400 million
The effect of capitalizing Company A’s operating leases on its return on beginning equity equals
A. An increase from 0.10 to 0.15 (rounded).
B. An increase from 0.10 to 0.13 (rounded).
C. A decrease from 0.10 to 0.07 (rounded).
D. A decrease from 0.10 to 0.05 (rounded).
Solution
accounting topic of leases is a popular Paper F7 exam area that could feature to varying degrees
in Questions 2, 3, 4 or 5 of the exam. This topic area is currently covered by IAS 17, Leases. IAS
17, Leases takes the concept of substance over form and applies it to the specific accounting area
of leases.
When applying this concept, it is often deemed necessary to account for the substance of a
transaction – ie its commercial reality, rather than its strict legal form. In other words, the legal
basis of a transaction can be used to hide the true nature of a transaction. It is argued that by
applying substance, the financial statements become more reliable and ensure that the lease is
faithfully represented.
WHY DO WE NEED TO APPLY SUBSTANCE TO A LEASE?
A lease agreement is a contract between two parties, the lessor and the lessee. The lessor is the
legal owner of the asset, the lessee obtains the right to use the asset in return for rental payments.
Historically, assets that were used but not owned were not shown on the statement of financial
position and therefore any associated liability was also left out of the statement – this was known
as ‘off balance sheet’ finance and was a way that companies were able to keep their liabilities
low, thus distorting gearing and other key financial ratios. This form of accounting did not
faithfully represent the transaction. In reality a company often effectively ‘owned’ these assets
and ‘owed a liability’.
Under modern day accounting the IASB framework states that an asset is ‘a resource controlled
by an entity as a result of past events and from which future economic benefits are expected to
flow to the entity’ and a liability is ‘a present obligation of the entity arising from past events, the
settlement of which is expected to result in an outflow fr.
Why is a visual analysis important in communicating findings in a bu.pdf
Why is a visual analysis important in communicating findings in a business research paper or
proposal? Give an example of a visual analysis that would be appropriate for your paper and
presentation (my research paper is regarding the decrease in ethics in nurses as well as having
usage of personal cell phones on the work floor). Describe acceptable report page design
methods to make your paper or presentation visually appealing.
Solution
The increase of development in IT has leaded to tremendous increase in the use of visual forms
of communication. From the early begins of usage of social anthropology, where novel
technologies of register and reproduction of images were gradually incorporated within social
anthropology in terms of ethnographic film and later visual anthropology, research with visual
technologies and on visual data has evolved into other fields in the social science.
Apart from social science, visual materials are found in a wide range of cultural spheres. They
are extensively used at various fields.
Videography, Photography plays very important roles in terms of gaining the full focus on
peculiarity of the visual and its specific symbolic structure.
The following terms will describe best about the Visual analysis usage at a greater extent. 1.
Involving, 2. Sharing and 3. Analyzing.
The three terms described above are exactly useful for the best communication of Business
Research Paper or Proposal.
Because, involving mostly related to successful participation in Trans disciplinary projects. The
photos can be analyzed as participatory acts. Sharing mostly involves in the combination of the
verbal data and visual data which leads to the easy understanding of presentation. Analysis
completely describes the different methods of analyses and show societal context.
Most of the design methods follow a simple procedure which gives a good pave to go ahead. One
of the design methods which is relevant to most of the design methods is described as follows.
Any methodology contains 1. Sample Design, 2. Data Collection Methods, 3. Data Processing
Procedures, 4. Response Rates, 5. Weighing and Variance Estimation.
All terminologies mentioned above are related to the easy understanding of the design in a
feature of successful survey and presents many issues that had to be addressed at the design
stage..
what usually happens to autoimmune antibody-producing clones dur.pdf
what usually happens to autoimmune antibody-producing clones during development?
what usually happens to autoimmune antibody-producing clones during development?
what usually happens to autoimmune antibody-producing clones during development?
Solution
The process of immune response against the antibodies is the same when the immune response
acts against the normal body cells. They bind with high affinity to self antigens..
Why platinum and its compounds display a variety of colorSoluti.pdf
Why platinum and its compounds display a variety of color?
Solution
Platinum and its compounds exhibit varieties of color because plantinum in its compounds have
d electrons in their respective orbitals which absorb wavelengths from Ultra violet and visible
light region of electromagnetic radiation and give off complimentary color and in this way
platinum generally appreas greyish white in color..
What is Vertical versus horizontal integration in healthcare W.pdf
What is Vertical versus horizontal integration in healthcare?
What is Vertical versus horizontal integration in healthcare?
Solution
in healthcare a vertical integration is the arrangement, a healthcare organization provide and
offer various services to the patients and needed people either directly or indirectlty, which
includes some support services or a broad range of patient care services. In this the care
providers think before and after the work or role he has done to the patients. The service
providers interacted and integrated with similar parties, exchange, share, and at the end benefit
all. Most of the care centers which offers variety of services join together, and exchange
information or patients also.
in horizontal integraion, one kind of services are offered by the care centers, all these joined as a
group for sharing the work. Here the care provider may look for other provider who do the same,
and join with them. It helps them to utilize their resoruces in optimum way and minimize
wastage. Forexample: these both can maintain a group of professional doctors, they can utilize
them in both places based on demand from patients..
What is the role of the financial system Name and describe two mark.pdf
What is the role of the financial system? Name and describe two markets that are part of the
financial system in the U.S. economy. Name and describe two financial intermediaries. What is
the difference between savings and investment (according to economists) and what role does the
financial system serve in bridging the gap between savings and investment? Explain why a
person would want to invest in stocks and bonds rather than keep their money in a savings
account.? ( 200 words)
Solution
Ans.
The financial system is concerned about credit, money and finance. It is a set of complex and
inter-mixed financial institutions ,financial markets, financial instruments and services.It is a
system that enables lenders and borrowers to exchange funds.It serves as a link between savers
and investors.Financial system promotes the process of capital formation and is a mechanism of
transfer of resources.Such as it channelise savings into business sector by offering loans.
Most important financial market is bonds market.In this market bond is like a contract paper with
maturity date and rate at which it is paid back to buyers.The money that companies get from
bonds can be further used for making profitable projects.When bond matures it is paid back to
purchaser with interest. The other market is of stocks market where stocks are issued by selling
shares to the public and trading is done with stockholders in stock exchange.It has higher returns
than bonds.
The two financial intermediaries are banks and mutual funds.Where banks are the financial
institutions that receives deposits and make loans.They pay interest on their deposits and charge
borrowers higher interst rate on their loans.Mutual funds are investment that uses money from
many investors to invest in bonds and stocks.
Savings are part of disposable income that is left after consumption of goods and
accumulated.Whereas investment means addition to stock of capital goods that adds to the future
productive capacity of the economy. Financial system is an important vehicle for economic
transformation.It bridges the gap between savings and investment through efficient mobilization
and allocation of surplus funds.It facilitates flow of funds from areas of surplus to deficit.
The money in savings account cannot be sold or traded to other investors such that they are not
negotiable, Whereas bonds are sold by investors to other investors before bonds reach there
maturity date.Money in savings account offers less return than bonds but bonds offer higher
returns which are considered as safe investment when borrowed from goverenment .so its better
to invest in stocks and bonds rather than keeping money in savings account..
What is the difference between a paired artery and an unpaired arter.pdf
What is the difference between a paired artery and an unpaired artery?
Solution
Paired Artery:
The abdominal aorta is the largest artery in the abdominal cavity. As part of the aorta, it is a
direct continuation of the descending aorta (of the thorax).
Anatomy Any of the muscular elastic tubes that form a branching system and that carry blood
away from the heart to the cells, tissues, and organs of the body. 2. A major route of
transportation into which local routes flow: Traffic was heavy on the central artery.
Unpaired Artery:
an unpaired artery that is formed by the union of the two vertebral arteries, runs forward within
the skull just under the pons, divides into the two posterior cerebral arteries, and supplies the
pons, cerebellum, posterior part of the cerebrum, and the inner ear..
What has Microsoft improved in Windows Server 2012 when it comes to .pdf
What has Microsoft improved in Windows Server 2012 when it comes to provisioning and
administrating a virtualized environment?
Solution
Windows Server 2012:
Windows server 2012 is the 6th version of windows server family. Microsoft introduced
windows server 2012 and BUILD 2011 of developer preview. Unlike windows 8, windows
server 2012 developer made provisions to MSDN subscribers. Microsoft improves, Windows
Update, Window Server Update Services, system center configuration manager and release an
update that extends the finishing date from 8th April 2012 to 15th January 2013.
Based on provisions: made to MSDN subscribers
Based on virtual environment: It is generally made for customers..
What are the steps for developing Developing a Project Scope Stateme.pdf
What are the steps for developing Developing a Project Scope Statement?
Solution
A scope statement has to be created in order to convince and get agreement from the client and
other stakeholders. With a scope statement you are more likely to decrease the changes of
miscommunication
The scope statement should be created by including the following points:.
True or False Questions1) Good technology transfer between pr.pdf
True or False Questions:
1) Good technology transfer between projects is one of the major advantages of projectized
organization. ( T or F)
2) Program Management is less comprehensive and inclusive than project management.( T or F)
3) The partnership type of organization is very difficult to organize as it has relatively large
restrictions. (T or F).
4) Forecasting estimates the results and planning provides strategies. (T or F)
Multiple choice questions:
1) The distribution that you use for networks in project considering probability of three times is
known as:
A) Normal distribution
B) Exponential distribution.
C. Binomial distribution
D) Negative exponential distribution
E. None of the above
2) The biggest Advantage/advantages of the exponential smoothing method of forecasting is:
A) No waiting period is required and not a large data is required even then reasonably good
forecasting.
B) It includes the benefits of the time series analysis of forecasting as it belongs to this category.
C) The value of alpha (smoothing constant )can be made to change or adapt to changed
circumstances.
D. All of the above
E. Only A and B.
3) The important disadvantage/disadvantages of MBO as observed are:
A. Time and paper work involved
B. Supervisor just assigns and does not negotiate
C. It leads to Gamesmanship of subordinates
D. It does not avoid the tendency of subordinates to focus on relatively few goals.
E. All of the above
4) The title \'section chief or foreman, is used for which of the following position?
A) team leader
B. middle manager
C. first-line manager
D. top manager
E. Aged worker
5) About the middle managers which of the following is true?
A. Make plan of intermediate range to achieve long terr goals
B. Integrate and coordinate the short range decisions and activities
C. Establish department policies together
D. Options A and B are true if taken together
E. options A, B, and C, are true if taken
True or False Questions:
1) Good technology transfer between projects is one of the major advantages of projectized
organization. ( T or F)
2) Program Management is less comprehensive and inclusive than project management.( T or F)
3) The partnership type of organization is very difficult to organize as it has relatively large
restrictions. (T or F).
4) Forecasting estimates the results and planning provides strategies. (T or F)
Multiple choice questions:
1) The distribution that you use for networks in project considering probability of three times is
known as:
A) Normal distribution
B) Exponential distribution.
C. Binomial distribution
D) Negative exponential distribution
E. None of the above
2) The biggest Advantage/advantages of the exponential smoothing method of forecasting is:
A) No waiting period is required and not a large data is required even then reasonably good
forecasting.
B) It includes the benefits of the time series analysis of forecasting as it belongs to this category.
C) The value of alpha (smoothing constant )can be made to chan.
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H&M The Challenges of Global Expansion and the Move to adopt Intern.pdf
H&M The Challenges of Global Expansion and the Move to adopt Intern.pdf
C++ ProgrammingYou are to develop a program to read Baseball playe.pdf
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