Using the provided table interface table.h and the sample linked lis.pdf
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Using the provided table interface table.h and the sample linked list code linkedList.c, complete an implementation of the Table ADT. Make sure that you apply the concepts of design by contract (DbC) to your implementation. Once you have fully implemented the table, create a main.c file that implements a testing framework for your table. Your table implementation must ensure that values inserted are unique, and internally sorted within a linked list. table.h linkedList.c Solution khsdg #include #include typedef enum BOOL { false, true } bool; // Linked list node definition typedef struct Node node; struct Node { int number; node *next; }; static node *top = NULL; // used to track where we are for the list traversal methods static node *traverseNode = NULL; // \"destroy\" will deallocate all nodes in a linked list object // and will set \"top\" to NULL. void destroy() { node *curr = top; node *temp = NULL; while ( curr != NULL ) { // flip order to see it blow up... temp = curr; curr = curr->next; free( temp ); } top = NULL; } // \"build\" will create an ordered linked list consisting // of the first \"size\" even integers. void build( int size ) { node *newNode = NULL; int i = 0; // make sure we don\'t have a list yet destroy(); for ( i=size ; i>0 ; i-- ) { newNode = malloc( sizeof( node ) ); newNode->number = i*2; newNode->next = top; top = newNode; } } // starts a list traversal by getting the data at top. // returns false if top == NULL. bool firstNode( int *item ) { bool result = false; if ( top ) { *item = top->number; traverseNode = top->next; result = true; } return result; } // gets the data at the current traversal node and increments the traversal. // returns false if we\'re at the end of the list. bool nextNode( int *item ) { bool result = false; if ( traverseNode ) { *item = traverseNode->number; traverseNode = traverseNode->next; result = true; } return result; } // \"print\" will output an object\'s entire linked list // to the standard output device -- one \"number\" per line. void print() { int value; if ( firstNode( &value ) ) { do { printf( \"%d\ \", value ); } while ( nextNode( &value ) ); } } int main( int argc, char *argv[] ) { build( 10 ); print(); destroy(); build( 5 ); build( 20 ); print(); destroy(); return 0; }.
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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.
Lab-2.4 101.pdf
The document contains C program code to implement three data structures using linked lists:
1) A stack using a singly linked list with push, pop, and display functions.
2) A queue using a singly linked list with enqueue, dequeue, and display functions.
3) A program to merge two sorted linked lists by comparing nodes and appending to the merged list.
Pseudocode and sample test cases are provided for each problem.
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.
could you implement this function please, im having issues with it..pdf
could you implement this function please, im having issues with it.
void makeList (const ListNode::value_type [],const size_t& count)
class ListNode
{
public:
typedef int value_type;
ListNode (value_type d = value_type(), ListNode* n = NULL) { datum = d; next = n; }
//Assessor
value_type getDatum () const { return datum; }
ListNode const* getNext () const { return next; }
//Mutator
void setDatum (const value_type& d) {datum = d; }
ListNode* getNext () { return next; }
void setNext (ListNode* new_link) {next = new_link; }
private:
value_type datum;
ListNode* next;
};
class LinkedList
{
public:
LinkedList ();
virtual ~LinkedList ();
void insertItem (ListNode::value_type);
void makeList (const ListNode::value_type [],const size_t& count);
void deleteList ();
//The following friend function is implemented in lablinklist.cpp
friend std::ostream& operator<<(std::ostream&, const LinkedList&);
private:
ListNode* head;
};
This is the pseudocode, but i still have a hard time undertanding it.
Creating a List (makeList(const ListNode::value_type [],const size_t& count)) This function
receives an array in the order that we want to add it to the linkedlist. Index 0 will be the head
node, index n will be the last node.
First, create a node initialized with a data value and a NULL pointer. Set the \"head\" to point to
the first node. Set up a current-pointer to the first node (or \"head\"). Get a data value for the next
node. While more nodes to add { Create a new node initialized with the data value and a NULL
pointer. Set the current-pointer link member (\"next\") to the new node. Set the current-pointer to
point to the new node. Get a data value for the next node. }
Thanks.
Solution
//linkedList.h
#ifndef LINKEDLIST_H
#define LINKEDLIST_H
#include
#include
class ListNode
{
public:
typedef int value_type;
ListNode(value_type d = value_type(), ListNode* n = NULL) { datum = d; next = n; }
//Assessor
value_type getDatum() const { return datum; }
ListNode const* getNext() const { return next; }
//Mutator
void setDatum(const value_type& d) { datum = d; }
ListNode* getNext() { return next; }
void setNext(ListNode* new_link) { next = new_link; }
private:
value_type datum;
ListNode* next;
};
class LinkedList
{
public:
LinkedList();
virtual ~LinkedList();
void insertItem(ListNode::value_type);
void makeList(const ListNode::value_type[], const size_t& count);
void deleteList();
//The following friend function is implemented in lablinklist.cpp
friend std::ostream& operator<<(std::ostream&, const LinkedList&);
private:
ListNode* head;
};
#endif
-------------------------------------------------------------
//linkedList.cpp
#include \"linkedlist.h\"
LinkedList::LinkedList()
{
head = NULL;
}
LinkedList::~LinkedList()
{
ListNode *tmp = head;
while (tmp != NULL)
{
head = head->getNext();
delete tmp;
tmp = head;
}
}
void LinkedList::insertItem(ListNode::value_type item)
{
ListNode *newNode,*cur = head;
newNode = new ListNode;
newNode->setDatum(item);
newNode->setNext(NULL);.
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#includeiostream #includecstdio #includecstdlib using na.pdf
#include
#include
#include
using namespace std;
struct node
{
int val;
struct node *next;
}*startNode;
// declaring class
class LinkedList
{
public:
node* create_node(int);
void insert_last();
void sort();
void search();
void reverse();
void display();
LinkedList()
{
startNode = NULL;
}
};
// main
main()
{
int userChoice, linkedListNodes, i;
LinkedList LinkList;
startNode = NULL;
while (1)
{
// displaying menu
cout<<\" 1)Insert\";
cout<<\"2) Remove\";
cout<<\"3) Search\";
cout<<\"4) Display\";
cout<<\"5) Reverse\";
cout<<\"6) Exit \";
// reading user choice
cout<<\"Enter your userChoice : \";
cin>>userChoice;
switch(userChoice)
{
case 1:
cout<<\"Inserting node to linked list: \";
LinkList.insert_begin();
cout;
break;
case 2:
cout<<\"Delete element from linked list: \";
LinkList.delete();
break;
case 3:
cout<<\"Searching data in linked list: \";
LinkList.search();
cout;
break;
case 4:
cout<<\"Displaying linked list data: \";
LinkList.display();
cout;
break;
case 5:
cout<<\"Linked list reversing: \";
LinkList.reverse();
cout;
break;
case 6:
cout<<\"Exiting...\";
exit(1);
break;
default:
cout<<\"Enter correct userChoice\";
}
}
}
// node creating to store into list
node *LinkedList::create_node(int val)
{
struct node *tempNode, *tempNode;
tempNode = new(struct node);
if (tempNode == NULL)
{
cout<<\"No memory allocated \";
return 0;
}
else
{
tempNode->val = val;
tempNode->next = NULL;
return tempNode;
}
}
/*
* Inserting Node at last
*/
void LinkedList::insert_last()
{
int val;
cout<<\"Enter the val to be inserted: \";
cin>>val;
struct node *tempNode, *tempNode;
tempNode = create_node(val);
tempNode = startNode;
while (tempNode->next != NULL)
{
tempNode = tempNode->next;
}
tempNode->next = NULL;
tempNode->next = tempNode;
cout<<\"Element Inserted\";
}
// deleting element
void LinkedList::delete()
{
int position, i, count = 0;
if (startNode == NULL)
{
cout<<\"Linkedlist empty\"<>position;
struct node *tempNode, *ptr;
tempNode = startNode;
if (position == 1)
{
startNode = tempNode->next;
}
else
{
while (tempNode != NULL)
{
tempNode = tempNode->next;
count++;
}
if (position > 0 && position <= count)
{
tempNode = startNode;
for (i = 1;i < position;i++)
{
ptr = tempNode;
tempNode = tempNode->next;
}
ptr->next = tempNode->next;
}
else
{
cout<<\"Enter correct position\";
}
}
}
// search data in linked list
void LinkedList::search()
{
int val, position = 0;
bool flagValue = false;
if (startNode == NULL)
{
cout<<\"List is empty\";
return;
}
cout<<\"Enter the val to search: \";
cin>>val;
struct node *tempNode;
tempNode = startNode;
while (tempNode != NULL)
{
position++;
if (tempNode->val == val)
{
flagValue = true;
cout<<\"Data \"<next;
}
if (!flagValue)
cout<<\"data not found\";
}
// linkedlist reverese
void LinkedList::reverse()
{
struct node *pointer1, *pointer2, *pointer3;
if (startNode == NULL)
{
cout<<\"List is empty\"<next == NULL)
{
return;
}
pointer1 = startNode;
pointer2 = pointer1->next;
pointer3 = pointer2->next;
pointer1->next = NULL;
pointer2->.
C code on linked list #include stdio.h #include stdlib.h.pdf
The C code defines functions to create and manipulate a linked list. It takes 5 user-input values, inserts them into an empty linked list, prints the list, takes another input value and inserts it, then prints the final list.
linked List.docx vhjgvjhvgjhjhbbjkhkjhkjh
This C code defines structures for nodes and linked lists. It provides functions for creating nodes, inserting nodes at the beginning or end of a linked list, deleting nodes with a specific value, searching for a node, and displaying the linked list. These functions are used in the main function to demonstrate manipulating a linked list, including inserting nodes, deleting nodes, and searching.
Recommended
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.
Lab-2.4 101.pdf
The document contains C program code to implement three data structures using linked lists:
1) A stack using a singly linked list with push, pop, and display functions.
2) A queue using a singly linked list with enqueue, dequeue, and display functions.
3) A program to merge two sorted linked lists by comparing nodes and appending to the merged list.
Pseudocode and sample test cases are provided for each problem.
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.
could you implement this function please, im having issues with it..pdf
could you implement this function please, im having issues with it.
void makeList (const ListNode::value_type [],const size_t& count)
class ListNode
{
public:
typedef int value_type;
ListNode (value_type d = value_type(), ListNode* n = NULL) { datum = d; next = n; }
//Assessor
value_type getDatum () const { return datum; }
ListNode const* getNext () const { return next; }
//Mutator
void setDatum (const value_type& d) {datum = d; }
ListNode* getNext () { return next; }
void setNext (ListNode* new_link) {next = new_link; }
private:
value_type datum;
ListNode* next;
};
class LinkedList
{
public:
LinkedList ();
virtual ~LinkedList ();
void insertItem (ListNode::value_type);
void makeList (const ListNode::value_type [],const size_t& count);
void deleteList ();
//The following friend function is implemented in lablinklist.cpp
friend std::ostream& operator<<(std::ostream&, const LinkedList&);
private:
ListNode* head;
};
This is the pseudocode, but i still have a hard time undertanding it.
Creating a List (makeList(const ListNode::value_type [],const size_t& count)) This function
receives an array in the order that we want to add it to the linkedlist. Index 0 will be the head
node, index n will be the last node.
First, create a node initialized with a data value and a NULL pointer. Set the \"head\" to point to
the first node. Set up a current-pointer to the first node (or \"head\"). Get a data value for the next
node. While more nodes to add { Create a new node initialized with the data value and a NULL
pointer. Set the current-pointer link member (\"next\") to the new node. Set the current-pointer to
point to the new node. Get a data value for the next node. }
Thanks.
Solution
//linkedList.h
#ifndef LINKEDLIST_H
#define LINKEDLIST_H
#include
#include
class ListNode
{
public:
typedef int value_type;
ListNode(value_type d = value_type(), ListNode* n = NULL) { datum = d; next = n; }
//Assessor
value_type getDatum() const { return datum; }
ListNode const* getNext() const { return next; }
//Mutator
void setDatum(const value_type& d) { datum = d; }
ListNode* getNext() { return next; }
void setNext(ListNode* new_link) { next = new_link; }
private:
value_type datum;
ListNode* next;
};
class LinkedList
{
public:
LinkedList();
virtual ~LinkedList();
void insertItem(ListNode::value_type);
void makeList(const ListNode::value_type[], const size_t& count);
void deleteList();
//The following friend function is implemented in lablinklist.cpp
friend std::ostream& operator<<(std::ostream&, const LinkedList&);
private:
ListNode* head;
};
#endif
-------------------------------------------------------------
//linkedList.cpp
#include \"linkedlist.h\"
LinkedList::LinkedList()
{
head = NULL;
}
LinkedList::~LinkedList()
{
ListNode *tmp = head;
while (tmp != NULL)
{
head = head->getNext();
delete tmp;
tmp = head;
}
}
void LinkedList::insertItem(ListNode::value_type item)
{
ListNode *newNode,*cur = head;
newNode = new ListNode;
newNode->setDatum(item);
newNode->setNext(NULL);.
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.
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#includeiostream #includecstdio #includecstdlib using na.pdf
#include
#include
#include
using namespace std;
struct node
{
int val;
struct node *next;
}*startNode;
// declaring class
class LinkedList
{
public:
node* create_node(int);
void insert_last();
void sort();
void search();
void reverse();
void display();
LinkedList()
{
startNode = NULL;
}
};
// main
main()
{
int userChoice, linkedListNodes, i;
LinkedList LinkList;
startNode = NULL;
while (1)
{
// displaying menu
cout<<\" 1)Insert\";
cout<<\"2) Remove\";
cout<<\"3) Search\";
cout<<\"4) Display\";
cout<<\"5) Reverse\";
cout<<\"6) Exit \";
// reading user choice
cout<<\"Enter your userChoice : \";
cin>>userChoice;
switch(userChoice)
{
case 1:
cout<<\"Inserting node to linked list: \";
LinkList.insert_begin();
cout;
break;
case 2:
cout<<\"Delete element from linked list: \";
LinkList.delete();
break;
case 3:
cout<<\"Searching data in linked list: \";
LinkList.search();
cout;
break;
case 4:
cout<<\"Displaying linked list data: \";
LinkList.display();
cout;
break;
case 5:
cout<<\"Linked list reversing: \";
LinkList.reverse();
cout;
break;
case 6:
cout<<\"Exiting...\";
exit(1);
break;
default:
cout<<\"Enter correct userChoice\";
}
}
}
// node creating to store into list
node *LinkedList::create_node(int val)
{
struct node *tempNode, *tempNode;
tempNode = new(struct node);
if (tempNode == NULL)
{
cout<<\"No memory allocated \";
return 0;
}
else
{
tempNode->val = val;
tempNode->next = NULL;
return tempNode;
}
}
/*
* Inserting Node at last
*/
void LinkedList::insert_last()
{
int val;
cout<<\"Enter the val to be inserted: \";
cin>>val;
struct node *tempNode, *tempNode;
tempNode = create_node(val);
tempNode = startNode;
while (tempNode->next != NULL)
{
tempNode = tempNode->next;
}
tempNode->next = NULL;
tempNode->next = tempNode;
cout<<\"Element Inserted\";
}
// deleting element
void LinkedList::delete()
{
int position, i, count = 0;
if (startNode == NULL)
{
cout<<\"Linkedlist empty\"<>position;
struct node *tempNode, *ptr;
tempNode = startNode;
if (position == 1)
{
startNode = tempNode->next;
}
else
{
while (tempNode != NULL)
{
tempNode = tempNode->next;
count++;
}
if (position > 0 && position <= count)
{
tempNode = startNode;
for (i = 1;i < position;i++)
{
ptr = tempNode;
tempNode = tempNode->next;
}
ptr->next = tempNode->next;
}
else
{
cout<<\"Enter correct position\";
}
}
}
// search data in linked list
void LinkedList::search()
{
int val, position = 0;
bool flagValue = false;
if (startNode == NULL)
{
cout<<\"List is empty\";
return;
}
cout<<\"Enter the val to search: \";
cin>>val;
struct node *tempNode;
tempNode = startNode;
while (tempNode != NULL)
{
position++;
if (tempNode->val == val)
{
flagValue = true;
cout<<\"Data \"<next;
}
if (!flagValue)
cout<<\"data not found\";
}
// linkedlist reverese
void LinkedList::reverse()
{
struct node *pointer1, *pointer2, *pointer3;
if (startNode == NULL)
{
cout<<\"List is empty\"<next == NULL)
{
return;
}
pointer1 = startNode;
pointer2 = pointer1->next;
pointer3 = pointer2->next;
pointer1->next = NULL;
pointer2->.
C code on linked list #include stdio.h #include stdlib.h.pdf
The C code defines functions to create and manipulate a linked list. It takes 5 user-input values, inserts them into an empty linked list, prints the list, takes another input value and inserts it, then prints the final list.
linked List.docx vhjgvjhvgjhjhbbjkhkjhkjh
This C code defines structures for nodes and linked lists. It provides functions for creating nodes, inserting nodes at the beginning or end of a linked list, deleting nodes with a specific value, searching for a node, and displaying the linked list. These functions are used in the main function to demonstrate manipulating a linked list, including inserting nodes, deleting nodes, and searching.
C Homework Help
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You can also call on +1 678 648 4277 for any assistance with C Homework.
How do you stop infinite loop Because I believe that it is making a.pdf
How do you stop infinite loop? Because I believe that it is making an infinite circular list.
c++ code:
Here is the list class:
#ifndef LIN_J_LIST
#define LIN_J_LIST
typedef unsigned int uint;
#include
#include
using namespace std;
/**
* a simplified generic singly linked list class to illustrate C++ concepts
* @author Jerry Lin
* @version 2/17/17
*/
template< typename Object >
class List
{
private:
/**
* A class to store data and provide a link to the next node in the list
*/
class Node
{
public:
/**
* The constructor
* @param data the data to be stored in this node
*/
explicit Node( const Object & data )
: data{ data }, next{ nullptr } {}
Object data;
Node * next;
};
public:
/**
* The constructor for an empty list
*/
List()
: size{ 0 }, first{ nullptr }, last{ nullptr } {}
/**
* the copy constructor-creates and copy the list
*/
List( List && rhs ) = delete;
List( const List & rhs )
{
count = 0;
size = 0;
if(rhs.size != 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first;
size++;
while(current->next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op.
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode->next;
size++;
}
}
// you document and implement this method
}
/**
* the operator= method-copies the list
*/
List & operator=( List && rhs) = delete;
List & operator=( const List & rhs )
{
count = 0;
size = 0;
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
if(rhs.size!= 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first; //create a temporary to store
size++;
while(current -> next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode -> next;
size++;
}
}
return *this;
// you document and implement this method
}
/**
* accessor
* @return count
*/
int get_count() const
{
return count;
}
/**
* The destructor that gets rid of everything that\'s in the list and
* resets it to empty. If the list is already empty, do nothing.
*/
~List()
{
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
}
/**
* Put a new element onto the beginning of the list
* @param item the data the new element will contain
*/
void push_front( const Object& item )
{
Node * new_node = new Node( item );//basic op.
if(is_empty())
{
last = new_node;
}
else
{
new_node->next = first;
}
first = new_node;
size++;
/* you complete the rest */
}
/**
* Remove the element that\'s at the front of the list. Causes an
* assertion error if the list is empty.
*/
void pop_front()
{
assert( !is_empty() );
Node * temp = first;
if( first == last )
{
first = last = nullptr;
}
else
{
first = first->next;
}
delete temp;
size--;
}
/**
* Accessor to return the da.
A)B) C++ program to create a Complete Binary tree from its Lin.pdf
A)
B)
// C++ program to create a Complete Binary tree from its Linked List
// Representation
#include
#include
#include
using namespace std;
// Linked list node
struct ListNode
{
int data;
ListNode* next;
};
// Binary tree node structure
struct BinaryTreeNode
{
int data;
BinaryTreeNode *left, *right;
};
// Function to insert a node at the beginning of the Linked List
void push(struct ListNode** head_ref, int new_data)
{
// allocate node and assign data
struct ListNode* new_node = new ListNode;
new_node->data = new_data;
// link the old list off the new node
new_node->next = (*head_ref);
// move the head to point to the new node
(*head_ref) = new_node;
}
// method to create a new binary tree node from the given data
BinaryTreeNode* newBinaryTreeNode(int data)
{
BinaryTreeNode *temp = new BinaryTreeNode;
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// converts a given linked list representing a complete binary tree into the
// linked representation of binary tree.
void convertList2Binary(ListNode *head, BinaryTreeNode* &root)
{
// queue to store the parent nodes
queue q;
// Base Case
if (head == NULL)
{
root = NULL; // Note that root is passed by reference
return;
}
// 1.) The first node is always the root node, and add it to the queue
root = newBinaryTreeNode(head->data);
q.push(root);
// advance the pointer to the next node
head = head->next;
// until the end of linked list is reached, do the following steps
while (head)
{
// 2.a) take the parent node from the q and remove it from q
BinaryTreeNode* parent = q.front();
q.pop();
// 2.c) take next two nodes from the linked list. We will add
// them as children of the current parent node in step 2.b. Push them
// into the queue so that they will be parents to the future nodes
BinaryTreeNode *leftChild = NULL, *rightChild = NULL;
leftChild = newBinaryTreeNode(head->data);
q.push(leftChild);
head = head->next;
if (head)
{
rightChild = newBinaryTreeNode(head->data);
q.push(rightChild);
head = head->next;
}
// 2.b) assign the left and right children of parent
parent->left = leftChild;
parent->right = rightChild;
}
}
// Utility function to traverse the binary tree after conversion
void inorderTraversal(BinaryTreeNode* root)
{
if (root)
{
inorderTraversal( root->left );
cout << root->data << \" \";
inorderTraversal( root->right );
}
}
// Driver program to test above functions
int main()
{
// create a linked list shown in above diagram
struct ListNode* head = NULL;
push(&head, 36); /* Last node of Linked List */
push(&head, 30);
push(&head, 25);
push(&head, 15);
push(&head, 12);
push(&head, 10); /* First node of Linked List */
BinaryTreeNode *root;
convertList2Binary(head, root);
cout << \"Inorder Traversal of the constructed Binary Tree is: \ \";
inorderTraversal(root);
return 0;
}
c)
#include
#include
using namespace std;
template
class bintree
{
bintree *left;
T data;
bintree *right;
public :
bintree()
{
left=right=NULL;
}
void create();
void preorder();
void inorder();
void pos.
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED
HELP FIXING THIS CODE?
THIS IS THE PROBLEM
Write a program that allows the user to enter two positive integers and outputs their sum. Sounds
simple? Here\'s the catch: The numbers may be any size! They may be way too large to store in a
single variable. You may not assume any particular maximum number of digits.
HERE IS THE CODE
#include
#include
struct node /* Linked list node*/
{
int data;
struct node* next;
};
struct node *newNode(int data) /* Function to create a new node with giving the data*/
{
struct node *new_node = (struct node *) malloc(sizeof(struct node));
new_node->data = data;
new_node->next = NULL;
return new_node;
}
void push(struct node** head_ref, int new_data) /* Function to insert a node at the beginning of
the Doubly Linked List*/
{
struct node* new_node = newNode(new_data); /*allocate a node*/
new_node->next = (*head_ref); /* link the old list off to the new node*/
(*head_ref) = new_node; /*move the head to point to the new node*/
}
/* Adding the contents of two linked lists and return the head node of resultant list*/
struct node* addTwoLists (struct node* first, struct node* second)
{
struct node* res = NULL; /* res is head node of the resultant list*/
struct node *temp, *prev = NULL;
int carry = 0, sum;
while (first != NULL || second != NULL) /*while both the lists exists*/
{
/* Calculate value of next digit in resultant list.
// The next digit is sum of following things
// (i) Carry
// (ii) Next digit of first list (if there is a next digit)
// (ii) Next digit of second list (if there is a next digit)*/
sum = carry + (first? first->data: 0) + (second? second->data: 0);
carry = (sum >= 10)? 1 : 0; /* update carry for next calulation*/
sum = sum % 10; /*update sum if it is greater than 10*/
temp = newNode(sum); /*Create a new node with sum as data*/
if(res == NULL) /*if this is the first node then set it as head of the resultant list*/
res = temp;
else /* If this is not the first node then connect it to the rest.*/
prev->next = temp;
prev = temp; /* Set prev for next insertion*/
if (first) first = first->next; /* Move first pointers to next node*/
if (second) second = second->next; /*Move second pointer to the next node*/
}
if (carry > 0)
temp->next = newNode(carry);
return res; /* return head of the resultant list*/
}
void printList(struct node *node) /* function to print a linked list*/
{
while(node != NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
printf(\"\ \");
}
int main(void) /* program to test above function*/
{
struct node* res = NULL;
struct node* first = NULL;
struct node* second = NULL;
printf(\"Enter first positive integer: \");
scanf(\"%d\",&first);
printList(first);
printf(\"Enter second positive integer: \");
scanf(\"%d\",&second);
printList(second);
/*Add the two lists and see result*/
res = addTwoLists(first, second);
printf(\"Sum is: %d\");
printList(res);
return 0;
}
Solution
#include
#include
typedef struct Node
{
int da.
Inspect the class declaration for a doubly-linked list node in Node-h-.pdf
Inspect the class declaration for a doubly-linked list node in Node.h. Access Node.h by clicking
on the orange arrow next to main.cpp at the top of the coding window. The Node class has three
member variables: a double data value, a pointer to the next node, and a pointer to the previous
node. Each member variable is protected. So code outside of the class must use the provided
getter and setter member functions to get or set a member variable. Node.h is read only, since no
changes are required. Step 2: Implement the Insert() member function A class for a sorted,
doubly-linked list is declared in SortedNumberList.h. Implement the SortedNumberList class's
Insert() member function. The function must create a new node with the parameter value, then
insert the node into the proper sorted position in the linked list. Ex: Suppose a
SortedNumberList's current list is 23 47.25 86, then Insert(33.5) is called. A new node with data
value 33.5 is created and inserted between 23 and 47.25, thus preserving the list's sorted order
and yielding: 23 35.5 47.25 86 Step 3: Test in develop mode Code in main() takes a space-
separated list of numbers and inserts each into a SortedNumberList. The list is displayed after
each insertion. Ex: If input is 77 15 -42 63.5 then output is: List after inserting 77: 77 List after
inserting 15: 15 77 List after inserting -42: -42 15 77 List after inserting 63.5: -42 15 63.5 77 Try
various program inputs, ensuring that each outputs a sorted list. Step 4: Implement the Remove()
member function Implement the SortedNumberList class's Remove() member function. The
function takes a parameter for the number to be removed from the list. If the number does not
exist in the list, the list is not changed and false is returned. Otherwise, the first instance of the
number is removed from the list and true is returned. Uncomment the commented-out part in
main() that reads a second input line and removes numbers from the list. Test in develop mode to
ensure that insertion and removal both work properly, then submit code for grading. Ex: If input
is 84 72 19 61 19 84 then output is: List after inserting 84: 84 List after inserting 72: 72 84 List
after inserting 19: 19 72 84 List after inserting 61: 19 61 72 84 List after removing 19: 61 72 84
List after removing 84: 61 72
main.cpp
#include
#include
#include
#include "Node.h"
#include "SortedNumberList.h"
using namespace std; void PrintList(SortedNumberList& list); vector SpaceSplit(string source);
int main(int argc, char *argv[]) { // Read the line of input numbers string inputLine; getline(cin,
inputLine); // Split on space character vector terms = SpaceSplit(inputLine); // Insert each value
and show the sorted list's contents after each insertion SortedNumberList list; for (auto term :
terms) { double number = stod(term); cout << "List after inserting " << number << ": " << endl;
list.Insert(number); PrintList(list); } /* // Read the input line with numbers to remove getline(cin,
inputLi.
C program to insert a node in doubly linked list
This C program implements functions to create and manipulate a doubly linked list. It allows the user to choose from a menu to create a list, insert nodes at the beginning, end, or a specified position, and display the list. Functions are defined to create a list by dynamically allocating nodes and linking them together, insert nodes at different positions by updating next and prev pointers, and display the list by traversing from head to tail.
Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
Assignment is:
\"Page 349-350 #4 and #5 Use the \"Linked List lab\" you have been working on in class and add
the two functions the questions are asking you to develop: divideMid and divideAt. Be sure to
include comments Use meaningful identifier names (constants where appropriate) Turn in .cpp
file AND Turn in a \"print-screen\' of your output (press \"print-screen\' on keyboard, then
\'paste\' in MS-Word)\"
How do you solve QUESTION #4 in the book data structures using c++ by D.S. Malik in Visiual
Studios using the linked list below with what is being asked? Please need help
Linked list :
#include
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *first, *last;
int num;
createList(first, last);
int choice;
while(true)
{
cout<<\"1. Insert Front.\ 2. Insert Last.\ 3. Delete Front.\ 4. Delete Last.\ 5. Print List.\ 6. Exit.\
\";
cout<<\"Enter your choice: \";
cin>>choice;
switch(choice)
{
case 1: insertFront(first); break;
case 2: insertBack(last); break;
case 3: deleteFirst(first); break;
case 4: deleteLast(last, first); break;
case 5: printList(first); break;
case 6: return 0;
default: cout<<\"Invalid menu option. Try again.\"<>number;
while (number != -999)
{
newNode = new nodeType; // create new node
newNode->info = number;
newNode->link = NULL;
if (first == NULL)
{
first = newNode;
last = newNode;
}
else
{
last->link = newNode;
last = newNode;
}
cout<<\"Enter an integer (-999 to stop): \";
cin>>number;
} // end of while-loop
} // end of build list function
void deleteFirst(nodeType*& first)
{
nodeType *temp;
temp= first;
first= temp->link;
delete temp;
return;
}
void deleteLast(nodeType*& last, nodeType* current)
{
nodeType *temp;
while(current->link != NULL)
{
temp=current;
current=current->link;
}
temp=last;
current->link=NULL;
delete temp;
last = current;
return;
}
void insertFront(nodeType*& front)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= front;
front= newNode;
return;
}
void insertBack(nodeType*& last)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= NULL;
last->link= newNode;
last = newNode;
return;
}
void printList(nodeType*& first)
{
cout<<\"Inside printList...printing linked list...\ \"<info << \" \";
current = current->link;
}
cout<
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *firs.
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
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.
C++Write a method Node Nodereverse() which reverses a list..pdf
C++
Write a method Node *Node::reverse() which reverses a list.
Solution
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Function to reverse the linked list */
static void reverse(struct node** head_ref)
{
struct node* prev = NULL;
struct node* current = *head_ref;
struct node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
/* Function to push a node */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print linked list */
void printList(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
printf(\"%d \", temp->data);
temp = temp->next;
}
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 85);
printf(\"Given linked list\ \");
printList(head);
reverse(&head);
printf(\"\ Reversed Linked list \ \");
printList(head);
getchar();
}.
There are a number of errors in the following program- All errors are.docx
There are a number of errors in the following program.
All errors are located in main() and structure definitions.
Function declarations and definitions are correct!
Locate all errors, fix them (as shown below), run the program and save its output
as a comment at the end of the source file.
Example:
int num = 10;
int *ptr;
num = &ptr; // <== Error: Comment the line and write the correct line below
// Write a short justification where appropriate
// num = &ptr; // Error #1
ptr = #
Name:
*/
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#define DUMMY_TRAILER '\177'
// octal ASCII code of the
// last character in the ASCII table
#define NUM_CITIES 10
typedef struct
{
char name[12];
int temperature[5];
} CITY;
// Stack and Queue Node
typedef struct node NODE;
struct node
{
CITY city;
node *next;
};
// Doubly Linked List Node
typedef struct d_node D_NODE;
struct d_node
{
CITY city;
NODE *forw;
NODE *back;
};
// Stack Functions
NODE *push(NODE *stack, const CITY *pStu);
NODE *pop(NODE **stack);
// Queue Functions
void enqueue(NODE **queue, NODE **rear, const CITY *pStu);
NODE *dequeue(NODE **queue, NODE **rear);
// Doubly Linked List Functions
D_NODE *init_list(void);
int insert(D_NODE *list, const CITY *pStu);
void traverse_forw(D_NODE *list);
void traverse_back(D_NODE *list);
// Other Functions
void printCity(const CITY *pCity);
int main (void)
{
CITY cList[NUM_CITIES] =
{
{"Cupertino", {88, 89, 87, 85, 89}},
{"Flagstaff", {81, 80, 88, 89, 89}},
{"Los Angeles", {87, 88, 89, 89, 90}},
{"Philadelphia", {96, 99, 99, 90, 95}},
{"Phoenix", {106, 109, 109, 100, 105}},
{"Portland", {89, 90, 85, 89, 90}},
{"Reno", {108, 105, 109, 100, 108}},
{"Salem", {85, 90, 85, 89, 90}},
{"Tucson", {107, 100, 109, 100, 108}},
{"Yreka", {101, 109, 100, 108, 109}}
};
NODE *stack = NULL;
NODE *top = NULL;
NODE *queue = NULL, *rear = NULL;
NODE *front;
D_NODE *list;
list = init_list();
// build stack and queue with data from an array of CITY structures
srand((unsigned int)time(NULL));
int count = rand() % 10;
for ( int n = 0; n < count; n++)
{
int i = rand() % NUM_CITIES;
int duplicate = insert(list, &cList[i]);
if(duplicate)
{
// already in the list!
push(stack, &cList[i]);
enqueue(&queue, &rear, cList[i]);
}
}
// display list
printf("\nLIST contents (forwards):\n");
traverse_forw(list);
printf("\nLIST contents (backwards):\n");
traverse_back(list);
// display stack
if (top)
{
printf("\nSTACK contents from top to bottom:\n");
while ((top = pop(stack)))
{
printCity(top->city);
}
}
else
printf ("Empty Stack!\n");
// display queue
if (front)
{
printf("\nQUEUE contents from front to rear:\n");
while ((front = dequeue( queue, rear)))
{
printCity(front->city);
}
}
else
printf ("Empty Queue!\n");
return 0;
}
/***************************************************
Displays the fileds of a CIS_CLASS structure
Pre pCls - a pointer to a CIS_CLASS structure
Post
*/
void printCity(const CITY *pCity)
{
printf("%-20s %3d\n", pCity->name, pCity->temperatur.
take the following code and give details of what each line of code i.pdf
take the following code and give details of what each line of code is doing
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<info<<\"->\";
temp = temp->next;
}
cout<<\"NULL\"<
Solution
ANSWER ::
ASSMUING THE TOTAL LINE IN CODE ARE 158*(INCLUDING COMMENTS)
SO...
PROGRAM STARTS WITH SPECIFING HEADER FILES AND GLOBAL VARIABLE
DECLARATION
THEN Node Declaration USING STRUCTURES,
THEN WE WANT TO DECLARE THE CLASS USING SINGLE LINKED LIST AND THE
PROTOTYPE IS PUBLIC
IN THE PUBLIC WE WANT TO CREATE THE NODE AS INTEGER DATATYPE
THE NODE CONSIST OF insert_begin();
insert_pos();-> INDICATES THE CURRENT POSSITION
insert_last();->INSERT AT LAST
delete_pos();->TO DELETE THE NUMBER AT CURRENT POSITION
sort();->START THE SORTING
search();->USED TO SEARCH
update();->UPDATE THE NODE
reverse();->REVERSE NUMBERS
display();->TO DISPLAY THE NODE
single_llist()->FIFO
WE ARE STARTING TAKING NULL VALUE
AND THEN TAKE USER DEFIND FUNCTION MAIN()
CREATE THE DATATYPES choice, nodes, element, position, i
OBJECT CREATION OF SLL IS SL
USING WHILE LOOP THE NODE INSERT AT THE BEGINING OR ENDING STAGE
WE ARE USING SWITCH CASES TO PERFORM SLL ON NODE
NODE CREATION AS
node *single_llist::create_node(int value)
ALSO DECLARE THE TEMP IS TAKEN AS NULL VALUE
INSERT THE ELEMENT AT THE BEGINING
USING IF CONDITION WE CAN INSERT THE NODE AT THE BEGINING
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
AND THEN INSERT THE NODE AT THE LAST
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
USING WHILE LOOP WE ARE INSERTING THE NODE AT LAST
USING THIS BELLOW CODE WE CAN SEARCH THE ELEMENTS IN THE NODE
struct node *temp;
if (start == NULL)
{
cout<<\"The List is Empty\"<info<<\"->\";
temp = temp->next;
}
cout.
Linkedlist
Linked lists are a data structure where each node contains a pointer to the next node in the list. Each node contains data and a pointer to the next node. The last node's pointer is set to NULL. Linked lists can dynamically grow and shrink during program execution. Nodes are implemented as structs containing data and a pointer. The head pointer points to the first node. Linked lists allow traversing the list by following node pointers until NULL is reached.
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
#includeiostream#includecstdio#includecstdlibusing names.pdf
This document defines a C++ class for a circular linked list. The class contains methods for creating nodes, adding elements to the beginning of the list, deleting elements, and displaying the list. A node struct is also defined containing an info field and a next pointer.
Linked lists
The document contains C++ code for implementing and testing various operations on linked lists, including adding/deleting nodes, sorting, splitting lists, reversing lists, and checking for circularity. It defines a Node struct with data and next pointer, and includes functions for initializing nodes, adding/deleting nodes, sorting, splitting, reversing lists, checking for circularity, and more. It also includes unit tests for various functions using the BOOST test framework.
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
The document provides code for implementing a linked list data structure in C++. It includes class definitions for a ListInterface, Node, and LinkedList class. The Node class represents individual nodes in the linked list. The LinkedList class implements the linked list and inherits from ListInterface. It uses Node objects to store and connect list elements. Helper functions and exceptions are also defined to support linked list operations like insertion, removal and accessing elements.
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.
In the class we extensively discussed a node class called IntNode in.pdf
In the class we extensively discussed a node class called IntNode in which each linked list node
contains an integer and each node is linked forward to the next element in the list. Implement a
node class called DoublyIntNode in which each node contains an integer number and each node
is linked not only forward to the next element in the list but also backward to the previous
element in the list. Your implementation of DoublyIntNode should adapt and implement all the
methods of IntNode including the following methods:
Constructor for DoublyIntNode
addNodeAfter, removeNodeAfter
getData, setData
getForeLink, setForeLink
getBackLink, setBackLink
listCopy
listCopyWithTail
listLength
listPart
listPosition
listSearch
IntNode and Node classes:
public class IntNode{
private int data;
private IntNode link;
// methods here
}
public class Node {
private E data;
private Node link;
// methods here
}
Solution
program
#include
using namespace std;
struct list
{
struct list *prev;
int data;
struct list *next;
}
*node = NULL, *first = NULL, *last = NULL, *node1 = NULL, *node2 = NULL;
class Dlist
{
public:
void insert_beg() {
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;
}
}
void insert_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;
}
}
void display() {
node = first;
cout << \"List of data in Linked list in Ascending order!\" << endl;
while(node != NULL) {
cout << node->data << endl;
node = node->next;
}
node = last;
cout << \"List of data in Linked list in Descending order!\" << endl;
while(node != NULL) {
cout << node->data << endl;
node = node->prev;
}
}
void del()
{
int count = 0, number, i;
node = node1 = node2 = first;
for(node = first; node != NULL; node = node->next)
cout << \"Enter value for the node:\" << endl;
count++;
display();
cout << count << \" nodes available here!\" << endl;
cout << \"Enter the node number which you want to delete:\" << endl;
cin >> number;
if(number != 1)
{
if(number < count && number > 0)
{
for(i = 2; i <= number; i++)
node = node->next;
for(i = 2; i <= number-1; i++)
node1 = node1->next;
for(i = 2; i <= number+1; i++)
node2 = node2->next;
node2->prev = node1;
node1->next = node2;
node->prev = NULL;
node->next = NULL;
node = NULL;
}
else if(number == count)
{
node = last;
last = node->prev;
last->next = NULL;
node = NULL;
}
else
cout << \"Invalid node.
Note Given Code modified as required and required met.pdf
Note:
Given Code modified as required and required methods added.
Code:
// class node
class Node
{
// declare the required variable
String data;
Node next;
// constructor
Node(){}
// constructor with arguments
Node(String data)
{
this.data = data;
next = null;
}
// method to get the data
public String getData()
{
return data;
}
// method to set the data
public void setData(String data)
{
this.data = data;
}
// method to get the next data
public Node getNext()
{
return next;
}
// method to set the next data
public void setNext(Node next)
{
this.next = next;
}
}
// class for a linked list
class MyLInkedList
{
Node head;alue into a list
// method to add the v
public void add(String num)
{
if (head == null)
{
head = new Node(num);
}
else{
Node temp = new Node(num);
Node current = head;
if (current != null) {
while (current.getNext() != null) {
current = current.getNext();
}
current.setNext(temp);
}
}
}
// method to search a value from a list
public void search(String num)
{
Node current = head;
boolean flag=true;
if (current != null)
{
while (current != null)
{
if(current.getData() == num)
{
System.out.println(\"Search value \"+
current.getData()+\" is available\");
flag=false;
}
current = current.getNext();
}
if (flag==true)
{
System.out.println(\"Search value \"+
current.getData()+\" is not available\");
}
}
else
{
System.out.println(\"Linked list is empty\");
}
}
// method to delete a value from a list
public void delete(String num)
{
//Node current = head;
boolean flag=true;
if (head == null)
{
System.out.println(\"Can\'t Delete, Linked list is
empty\");
return;
}
if(head.getData() == num)
{
head=head.getNext();
return;
}
Node temp = new Node(num);
Node cr = head;
Node prev=null;
while ((cr != null) && !(cr.getData() == num))
{
prev=cr;
cr = cr.getNext();
}
prev.setNext(cr.getNext());
}
// method to print the list
public void print()
{
Node current = head;
if (current != null) {
while (current != null) {
System.out.println(current.getData());
current = current.getNext();
}
}
}
}
// method to test the class
public class MyLInkedListTester
{
// main method to perform the linked list operatrions
public static void main(String args[])
{
MyLInkedList llist = new MyLInkedList();
// insert the values into list
llist.add(\"1\");
llist.add(\"2\");
llist.add(\"3\");
llist.add(\"4\");
// print the list values
llist.print();
// sear the values in a list
llist.search(\"1\");
llist.print();
// delete the value from a list
llist.delete(\"4\");
System.out.println(\"List after deletion:\");
llist.print();
}
}
Result:
1
2
3
4
Search value 1 is available
1
2
3
4
List after deletion:
1
2
3
Solution
Note:
Given Code modified as required and required methods added.
Code:
// class node
class Node
{
// declare the required variable
String data;
Node next;
// constructor
Node(){}
// constructor with arguments
Node(String data)
{
this.data = data;
next = null;
}
// method to get the data
public String getData()
{
return data;
}
// method to set the data
public void se.
JAVA ProgrammingDesign a class Message that models an e-mail messa.pdf
JAVA Programming
Design a class Message that models an e-mail message. A message has a recipient, a sender, and
a message text. Support the following methods:
a. A constructor that takes the sender and recipient
b. A method append that appends a line of text to the message body
c. A method toString that makes the message into one long string like this: “From: Fname
Lname\ To: Some Recipient\ …”
Write a driver program that uses this class to make a message and print it.
Solution
MessageTest.java
import java.util.Scanner;
public class MessageTest {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter Recipient Name: \");
String recipientName = scan.nextLine();
System.out.println(\"Enter Sender Name: \");
String senderName = scan.nextLine();
System.out.println(\"Enter the body text: \");
String text = scan.nextLine();
Message m = new Message(recipientName, senderName);
m.append(text);
System.out.println(m.toString());
}
}
Message.java
public class Message {
private String recipient;
private String sender;
private String text;
public Message(String r, String s){
this.recipient = r;
this.sender = s;
text = \"\";
}
public void append(String t){
text = text + t +\"\ \";
}
public String toString(){
return \"From: \"+sender+\"\ \ \"+text+\"\ To: \"+recipient+\"\ \";
}
}.
In which of the following structures are spores formed (Select all .pdf
In which of the following structures are spores formed? (Select all that apply.)ovule conespollen
conesNone of the answer options is correct.carpelsstamensIn which of the following structures
are spores formed? (Select all that apply.)ovule conespollen conesNone of the answer options is
correct.carpelsstamens
Solution
SPORES:it is very light and carried great distances survive in un favourable droughts and
temperatures.A spore is a single cell that develop in to exact copy of plant from which it came.it
is very small,spores core contain cytoplasm and dna.
spores are formed in carpels and stamens.the stamens and carpels of flower contain sprongia,in
which spores are developed..
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How do you stop infinite loop Because I believe that it is making a.pdf
How do you stop infinite loop? Because I believe that it is making an infinite circular list.
c++ code:
Here is the list class:
#ifndef LIN_J_LIST
#define LIN_J_LIST
typedef unsigned int uint;
#include
#include
using namespace std;
/**
* a simplified generic singly linked list class to illustrate C++ concepts
* @author Jerry Lin
* @version 2/17/17
*/
template< typename Object >
class List
{
private:
/**
* A class to store data and provide a link to the next node in the list
*/
class Node
{
public:
/**
* The constructor
* @param data the data to be stored in this node
*/
explicit Node( const Object & data )
: data{ data }, next{ nullptr } {}
Object data;
Node * next;
};
public:
/**
* The constructor for an empty list
*/
List()
: size{ 0 }, first{ nullptr }, last{ nullptr } {}
/**
* the copy constructor-creates and copy the list
*/
List( List && rhs ) = delete;
List( const List & rhs )
{
count = 0;
size = 0;
if(rhs.size != 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first;
size++;
while(current->next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op.
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode->next;
size++;
}
}
// you document and implement this method
}
/**
* the operator= method-copies the list
*/
List & operator=( List && rhs) = delete;
List & operator=( const List & rhs )
{
count = 0;
size = 0;
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
if(rhs.size!= 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first; //create a temporary to store
size++;
while(current -> next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode -> next;
size++;
}
}
return *this;
// you document and implement this method
}
/**
* accessor
* @return count
*/
int get_count() const
{
return count;
}
/**
* The destructor that gets rid of everything that\'s in the list and
* resets it to empty. If the list is already empty, do nothing.
*/
~List()
{
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
}
/**
* Put a new element onto the beginning of the list
* @param item the data the new element will contain
*/
void push_front( const Object& item )
{
Node * new_node = new Node( item );//basic op.
if(is_empty())
{
last = new_node;
}
else
{
new_node->next = first;
}
first = new_node;
size++;
/* you complete the rest */
}
/**
* Remove the element that\'s at the front of the list. Causes an
* assertion error if the list is empty.
*/
void pop_front()
{
assert( !is_empty() );
Node * temp = first;
if( first == last )
{
first = last = nullptr;
}
else
{
first = first->next;
}
delete temp;
size--;
}
/**
* Accessor to return the da.
A)B) C++ program to create a Complete Binary tree from its Lin.pdf
A)
B)
// C++ program to create a Complete Binary tree from its Linked List
// Representation
#include
#include
#include
using namespace std;
// Linked list node
struct ListNode
{
int data;
ListNode* next;
};
// Binary tree node structure
struct BinaryTreeNode
{
int data;
BinaryTreeNode *left, *right;
};
// Function to insert a node at the beginning of the Linked List
void push(struct ListNode** head_ref, int new_data)
{
// allocate node and assign data
struct ListNode* new_node = new ListNode;
new_node->data = new_data;
// link the old list off the new node
new_node->next = (*head_ref);
// move the head to point to the new node
(*head_ref) = new_node;
}
// method to create a new binary tree node from the given data
BinaryTreeNode* newBinaryTreeNode(int data)
{
BinaryTreeNode *temp = new BinaryTreeNode;
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// converts a given linked list representing a complete binary tree into the
// linked representation of binary tree.
void convertList2Binary(ListNode *head, BinaryTreeNode* &root)
{
// queue to store the parent nodes
queue q;
// Base Case
if (head == NULL)
{
root = NULL; // Note that root is passed by reference
return;
}
// 1.) The first node is always the root node, and add it to the queue
root = newBinaryTreeNode(head->data);
q.push(root);
// advance the pointer to the next node
head = head->next;
// until the end of linked list is reached, do the following steps
while (head)
{
// 2.a) take the parent node from the q and remove it from q
BinaryTreeNode* parent = q.front();
q.pop();
// 2.c) take next two nodes from the linked list. We will add
// them as children of the current parent node in step 2.b. Push them
// into the queue so that they will be parents to the future nodes
BinaryTreeNode *leftChild = NULL, *rightChild = NULL;
leftChild = newBinaryTreeNode(head->data);
q.push(leftChild);
head = head->next;
if (head)
{
rightChild = newBinaryTreeNode(head->data);
q.push(rightChild);
head = head->next;
}
// 2.b) assign the left and right children of parent
parent->left = leftChild;
parent->right = rightChild;
}
}
// Utility function to traverse the binary tree after conversion
void inorderTraversal(BinaryTreeNode* root)
{
if (root)
{
inorderTraversal( root->left );
cout << root->data << \" \";
inorderTraversal( root->right );
}
}
// Driver program to test above functions
int main()
{
// create a linked list shown in above diagram
struct ListNode* head = NULL;
push(&head, 36); /* Last node of Linked List */
push(&head, 30);
push(&head, 25);
push(&head, 15);
push(&head, 12);
push(&head, 10); /* First node of Linked List */
BinaryTreeNode *root;
convertList2Binary(head, root);
cout << \"Inorder Traversal of the constructed Binary Tree is: \ \";
inorderTraversal(root);
return 0;
}
c)
#include
#include
using namespace std;
template
class bintree
{
bintree *left;
T data;
bintree *right;
public :
bintree()
{
left=right=NULL;
}
void create();
void preorder();
void inorder();
void pos.
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED
HELP FIXING THIS CODE?
THIS IS THE PROBLEM
Write a program that allows the user to enter two positive integers and outputs their sum. Sounds
simple? Here\'s the catch: The numbers may be any size! They may be way too large to store in a
single variable. You may not assume any particular maximum number of digits.
HERE IS THE CODE
#include
#include
struct node /* Linked list node*/
{
int data;
struct node* next;
};
struct node *newNode(int data) /* Function to create a new node with giving the data*/
{
struct node *new_node = (struct node *) malloc(sizeof(struct node));
new_node->data = data;
new_node->next = NULL;
return new_node;
}
void push(struct node** head_ref, int new_data) /* Function to insert a node at the beginning of
the Doubly Linked List*/
{
struct node* new_node = newNode(new_data); /*allocate a node*/
new_node->next = (*head_ref); /* link the old list off to the new node*/
(*head_ref) = new_node; /*move the head to point to the new node*/
}
/* Adding the contents of two linked lists and return the head node of resultant list*/
struct node* addTwoLists (struct node* first, struct node* second)
{
struct node* res = NULL; /* res is head node of the resultant list*/
struct node *temp, *prev = NULL;
int carry = 0, sum;
while (first != NULL || second != NULL) /*while both the lists exists*/
{
/* Calculate value of next digit in resultant list.
// The next digit is sum of following things
// (i) Carry
// (ii) Next digit of first list (if there is a next digit)
// (ii) Next digit of second list (if there is a next digit)*/
sum = carry + (first? first->data: 0) + (second? second->data: 0);
carry = (sum >= 10)? 1 : 0; /* update carry for next calulation*/
sum = sum % 10; /*update sum if it is greater than 10*/
temp = newNode(sum); /*Create a new node with sum as data*/
if(res == NULL) /*if this is the first node then set it as head of the resultant list*/
res = temp;
else /* If this is not the first node then connect it to the rest.*/
prev->next = temp;
prev = temp; /* Set prev for next insertion*/
if (first) first = first->next; /* Move first pointers to next node*/
if (second) second = second->next; /*Move second pointer to the next node*/
}
if (carry > 0)
temp->next = newNode(carry);
return res; /* return head of the resultant list*/
}
void printList(struct node *node) /* function to print a linked list*/
{
while(node != NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
printf(\"\ \");
}
int main(void) /* program to test above function*/
{
struct node* res = NULL;
struct node* first = NULL;
struct node* second = NULL;
printf(\"Enter first positive integer: \");
scanf(\"%d\",&first);
printList(first);
printf(\"Enter second positive integer: \");
scanf(\"%d\",&second);
printList(second);
/*Add the two lists and see result*/
res = addTwoLists(first, second);
printf(\"Sum is: %d\");
printList(res);
return 0;
}
Solution
#include
#include
typedef struct Node
{
int da.
Inspect the class declaration for a doubly-linked list node in Node-h-.pdf
Inspect the class declaration for a doubly-linked list node in Node.h. Access Node.h by clicking
on the orange arrow next to main.cpp at the top of the coding window. The Node class has three
member variables: a double data value, a pointer to the next node, and a pointer to the previous
node. Each member variable is protected. So code outside of the class must use the provided
getter and setter member functions to get or set a member variable. Node.h is read only, since no
changes are required. Step 2: Implement the Insert() member function A class for a sorted,
doubly-linked list is declared in SortedNumberList.h. Implement the SortedNumberList class's
Insert() member function. The function must create a new node with the parameter value, then
insert the node into the proper sorted position in the linked list. Ex: Suppose a
SortedNumberList's current list is 23 47.25 86, then Insert(33.5) is called. A new node with data
value 33.5 is created and inserted between 23 and 47.25, thus preserving the list's sorted order
and yielding: 23 35.5 47.25 86 Step 3: Test in develop mode Code in main() takes a space-
separated list of numbers and inserts each into a SortedNumberList. The list is displayed after
each insertion. Ex: If input is 77 15 -42 63.5 then output is: List after inserting 77: 77 List after
inserting 15: 15 77 List after inserting -42: -42 15 77 List after inserting 63.5: -42 15 63.5 77 Try
various program inputs, ensuring that each outputs a sorted list. Step 4: Implement the Remove()
member function Implement the SortedNumberList class's Remove() member function. The
function takes a parameter for the number to be removed from the list. If the number does not
exist in the list, the list is not changed and false is returned. Otherwise, the first instance of the
number is removed from the list and true is returned. Uncomment the commented-out part in
main() that reads a second input line and removes numbers from the list. Test in develop mode to
ensure that insertion and removal both work properly, then submit code for grading. Ex: If input
is 84 72 19 61 19 84 then output is: List after inserting 84: 84 List after inserting 72: 72 84 List
after inserting 19: 19 72 84 List after inserting 61: 19 61 72 84 List after removing 19: 61 72 84
List after removing 84: 61 72
main.cpp
#include
#include
#include
#include "Node.h"
#include "SortedNumberList.h"
using namespace std; void PrintList(SortedNumberList& list); vector SpaceSplit(string source);
int main(int argc, char *argv[]) { // Read the line of input numbers string inputLine; getline(cin,
inputLine); // Split on space character vector terms = SpaceSplit(inputLine); // Insert each value
and show the sorted list's contents after each insertion SortedNumberList list; for (auto term :
terms) { double number = stod(term); cout << "List after inserting " << number << ": " << endl;
list.Insert(number); PrintList(list); } /* // Read the input line with numbers to remove getline(cin,
inputLi.
C program to insert a node in doubly linked list
This C program implements functions to create and manipulate a doubly linked list. It allows the user to choose from a menu to create a list, insert nodes at the beginning, end, or a specified position, and display the list. Functions are defined to create a list by dynamically allocating nodes and linking them together, insert nodes at different positions by updating next and prev pointers, and display the list by traversing from head to tail.
Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
Assignment is:
\"Page 349-350 #4 and #5 Use the \"Linked List lab\" you have been working on in class and add
the two functions the questions are asking you to develop: divideMid and divideAt. Be sure to
include comments Use meaningful identifier names (constants where appropriate) Turn in .cpp
file AND Turn in a \"print-screen\' of your output (press \"print-screen\' on keyboard, then
\'paste\' in MS-Word)\"
How do you solve QUESTION #4 in the book data structures using c++ by D.S. Malik in Visiual
Studios using the linked list below with what is being asked? Please need help
Linked list :
#include
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *first, *last;
int num;
createList(first, last);
int choice;
while(true)
{
cout<<\"1. Insert Front.\ 2. Insert Last.\ 3. Delete Front.\ 4. Delete Last.\ 5. Print List.\ 6. Exit.\
\";
cout<<\"Enter your choice: \";
cin>>choice;
switch(choice)
{
case 1: insertFront(first); break;
case 2: insertBack(last); break;
case 3: deleteFirst(first); break;
case 4: deleteLast(last, first); break;
case 5: printList(first); break;
case 6: return 0;
default: cout<<\"Invalid menu option. Try again.\"<>number;
while (number != -999)
{
newNode = new nodeType; // create new node
newNode->info = number;
newNode->link = NULL;
if (first == NULL)
{
first = newNode;
last = newNode;
}
else
{
last->link = newNode;
last = newNode;
}
cout<<\"Enter an integer (-999 to stop): \";
cin>>number;
} // end of while-loop
} // end of build list function
void deleteFirst(nodeType*& first)
{
nodeType *temp;
temp= first;
first= temp->link;
delete temp;
return;
}
void deleteLast(nodeType*& last, nodeType* current)
{
nodeType *temp;
while(current->link != NULL)
{
temp=current;
current=current->link;
}
temp=last;
current->link=NULL;
delete temp;
last = current;
return;
}
void insertFront(nodeType*& front)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= front;
front= newNode;
return;
}
void insertBack(nodeType*& last)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= NULL;
last->link= newNode;
last = newNode;
return;
}
void printList(nodeType*& first)
{
cout<<\"Inside printList...printing linked list...\ \"<info << \" \";
current = current->link;
}
cout<
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *firs.
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
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.
C++Write a method Node Nodereverse() which reverses a list..pdf
C++
Write a method Node *Node::reverse() which reverses a list.
Solution
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Function to reverse the linked list */
static void reverse(struct node** head_ref)
{
struct node* prev = NULL;
struct node* current = *head_ref;
struct node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
/* Function to push a node */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print linked list */
void printList(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
printf(\"%d \", temp->data);
temp = temp->next;
}
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 85);
printf(\"Given linked list\ \");
printList(head);
reverse(&head);
printf(\"\ Reversed Linked list \ \");
printList(head);
getchar();
}.
There are a number of errors in the following program- All errors are.docx
There are a number of errors in the following program.
All errors are located in main() and structure definitions.
Function declarations and definitions are correct!
Locate all errors, fix them (as shown below), run the program and save its output
as a comment at the end of the source file.
Example:
int num = 10;
int *ptr;
num = &ptr; // <== Error: Comment the line and write the correct line below
// Write a short justification where appropriate
// num = &ptr; // Error #1
ptr = #
Name:
*/
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#define DUMMY_TRAILER '\177'
// octal ASCII code of the
// last character in the ASCII table
#define NUM_CITIES 10
typedef struct
{
char name[12];
int temperature[5];
} CITY;
// Stack and Queue Node
typedef struct node NODE;
struct node
{
CITY city;
node *next;
};
// Doubly Linked List Node
typedef struct d_node D_NODE;
struct d_node
{
CITY city;
NODE *forw;
NODE *back;
};
// Stack Functions
NODE *push(NODE *stack, const CITY *pStu);
NODE *pop(NODE **stack);
// Queue Functions
void enqueue(NODE **queue, NODE **rear, const CITY *pStu);
NODE *dequeue(NODE **queue, NODE **rear);
// Doubly Linked List Functions
D_NODE *init_list(void);
int insert(D_NODE *list, const CITY *pStu);
void traverse_forw(D_NODE *list);
void traverse_back(D_NODE *list);
// Other Functions
void printCity(const CITY *pCity);
int main (void)
{
CITY cList[NUM_CITIES] =
{
{"Cupertino", {88, 89, 87, 85, 89}},
{"Flagstaff", {81, 80, 88, 89, 89}},
{"Los Angeles", {87, 88, 89, 89, 90}},
{"Philadelphia", {96, 99, 99, 90, 95}},
{"Phoenix", {106, 109, 109, 100, 105}},
{"Portland", {89, 90, 85, 89, 90}},
{"Reno", {108, 105, 109, 100, 108}},
{"Salem", {85, 90, 85, 89, 90}},
{"Tucson", {107, 100, 109, 100, 108}},
{"Yreka", {101, 109, 100, 108, 109}}
};
NODE *stack = NULL;
NODE *top = NULL;
NODE *queue = NULL, *rear = NULL;
NODE *front;
D_NODE *list;
list = init_list();
// build stack and queue with data from an array of CITY structures
srand((unsigned int)time(NULL));
int count = rand() % 10;
for ( int n = 0; n < count; n++)
{
int i = rand() % NUM_CITIES;
int duplicate = insert(list, &cList[i]);
if(duplicate)
{
// already in the list!
push(stack, &cList[i]);
enqueue(&queue, &rear, cList[i]);
}
}
// display list
printf("\nLIST contents (forwards):\n");
traverse_forw(list);
printf("\nLIST contents (backwards):\n");
traverse_back(list);
// display stack
if (top)
{
printf("\nSTACK contents from top to bottom:\n");
while ((top = pop(stack)))
{
printCity(top->city);
}
}
else
printf ("Empty Stack!\n");
// display queue
if (front)
{
printf("\nQUEUE contents from front to rear:\n");
while ((front = dequeue( queue, rear)))
{
printCity(front->city);
}
}
else
printf ("Empty Queue!\n");
return 0;
}
/***************************************************
Displays the fileds of a CIS_CLASS structure
Pre pCls - a pointer to a CIS_CLASS structure
Post
*/
void printCity(const CITY *pCity)
{
printf("%-20s %3d\n", pCity->name, pCity->temperatur.
take the following code and give details of what each line of code i.pdf
take the following code and give details of what each line of code is doing
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<info<<\"->\";
temp = temp->next;
}
cout<<\"NULL\"<
Solution
ANSWER ::
ASSMUING THE TOTAL LINE IN CODE ARE 158*(INCLUDING COMMENTS)
SO...
PROGRAM STARTS WITH SPECIFING HEADER FILES AND GLOBAL VARIABLE
DECLARATION
THEN Node Declaration USING STRUCTURES,
THEN WE WANT TO DECLARE THE CLASS USING SINGLE LINKED LIST AND THE
PROTOTYPE IS PUBLIC
IN THE PUBLIC WE WANT TO CREATE THE NODE AS INTEGER DATATYPE
THE NODE CONSIST OF insert_begin();
insert_pos();-> INDICATES THE CURRENT POSSITION
insert_last();->INSERT AT LAST
delete_pos();->TO DELETE THE NUMBER AT CURRENT POSITION
sort();->START THE SORTING
search();->USED TO SEARCH
update();->UPDATE THE NODE
reverse();->REVERSE NUMBERS
display();->TO DISPLAY THE NODE
single_llist()->FIFO
WE ARE STARTING TAKING NULL VALUE
AND THEN TAKE USER DEFIND FUNCTION MAIN()
CREATE THE DATATYPES choice, nodes, element, position, i
OBJECT CREATION OF SLL IS SL
USING WHILE LOOP THE NODE INSERT AT THE BEGINING OR ENDING STAGE
WE ARE USING SWITCH CASES TO PERFORM SLL ON NODE
NODE CREATION AS
node *single_llist::create_node(int value)
ALSO DECLARE THE TEMP IS TAKEN AS NULL VALUE
INSERT THE ELEMENT AT THE BEGINING
USING IF CONDITION WE CAN INSERT THE NODE AT THE BEGINING
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
AND THEN INSERT THE NODE AT THE LAST
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
USING WHILE LOOP WE ARE INSERTING THE NODE AT LAST
USING THIS BELLOW CODE WE CAN SEARCH THE ELEMENTS IN THE NODE
struct node *temp;
if (start == NULL)
{
cout<<\"The List is Empty\"<info<<\"->\";
temp = temp->next;
}
cout.
Linkedlist
Linked lists are a data structure where each node contains a pointer to the next node in the list. Each node contains data and a pointer to the next node. The last node's pointer is set to NULL. Linked lists can dynamically grow and shrink during program execution. Nodes are implemented as structs containing data and a pointer. The head pointer points to the first node. Linked lists allow traversing the list by following node pointers until NULL is reached.
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
#includeiostream#includecstdio#includecstdlibusing names.pdf
This document defines a C++ class for a circular linked list. The class contains methods for creating nodes, adding elements to the beginning of the list, deleting elements, and displaying the list. A node struct is also defined containing an info field and a next pointer.
Linked lists
The document contains C++ code for implementing and testing various operations on linked lists, including adding/deleting nodes, sorting, splitting lists, reversing lists, and checking for circularity. It defines a Node struct with data and next pointer, and includes functions for initializing nodes, adding/deleting nodes, sorting, splitting, reversing lists, checking for circularity, and more. It also includes unit tests for various functions using the BOOST test framework.
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
The document provides code for implementing a linked list data structure in C++. It includes class definitions for a ListInterface, Node, and LinkedList class. The Node class represents individual nodes in the linked list. The LinkedList class implements the linked list and inherits from ListInterface. It uses Node objects to store and connect list elements. Helper functions and exceptions are also defined to support linked list operations like insertion, removal and accessing elements.
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.
In the class we extensively discussed a node class called IntNode in.pdf
In the class we extensively discussed a node class called IntNode in which each linked list node
contains an integer and each node is linked forward to the next element in the list. Implement a
node class called DoublyIntNode in which each node contains an integer number and each node
is linked not only forward to the next element in the list but also backward to the previous
element in the list. Your implementation of DoublyIntNode should adapt and implement all the
methods of IntNode including the following methods:
Constructor for DoublyIntNode
addNodeAfter, removeNodeAfter
getData, setData
getForeLink, setForeLink
getBackLink, setBackLink
listCopy
listCopyWithTail
listLength
listPart
listPosition
listSearch
IntNode and Node classes:
public class IntNode{
private int data;
private IntNode link;
// methods here
}
public class Node {
private E data;
private Node link;
// methods here
}
Solution
program
#include
using namespace std;
struct list
{
struct list *prev;
int data;
struct list *next;
}
*node = NULL, *first = NULL, *last = NULL, *node1 = NULL, *node2 = NULL;
class Dlist
{
public:
void insert_beg() {
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;
}
}
void insert_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;
}
}
void display() {
node = first;
cout << \"List of data in Linked list in Ascending order!\" << endl;
while(node != NULL) {
cout << node->data << endl;
node = node->next;
}
node = last;
cout << \"List of data in Linked list in Descending order!\" << endl;
while(node != NULL) {
cout << node->data << endl;
node = node->prev;
}
}
void del()
{
int count = 0, number, i;
node = node1 = node2 = first;
for(node = first; node != NULL; node = node->next)
cout << \"Enter value for the node:\" << endl;
count++;
display();
cout << count << \" nodes available here!\" << endl;
cout << \"Enter the node number which you want to delete:\" << endl;
cin >> number;
if(number != 1)
{
if(number < count && number > 0)
{
for(i = 2; i <= number; i++)
node = node->next;
for(i = 2; i <= number-1; i++)
node1 = node1->next;
for(i = 2; i <= number+1; i++)
node2 = node2->next;
node2->prev = node1;
node1->next = node2;
node->prev = NULL;
node->next = NULL;
node = NULL;
}
else if(number == count)
{
node = last;
last = node->prev;
last->next = NULL;
node = NULL;
}
else
cout << \"Invalid node.
Note Given Code modified as required and required met.pdf
Note:
Given Code modified as required and required methods added.
Code:
// class node
class Node
{
// declare the required variable
String data;
Node next;
// constructor
Node(){}
// constructor with arguments
Node(String data)
{
this.data = data;
next = null;
}
// method to get the data
public String getData()
{
return data;
}
// method to set the data
public void setData(String data)
{
this.data = data;
}
// method to get the next data
public Node getNext()
{
return next;
}
// method to set the next data
public void setNext(Node next)
{
this.next = next;
}
}
// class for a linked list
class MyLInkedList
{
Node head;alue into a list
// method to add the v
public void add(String num)
{
if (head == null)
{
head = new Node(num);
}
else{
Node temp = new Node(num);
Node current = head;
if (current != null) {
while (current.getNext() != null) {
current = current.getNext();
}
current.setNext(temp);
}
}
}
// method to search a value from a list
public void search(String num)
{
Node current = head;
boolean flag=true;
if (current != null)
{
while (current != null)
{
if(current.getData() == num)
{
System.out.println(\"Search value \"+
current.getData()+\" is available\");
flag=false;
}
current = current.getNext();
}
if (flag==true)
{
System.out.println(\"Search value \"+
current.getData()+\" is not available\");
}
}
else
{
System.out.println(\"Linked list is empty\");
}
}
// method to delete a value from a list
public void delete(String num)
{
//Node current = head;
boolean flag=true;
if (head == null)
{
System.out.println(\"Can\'t Delete, Linked list is
empty\");
return;
}
if(head.getData() == num)
{
head=head.getNext();
return;
}
Node temp = new Node(num);
Node cr = head;
Node prev=null;
while ((cr != null) && !(cr.getData() == num))
{
prev=cr;
cr = cr.getNext();
}
prev.setNext(cr.getNext());
}
// method to print the list
public void print()
{
Node current = head;
if (current != null) {
while (current != null) {
System.out.println(current.getData());
current = current.getNext();
}
}
}
}
// method to test the class
public class MyLInkedListTester
{
// main method to perform the linked list operatrions
public static void main(String args[])
{
MyLInkedList llist = new MyLInkedList();
// insert the values into list
llist.add(\"1\");
llist.add(\"2\");
llist.add(\"3\");
llist.add(\"4\");
// print the list values
llist.print();
// sear the values in a list
llist.search(\"1\");
llist.print();
// delete the value from a list
llist.delete(\"4\");
System.out.println(\"List after deletion:\");
llist.print();
}
}
Result:
1
2
3
4
Search value 1 is available
1
2
3
4
List after deletion:
1
2
3
Solution
Note:
Given Code modified as required and required methods added.
Code:
// class node
class Node
{
// declare the required variable
String data;
Node next;
// constructor
Node(){}
// constructor with arguments
Node(String data)
{
this.data = data;
next = null;
}
// method to get the data
public String getData()
{
return data;
}
// method to set the data
public void se.
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A)B) C++ program to create a Complete Binary tree from its Lin.pdf
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THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf
Inspect the class declaration for a doubly-linked list node in Node-h-.pdf
Inspect the class declaration for a doubly-linked list node in Node-h-.pdf
Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned in.pdf
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 .pdf
C++Write a method Node Nodereverse() which reverses a list..pdf
C++Write a method Node Nodereverse() which reverses a list..pdf
There are a number of errors in the following program- All errors are.docx
There are a number of errors in the following program- All errors are.docx
take the following code and give details of what each line of code i.pdf
take the following code and give details of what each line of code i.pdf
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
#includeiostream#includecstdio#includecstdlibusing names.pdf
#includeiostream#includecstdio#includecstdlibusing names.pdf
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
PROBLEM STATEMENTIn this assignment, you will complete DoubleEnde.pdf
PROBLEM STATEMENTIn this assignment, you will complete DoubleEnde.pdf
In the class we extensively discussed a node class called IntNode in.pdf
In the class we extensively discussed a node class called IntNode in.pdf
Note Given Code modified as required and required met.pdf
Note Given Code modified as required and required met.pdf
More from connellalykshamesb60
JAVA ProgrammingDesign a class Message that models an e-mail messa.pdf
JAVA Programming
Design a class Message that models an e-mail message. A message has a recipient, a sender, and
a message text. Support the following methods:
a. A constructor that takes the sender and recipient
b. A method append that appends a line of text to the message body
c. A method toString that makes the message into one long string like this: “From: Fname
Lname\ To: Some Recipient\ …”
Write a driver program that uses this class to make a message and print it.
Solution
MessageTest.java
import java.util.Scanner;
public class MessageTest {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter Recipient Name: \");
String recipientName = scan.nextLine();
System.out.println(\"Enter Sender Name: \");
String senderName = scan.nextLine();
System.out.println(\"Enter the body text: \");
String text = scan.nextLine();
Message m = new Message(recipientName, senderName);
m.append(text);
System.out.println(m.toString());
}
}
Message.java
public class Message {
private String recipient;
private String sender;
private String text;
public Message(String r, String s){
this.recipient = r;
this.sender = s;
text = \"\";
}
public void append(String t){
text = text + t +\"\ \";
}
public String toString(){
return \"From: \"+sender+\"\ \ \"+text+\"\ To: \"+recipient+\"\ \";
}
}.
In which of the following structures are spores formed (Select all .pdf
In which of the following structures are spores formed? (Select all that apply.)ovule conespollen
conesNone of the answer options is correct.carpelsstamensIn which of the following structures
are spores formed? (Select all that apply.)ovule conespollen conesNone of the answer options is
correct.carpelsstamens
Solution
SPORES:it is very light and carried great distances survive in un favourable droughts and
temperatures.A spore is a single cell that develop in to exact copy of plant from which it came.it
is very small,spores core contain cytoplasm and dna.
spores are formed in carpels and stamens.the stamens and carpels of flower contain sprongia,in
which spores are developed..
How does Windows managestore user names and passwords How does it .pdf
How does Windows manage/store user names and passwords? How does it differ from Unix?
What privileges does the \"root\" UID have in Linux?
Solution
1) the usernames and passwords in windows are hashed using \"NTLMv2\"algorithm which is a
part of the suite of windows security protocols and finally encrypted credentials are stored in
SAM (security account manager) present in the following directory
C:\\windows\\system32\\config\\SAM
2) the user credential storage of windows differs from unix as said below :
In old unix variants ,passwords are stored in etc/passwd in encrypted form.
But in newer versions ,storage is etc/shadow which ensures no one can access the passwords
except the root.
3)privileges of root in linux :
- ability to modify OS
- grant and revoke permissions
- can access passwords file
- read and modify configuration files.
How to use class set a function(double money(); receive a number and .pdf
How to use class set a function(double money(); receive a number and then output with this
number*2) and use it in the main, cpp?
Solution
//Here number is a part of class definition.
//As your function definition money does not include any number variable i am assuming it to be
part of class.
#include
using namespace std;
class Program
{
private:
double number;
public:
Program()
{
number = 0;
}
Program(int number)
{
this->number = number;
}
double money()
{
return 2*number;
}
};
int main()
{
Program p(10);
cout << p.money() << endl;
return 0;
}.
For each of the following descriptions indicate which best describes.pdf
For each of the following descriptions indicate which best describes proteins or protein
sequences that are:
(S) similar; (H) homolgous; (O) orthologous; (P) paralogs; (D) domains; (F) a superfamily
____ sequences that have a shared evolutionary ancestry
____ sequences that have a certain statistically significant % of identical amino acids when
aligned
____ proteins that have a very high % of identical amino acids when aligned and which are
found in the same
Organism
____ sequences that are small but share structural and functional similarity
____ sequences that have a shared ancestral sequence that changed during a speciation event
A given protein has been monoubiquitylated. What is the expected result?
a. the protein will be activated to turn on transcription
b. the subcellular localization of the protein will change
c. the protein will be activated to repair the DNA
d. the protein will be activated to phosphorylate a substrate
e. any of the above
f. none of the above
The synthetic toxin 2,4-dinitrophenol can uncouple ATP synthesis from mitochondrial
respiration by decreasing the permeability of the inner membrane to protons. What would be the
effect of dinitrophenol treatment on the amount of ATP produced by mitochondria and on the
rate of ATP transport across the inner membrane, respectively?
a. positive, positive
b. positive, negative
c. negative, positive
d. negative, negative
Which of the following requires the mitochondria to create contact points with the ER?
a. mt fission
b. mt fusion
c. mt biogenesis of more mitochondria
d. autophagy of mitochondria (Mitophagy)
e. apoptosis
Solution
S = proteins that have a very high % of identical amino acids when aligned and which are found
in the same Organism
H = sequences that have a shared evolutionary ancestry
O = sequences that have a shared ancestral sequence that changed during a speciation event
P = sequences that are small but share structural and functional similarity
D = Discription not mentioned in the question (A domain is a well-structured 3D, independent
and fully functional part of a protein. Mostly various domains combine to form a proetin to
execute a specific function in cells e.g. enzymes)
F = sequences that have a certain statistically significant % of identical amino acids when
aligned.
Explain how an Annelidian differs from a MolluscanSolutionMoll.pdf
Explain how an Annelidian differs from a Molluscan?
Solution
Mollusks are known to very diverse including 110,000 named species and some examples are
clams, oysters, snails, octopi etc. These mollusks are known to have distant common ancestor
with the annelid worms (trochophore larva) and it has been believed that both mollusks and
annelids most probably might have evolved from free-living flatworms.
The main difference between mollusks and annelids are
Only one heart is seen in Mollusca, while Annelids have five hearts
Open circulatory system is found in Mollusca can have open circulatory system but annelids
have closed circulatory system
Mollusks are marine creatures while Nematoda and Annelids are known to inhabit various
ecologies, marine (tidal zone and hydrothermal vents), fresh water and terrestrial.
Mollusks have a more complex system when compared to annelids
Either externally or internally we can spot shells in Mollusks (snails, scallops, and squids), but
shells are absent in Annelids..
DNA s composed of nucleotides that are joined together by covalent bo.pdf
DNA s composed of nucleotides that are joined together by covalent bonds between which two
parts of each nucleotide? Between two sugars Between the sugar and the organic base.
Between the sugar and the phosphate group. Between the phosphate group and the organic base.
Solution
DNA is composed of nucleotides that are joined together by covalent bonds between which wo
parts of each nucleotide:
Betweeen the sugar and phosphate group.......................
Discuss some of the challenges involved in ensuring that critical kn.pdf
Discuss some of the challenges involved in ensuring that critical knowledge is transferred both to
present-day employees as well as future ones. (1 ~ 2 Paragraphs)
Solution
The challenges involved in ensuring both present and future employees are of many kinds.
Ensuring critical knowledge, the main is to provide security . Providing the right level of security
is the main part and shouldnot access from other systems. Employees come from different places
, cultures, and habits making them united in the office for working together can be a challenge.
The next comes, keeping up with technology , in which the knowledge could be shared to others
as easy and faster.Constant changing of the structures mean learning how to be smart and quick.
Interpretation effectively is that acquiring the correct and valuable information with in time
which can be used by other departments. Here comes the, last main one i.e., impressing the
person who works actively and effectively . By doing these kind of works not only the purticular
employee but also other persons can be given a way for encouraing the work .
To overcome the knowledge challenge facilities , can be shared by data bases and technological
aids.
Knowledge management contributes towards streamlining the ideas problems, projects and
deployment driving towards productivity.
But, it’s more than just knowing everything . It’s creating a synthesis between the people and the
information to the point that the whole is more than the sum of the parts..
Biologically, the term is synonymous with [Choose the term race. refe.pdf
Biologically, the term is synonymous with [Choose the term race. refers to the position of a
particular gene Choose on a chromosome. Traits that provide a benefit in particular [Choose
environments typically create when their prevalence within a population is mapped onto the
landscape. If two different alleles phenotypically result in a trait [Choose J that appears to be
blended, the alleles are said to exhibit can lead to speciation events. Choose s one of the four
fields of anthropology. Choose The structure of DNA was first discovered by [Choose Francis
Crick, Rosalind Franklin, and A type blood and B type blood are Both Choose appear in the
phenotype if they are in the genotype. states that only one of the two alleles (choose l present in
the parent will be present in the gamete produced by that parent.
Solution
1. Clines
2. Locus
3. Subspecies
4. Incomplete dominance
5. Disruptive selection
6. Linguistics
7. James Watson
8. Codominant
9. Law of segregation.
Anyone good with linguistics Are the following headlines transitive.pdf
Anyone good with linguistics? Are the following headlines transitive or intransitive?
\"Clinton Says Inaction By Russia May Lead to Syrian Civil War\"
\"Russian Church Is a Strong Voice In Opposing Intervention in Syria\"
\"Putin Reject Intervention But Fears Civil War In Syria\"
\"Syria: Heavy Clashes Reported\"
\"Syria Bars 17 Western Diplomats and Allows Increased Aid Agency Presence\"
Solution
Let us look at the dictionary meaning of transitive and intransitive.
Transitive verb 1. to lead astray or deceive; especially : SEDUCE 2 : to deliver to an enemy by
treachery 3 : to fail or desert especially in time of need 4 a : to reveal unintentionally b : SHOW,
INDICATE c : to disclose in violation of confidence intransitive verb : to prove false.
Intransitive verb is to practice deceit; also : to give a false impression
Looking at the headlines
\"Clinton Says Inaction By Russia May Lead to Syrian Civil War\" - TRANSITIVE
\"Russian Church Is a Strong Voice In Opposing Intervention in Syria\" - INTRANSITIVE
\"Putin Reject Intervention But Fears Civil War In Syria\" - TRANSISITVE
\"Syria: Heavy Clashes Reported\" - INTRANSITIVE
\"Syria Bars 17 Western Diplomats and Allows Increased Aid Agency Presence\" -
TRANSITIVE.
A mother and father both have normal karyotypes. The father is color .pdf
A mother and father both have normal karyotypes. The father is color blind, but the mother is
not. Color blindness is determined by a recessive mutation on the X chromosome. Their first
child is a color-blind girl with an XX karyotype. The family consults a genetic counselor because
their second child is a color blind boy with Klinefelter\'s syndrome. Klinefelter\'s syndrome
results when a child has the sex chromosome constitution XXY. Which of the following
statements is true about the nondisjunction event that led to the boy with Klinefelter\'s syndrome
(assume no recombination)? Only during the first meiotic division in Dad. Only during the
second meiotic division in Dad. Only during the first meiotic division in Mom. Only during the
second meiotic division in Mom. Either a or d. Either a orb.
Solution
ans-e
Non-disjunction (failure of homologous chromosomes) at paternal meiosis I produces XY
secondary spermatocytes and a 24,XY gamete. Fertilization with a 23,X ovum produce a
47,XXY individual (Klinefelter syndrome). Only nondisjunction at paternal meiosis II produces
a 24,YY gamete that yields a 47,XYY individual after fertilization..
A man has a condition where all of his gametes undergo nondisjunctio.pdf
A man has a condition where all his sperm undergo nondisjunction of sex chromosomes in meiosis I, producing sperm that are either XX or XY. He mates with a woman with normal eggs. There is a 50% (1/2) chance that fertilization between one of his XY sperm and her normal X egg would result in a zygote and child with Klinefelter syndrome (XXY).
____ The pigment produced by S. marvel scenes would appear first on t.pdf
____ The pigment produced by S. marvel scenes would appear first on the edge of colony
because these bacteria reach stationary phase more quickly than those in the middle. ____
Typhoid Mary was considered to have an acute infection with the organism that causes
tuberculosis; her gall bladder, where bile is made, contained the organism.
Solution
1. True, this organism produce red pigmentation peviously this organism was considered as non
pathogenic organism but now some harmfull activity are also known
2.True, the tuberculosis in the abdominal part is most common type of extra pulmonary
tuberculosis.
Write the advantages of using JavaFX with compare to Swing and AWT..pdf
Write the advantages of using JavaFX with compare to Swing and AWT.
Solution
The advantages of using java fx with compared to swing and awt are:
1.The first advantage is that java fx works similarly to a pdf.The java fx is basically a tree/node
struction which has various values that are inherited from the nodes and the swing is basically
iterative and has a flat structure .The awt is similar to swing as well.The PDF is built around a
tree structures which makes it easier to map onto JavaFx as compared to swing.It can also handle
xforms.
2.The second advantage is that JavaFx has more features as compared to swing and awt.For
example in order to create the shapes it is easier using javafx with variable line as compared to
swings.
3.The third advantage is JavaFx uses a more compact and a simpler code.The code is short as
compared to swing.JavaFx takes low level tedium out of the code which means we do not have
to implement paint() and print() methods in javafx which we have to do in swing.
4.It is an integral part of java8 and a lot of focus is going on to make it compatible and run on
android devices and rasberry pi operating systems.It can also interact with the existign swing
applications.It can be run on various devices.
5.The new developments in java will take place in javafx and it will make it user friendly with
easier coding..
Write a program in c++ that produces a bar chart showing the populat.pdf
Write a program in c++ that produces a bar chart showing the population growth of Prairieville, a
small town in the Midwest, at 20-year intervals during the past 100 years.
The program will take input from user for population figures (rounded to the nearest 1,000
people) for 1900, 1920, 1940, 1960, 1980, and 2000. For each year it should display the date and
a bar consisting of one asterisk for each 1,000 people.
Here is an example of how the chart might begin:
PRAIRIEVILLE POPULATION GROWTH
(each * represents 1,000 people)
1900 **
1920 ****
1940 *****
After the bar chart, print out the maximum, minimum and average number of population.
Solution
#include
using namespace std;
int main()
{
int year[6]={1900,1920,1940,1960,1980,2000};
long population[6],calc[6];
int i,j;
cout<<\"Enter the population for years\";
for(i=0;i<6;i++)
{
cin>>population[i]; //input years
}
cout << endl << \"PRAIRIEVILLE POPULATION GROWTH\";
cout << endl << \"(each * represents 1,000 people) \" << endl;
for(i=0;i<6;i++)
{
calc[i]= population[i]/1000; //calcualte population per 1000 persons
}
for(i=0;i<6;i++)
{
cout << \"\ \"<.
Write a 2 page essay describing what life was like in the Great Depr.pdf
Write a 2 page essay describing what life was like in the Great Depression 1929-1941 with a
focus on the important events and data which characterized economic conditions of that time
period. Cite at lease 4 outside sources of supporting information/data from the internet.
Solution
CAUSES
EFFECTS.
Why do silver nanoparticles of 10 nanometers in diameter have lower m.pdf
Why do silver nanoparticles of 10 nanometers in diameter have lower melting temperature than
the same silver but in the form of a 1/4 inch rod?
Solution
Firstly in case of spherical there is low volume to surface area so it has high heat absorbing
capacity so low melting point but in case of cylindrical surface to volume ratio is quite large
Secondly in case of spherical nanoparticle browniam motion is very high because of which high
heat transfer coeeficient is achieved and so low melting point in case of nanoparticle.
Why Computer System Management is so critical for the companies (50.pdf
Why Computer System Management is so critical for the companies? (500-600 words)
Solution
Managing computer systems is very difficult for the company so it appoints managers for it who
have the following responsibilities to carry on:
Computer and information systems managers, often called information technology (IT)
managers or IT project managers, plan, coordinate, and direct computer-related activities in an
organization. They help determine the information technology goals of an organization and are
responsible for implementing computer systems to meet those goals.
Duties:
Computer and information systems managers typically do the following:
Analyze their organization’s computer needs and recommend possible upgrades for top
executives to consider.
Plan and direct the installation and maintenance of computer hardware and software.
Ensure the security of an organization’s network and electronic documents.
Assess the costs and benefits of new projects and justify funding on projects to top executives.
Learn about new technology and look for ways to upgrade their organization’s computer
systems.
Determine short- and long-term personnel needs for their department.
Plan and direct the work of other IT professionals, including computer systems analysts,
software developers,information security analysts, andcomputer support specialists.
Negotiate with vendors to get the highest level of service for their organization’s technology
Few managers carry out all of these duties. There are various types of computer and information
systems managers, and the specific duties of each are determined by the size and structure of the
firm. Smaller firms may not employ every type of manager.
The following are examples of types of computer and information systems managers:
Chief information officers (CIOs) are responsible for the overall technology strategy of their
organizations.
They help determine the technology or information goals of an organization and then oversee
implementation of technology to meet those goals.
CIOs may focus on a specific area, such as electronic data processing or information systems,
but CIOs tend to focus more on long-term or big picture issues. At small organizations a CIO has
more direct control over the IT department, and at larger organizations other managers under the
CIO may handle the day-to-day activities of the IT department.
CIOs who do not have technical expertise and who focus solely on a company’s business aspects
are included in a separate profile on top executives.
Chief technology officers (CTOs) evaluate new technology and determine how it can help their
organization.
When both CIOs and CTOs are present, the CTO usually has more technical expertise.
The CTO is responsible for designing and recommending the appropriate technology solutions to
support the policies and directives issued by the CIO. CTOs also work with different
departments to implement the organization’s technology plans.
The CTO usually reports directl.
Which of the following statements is not true about facilitated diff.pdf
Which of the following statements is not true about facilitated diffusion?
a. It requires specific transport molecules in the plasma membrane.
b. A transporter protein undergoes a change in shape to accomplish the transfer.
c.A substrate molecule is transported from an area of high concentration to low concentration.
d. It requires ATP.
e. It is driven by osmotic pressure.
Solution
B. A transporter protein undergoes a change in shape to accomplish the transfer.
Facilitated diffusion is the process of spontaneous passive transport of molecules or ions across a
cell\'s membrane via specific transmembrane integral proteins. Larger molecules are transported
by transmembrane carrier proteins, such as permeases, that change their conformation as the
molecules are carried across (e.g. glucose or amino acids)..
Which is not a hypothesis for the origin of virusesThey originated .pdf
Which is not a hypothesis for the origin of viruses?They originated in plasmids and transposable
elements.They originated in symbiotic bacteria.They originated with the origin of life.They
originated from cell membrane proteins.
Solution
Anwer: They originated from cell membrane proteins
Reason:
Viruses might have not originated from cell membrane proteins because according virus-first
hypothesis enzyme that synthesizes cell wall or cell membrane may generated a virus first. The
other hypothesis are that viruses might have originated in plasmids and transposable elements
(progressive hypothesis of viral evolution) or originated in symbiotic bacteria or originated with
the origin of life..
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- 1. Using the provided table interface table.h and the sample linked list code linkedList.c, complete an implementation of the Table ADT. Make sure that you apply the concepts of design by contract (DbC) to your implementation. Once you have fully implemented the table, create a main.c file that implements a testing framework for your table. Your table implementation must ensure that values inserted are unique, and internally sorted within a linked list. table.h linkedList.c Solution khsdg #include #include typedef enum BOOL { false, true } bool; // Linked list node definition typedef struct Node node; struct Node { int number; node *next; }; static node *top = NULL; // used to track where we are for the list traversal methods static node *traverseNode = NULL; // "destroy" will deallocate all nodes in a linked list object // and will set "top" to NULL. void destroy() { node *curr = top; node *temp = NULL; while ( curr != NULL ) {
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- 4. }