could you implement this function please, im having issues with it..pdf
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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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![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&);](https://image.slidesharecdn.com/couldyouimplementthisfunctionpleaseimhavingissueswithit-230904003521-33cbcc15/85/could-you-implement-this-function-please-im-having-issues-with-it-pdf-1-320.jpg)
![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;](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
Consider a double-linked linked list implementation with the followin.pdf
Consider a double-linked linked list implementation with the following node: struct Node {int
data; Node *prev; Node *next;} Write a copyList method that is not a member of any class.
The method should take a head pointer and return another pointer. Do not modify the input.
Solution
struct Node {
Node *prev; // previous node
Node *next; // next node
int data; // stored value
};
#include
#include \"List.h\" // std: #include
using namespace std;
typedef DataList ; // std: typedef list Data;
int main() {
Data k;
// back stuff
k.push_back(5);
k.push_back(6);
cout << k.back() << endl;
k.pop_back();
// front stuff
k.push_front(4);
k.push_front(3);
cout << k.front() << endl;
k.pop_front();
// forward iterator
Data::iterator pos;
for (pos = k.begin(); pos != k.end(); ++pos)
cout << *pos << endl;
// output and delete list
while (!k.empty()) {
cout << k.front() << endl;
k.pop_front();
}
k.push_front(5);
k.push_front(6);
// remove and erase
k.remove(5);
pos = k.begin();
k.erase(pos);
k.push_front(5);
k.push_front(6);
// copy constructor
Data l = k;
// assignment operator
Data m;
m = k;
return 0;
}
// List.h
struct Node;
classIterator List;
class List {
public:
typedef ListIterator iterator;
// constructor
List();
// destructor
virtual ~List();
// copy constructor
List(const List& k);
// assignment operator
List& operator=(const List& k);
// insert value in front of list
void push_front(double data);
// insert value in back of list
void push_back(double data);
// delete value from front of list
void pop_front();
// delete value from back of list
void pop_back();
// return value on front of list
double front() const;
// return value on back of list
double back() const;
// delete value specified by iterator
void erase(const iterator& i);
// delete all nodes with specified value
void remove(double data);
// return true if list is empty
bool empty() const;
// return reference to first element in list
iterator begin() const;
// return reference to one past last element in list
iterator end() const;
private:
Node *head; // head of list
};
class ListIterator {
public:
// default constructor
ListIterator() {
i = 0;
}
// construct iterator for given pointer (used for begin/end)
ListIterator(Node *p) {
i = p;
}
// convert iterator to Node*
operator Node*() const {
return i;
}
// test two iterators for not equal
bool operator!=(const ListIterator& k) const {
return i != k.i;
}
// preincrement operator
ListIterator& operator++() {
i = i->next;
return *this;
}
// return value associated with iterator
double& operator*() const {
return i->data;
}
private:
Node *i; // current value of iterator
};
list.cpp
// delete list
static void deleteList(Node *head) {
Node *p = head->next;
while (p != head) {
Node *next = p->next;
delete p;
p = next;
}
delete head;
}
// copy list
static void copyList(const Node *from, Node *&to) {
// create dummy header
to = new Node;
to->next = to->prev = to;
// copy nodes
for (Node *p = from->next; p != from; p = p->next) {
Node *t = new Node;
t.
maincpp Build and procees a sorted linked list of Patie.pdf
main.cpp
/*
Build and procees a sorted linked list of Patient objects.
The list is sorted in ascending order by name.
Assume that the name is unique.
*/
#include <iostream>
#include <iomanip>
#include <fstream>
#include <sstream>
#include <string>
#include <cctype>
#include "LinkedList.h"
using namespace std;
void buildList(const string &filename, LinkedList &list);
void deleteManager(LinkedList &list);
void searchManager(const LinkedList &list);
void displayManager(const LinkedList &list);
int main() {
string inputFileName;
LinkedList list;
cout << "Enter file name: ";
getline(cin, inputFileName);
buildList(inputFileName, list);
displayManager(list);
searchManager(list);
deleteManager(list);
displayManager(list);
return 0;
}
/*
This function reads data from a file and inserts them
into a sorted linked list. The list is sorted in ascending order by name
*/
void buildList(const string &filename, LinkedList &list) {
ifstream inFile(filename);
cout << "Reading data from "" << filename << ""n";
if(!inFile) {
cout << "Error opening the input file: ""<< filename << """ << endl;
exit(EXIT_FAILURE);
}
string id;
int year;
string name;
int amount;
string line;
while (getline(inFile, line)) {
stringstream temp(line);
temp >> id >> year;
temp.ignore();
getline(temp, name, ';');
temp >> amount;
// create a Sales object and initialize it with data from file
/* Write your code here */
// call insert to insert this new Sales object into the sorted list
/* Write your code here */
}
inFile.close();
}
/*
Delete manager: delete items from the list until the user enters Q to quit
deleting
Input Parameter: list
*/
void deleteManager(LinkedList &list) {
string target = "";
cout << " Delete" << endl;
cout << "========" << endl;
while(target != "Q") {
cout << "Enter a name (or Q to stop deleting):" << endl;
getline(cin, target);
target[0] = toupper(target[0]);
if(target != "Q") {
if(list.deleteNode(target))
cout << target << " has been deleted!" << endl;
else
cout << target << " not found!" << endl;
}
}
cout << "___________________END DELETE SECTION_____" << endl;
}
/*
Search manager: search the list until the user enters Q to quit searching
Input Parameter: list
*/
void searchManager(const LinkedList &list) {
string target = "";
Sales obj;
cout << " Search" << endl;
cout << "========" << endl;
/* Write your code here */
cout << "___________________END SEARCH SECTION _____" << endl;
}
/*
Display manager: all, average, or year hired reports including header and footer where
appropriate,
depending on the user's choice;
displays the number of nodes (always)
Input Parameter: list
*/
void displayManager(const LinkedList &list) {
// Sub-functions of displayManager()
void showMenu(void);
string getOption(void);
void showHeader(string line);
string line = "==== ==================== =============n";
string option;
showMenu();
option = getOption();
while(option[0] != 'Q') {
switch (option[0]) {
case 'A':
showHeader(line);
list.displayList();
cout << line;
break;
case 'G':
cou.
Write a program to implement below operations with both singly and d.pdf
Write a program to implement below operations with both singly and doubly linked list. Use
inheritance to define singly linked list and doubly linked list. Add Node() Delete Node() Update
Node() Delete All() Size()
Solution
#include
using namespace std;
class List;
class Node
{
friend class List;
protected:
int Data;
Node *Link;
};
class List:public Node
{
friend class Node;
private:
Node *Start;
public:
List()
{
Start=NULL;
}
Node Insert(int x);
Node Delete(int &x);
void Display();
};
Node List::Insert(int x)
{
Node *New=new Node;
New->Data=x;
New->Link=Start;
Start=New;
return *this;
}
Node List::Delete(int &x)//deletes first node
{
Node *Del=Start;
Start=Start->Link;
delete Del;
return *this;
}
void List::Display()
{
Node *P=Start;
while(P!=NULL)
{
cout<Data<<\" \";
P=P->Link;
}
}
int main()
{
List L;
L.Insert(4);
L.Insert(64);
L.Insert(3);
L.Insert(90);
L.Insert(66);
L.Insert(13);
L.Display();
}
Double Linked List.
#include stdafx.h #include iostream using namespace std;vo.docx
#include stdafx.h #include iostream using namespace std;vo
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();
}.
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.
Write a C++ function that delete nodes in a doubly linkedlist- It shou.docx
Write a C++ function that delete nodes in a doubly linkedlist.
It should delete all nodes that have that specific value. Finally, return the head of the linkedlist.
node * DN(node *head, int value) {
}
The linkedlist structure:
struct n {
int value;
node *next;
node *previous;
};
Solution
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Function to delete the entire linked list */
void deleteList(struct node** head_ref)
{
/* deref head_ref to get the real head */
struct node* current = *head_ref;
struct node* next;
while (current != NULL)
{
next = current->next;
free(current);
current = next;
}
/* deref head_ref to affect the real head back
in the caller. */
*head_ref = NULL;
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
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)Â Â Â = new_node;
}
/* Drier program to test count function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
printf(\"\ Deleting linked list\");
deleteList(&head);
printf(\"\ Linked list deleted\");
}
.
Using the provided table interface table.h and the sample linked lis.pdf
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;
}.
Recommended
Consider a double-linked linked list implementation with the followin.pdf
Consider a double-linked linked list implementation with the following node: struct Node {int
data; Node *prev; Node *next;} Write a copyList method that is not a member of any class.
The method should take a head pointer and return another pointer. Do not modify the input.
Solution
struct Node {
Node *prev; // previous node
Node *next; // next node
int data; // stored value
};
#include
#include \"List.h\" // std: #include
using namespace std;
typedef DataList ; // std: typedef list Data;
int main() {
Data k;
// back stuff
k.push_back(5);
k.push_back(6);
cout << k.back() << endl;
k.pop_back();
// front stuff
k.push_front(4);
k.push_front(3);
cout << k.front() << endl;
k.pop_front();
// forward iterator
Data::iterator pos;
for (pos = k.begin(); pos != k.end(); ++pos)
cout << *pos << endl;
// output and delete list
while (!k.empty()) {
cout << k.front() << endl;
k.pop_front();
}
k.push_front(5);
k.push_front(6);
// remove and erase
k.remove(5);
pos = k.begin();
k.erase(pos);
k.push_front(5);
k.push_front(6);
// copy constructor
Data l = k;
// assignment operator
Data m;
m = k;
return 0;
}
// List.h
struct Node;
classIterator List;
class List {
public:
typedef ListIterator iterator;
// constructor
List();
// destructor
virtual ~List();
// copy constructor
List(const List& k);
// assignment operator
List& operator=(const List& k);
// insert value in front of list
void push_front(double data);
// insert value in back of list
void push_back(double data);
// delete value from front of list
void pop_front();
// delete value from back of list
void pop_back();
// return value on front of list
double front() const;
// return value on back of list
double back() const;
// delete value specified by iterator
void erase(const iterator& i);
// delete all nodes with specified value
void remove(double data);
// return true if list is empty
bool empty() const;
// return reference to first element in list
iterator begin() const;
// return reference to one past last element in list
iterator end() const;
private:
Node *head; // head of list
};
class ListIterator {
public:
// default constructor
ListIterator() {
i = 0;
}
// construct iterator for given pointer (used for begin/end)
ListIterator(Node *p) {
i = p;
}
// convert iterator to Node*
operator Node*() const {
return i;
}
// test two iterators for not equal
bool operator!=(const ListIterator& k) const {
return i != k.i;
}
// preincrement operator
ListIterator& operator++() {
i = i->next;
return *this;
}
// return value associated with iterator
double& operator*() const {
return i->data;
}
private:
Node *i; // current value of iterator
};
list.cpp
// delete list
static void deleteList(Node *head) {
Node *p = head->next;
while (p != head) {
Node *next = p->next;
delete p;
p = next;
}
delete head;
}
// copy list
static void copyList(const Node *from, Node *&to) {
// create dummy header
to = new Node;
to->next = to->prev = to;
// copy nodes
for (Node *p = from->next; p != from; p = p->next) {
Node *t = new Node;
t.
maincpp Build and procees a sorted linked list of Patie.pdf
main.cpp
/*
Build and procees a sorted linked list of Patient objects.
The list is sorted in ascending order by name.
Assume that the name is unique.
*/
#include <iostream>
#include <iomanip>
#include <fstream>
#include <sstream>
#include <string>
#include <cctype>
#include "LinkedList.h"
using namespace std;
void buildList(const string &filename, LinkedList &list);
void deleteManager(LinkedList &list);
void searchManager(const LinkedList &list);
void displayManager(const LinkedList &list);
int main() {
string inputFileName;
LinkedList list;
cout << "Enter file name: ";
getline(cin, inputFileName);
buildList(inputFileName, list);
displayManager(list);
searchManager(list);
deleteManager(list);
displayManager(list);
return 0;
}
/*
This function reads data from a file and inserts them
into a sorted linked list. The list is sorted in ascending order by name
*/
void buildList(const string &filename, LinkedList &list) {
ifstream inFile(filename);
cout << "Reading data from "" << filename << ""n";
if(!inFile) {
cout << "Error opening the input file: ""<< filename << """ << endl;
exit(EXIT_FAILURE);
}
string id;
int year;
string name;
int amount;
string line;
while (getline(inFile, line)) {
stringstream temp(line);
temp >> id >> year;
temp.ignore();
getline(temp, name, ';');
temp >> amount;
// create a Sales object and initialize it with data from file
/* Write your code here */
// call insert to insert this new Sales object into the sorted list
/* Write your code here */
}
inFile.close();
}
/*
Delete manager: delete items from the list until the user enters Q to quit
deleting
Input Parameter: list
*/
void deleteManager(LinkedList &list) {
string target = "";
cout << " Delete" << endl;
cout << "========" << endl;
while(target != "Q") {
cout << "Enter a name (or Q to stop deleting):" << endl;
getline(cin, target);
target[0] = toupper(target[0]);
if(target != "Q") {
if(list.deleteNode(target))
cout << target << " has been deleted!" << endl;
else
cout << target << " not found!" << endl;
}
}
cout << "___________________END DELETE SECTION_____" << endl;
}
/*
Search manager: search the list until the user enters Q to quit searching
Input Parameter: list
*/
void searchManager(const LinkedList &list) {
string target = "";
Sales obj;
cout << " Search" << endl;
cout << "========" << endl;
/* Write your code here */
cout << "___________________END SEARCH SECTION _____" << endl;
}
/*
Display manager: all, average, or year hired reports including header and footer where
appropriate,
depending on the user's choice;
displays the number of nodes (always)
Input Parameter: list
*/
void displayManager(const LinkedList &list) {
// Sub-functions of displayManager()
void showMenu(void);
string getOption(void);
void showHeader(string line);
string line = "==== ==================== =============n";
string option;
showMenu();
option = getOption();
while(option[0] != 'Q') {
switch (option[0]) {
case 'A':
showHeader(line);
list.displayList();
cout << line;
break;
case 'G':
cou.
Write a program to implement below operations with both singly and d.pdf
Write a program to implement below operations with both singly and doubly linked list. Use
inheritance to define singly linked list and doubly linked list. Add Node() Delete Node() Update
Node() Delete All() Size()
Solution
#include
using namespace std;
class List;
class Node
{
friend class List;
protected:
int Data;
Node *Link;
};
class List:public Node
{
friend class Node;
private:
Node *Start;
public:
List()
{
Start=NULL;
}
Node Insert(int x);
Node Delete(int &x);
void Display();
};
Node List::Insert(int x)
{
Node *New=new Node;
New->Data=x;
New->Link=Start;
Start=New;
return *this;
}
Node List::Delete(int &x)//deletes first node
{
Node *Del=Start;
Start=Start->Link;
delete Del;
return *this;
}
void List::Display()
{
Node *P=Start;
while(P!=NULL)
{
cout<Data<<\" \";
P=P->Link;
}
}
int main()
{
List L;
L.Insert(4);
L.Insert(64);
L.Insert(3);
L.Insert(90);
L.Insert(66);
L.Insert(13);
L.Display();
}
Double Linked List.
#include stdafx.h #include iostream using namespace std;vo.docx
#include stdafx.h #include iostream using namespace std;vo
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();
}.
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.
Write a C++ function that delete nodes in a doubly linkedlist- It shou.docx
Write a C++ function that delete nodes in a doubly linkedlist.
It should delete all nodes that have that specific value. Finally, return the head of the linkedlist.
node * DN(node *head, int value) {
}
The linkedlist structure:
struct n {
int value;
node *next;
node *previous;
};
Solution
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Function to delete the entire linked list */
void deleteList(struct node** head_ref)
{
/* deref head_ref to get the real head */
struct node* current = *head_ref;
struct node* next;
while (current != NULL)
{
next = current->next;
free(current);
current = next;
}
/* deref head_ref to affect the real head back
in the caller. */
*head_ref = NULL;
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
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)Â Â Â = new_node;
}
/* Drier program to test count function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
printf(\"\ Deleting linked list\");
deleteList(&head);
printf(\"\ Linked list deleted\");
}
.
Using the provided table interface table.h and the sample linked lis.pdf
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;
}.
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.
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
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.
C++ Please write the whole code that is needed for this assignment- wr.docx
Brianca plans to save $5,000, $1,000, and $42,000 a year over the next three years, respectively. How much would you need to deposit in one lump sum today to have the same amount as Brianca three years from now if you both earn 10.9 percent, compounded annually? $36,115 $35,192 $43,282 $41,635 $35,372
.
For this micro assignment, you must implement two Linked List functi.docx
For this micro assignment, you must implement two Linked List functions. We will use the following
example Linked List:
2 0 -1 5 7
getElementAt(index)
This function should return the element (i.e. value) of the Nth item inside the linked list. For example,
on the Linked List above, getElementAt(0) should return 2; getElementAt(3) should return 5.
addElementAt(value, location)
This function should insert a new value at the given location. Note that the location supplied must be
within bounds of the LinkedList. For example, we cannot call addElementAt(4, 11) on the above Linked
List because 11 is beyond the size of the Linked List.
Here are some examples. If we call addElementAt(0, 1), the above Linked List would now look like:
1 2 0 -1 5 7
If we again call addElementAt(2, 123), we would get:
1 2 123 0 -1 5 7
Grading
Your submission will be graded based on the following:
1. [7] Your solution does not cause any runtime issues and your file passes all test cases
2. [3] Your code contains good style. For example,
You provide meaningful variable names
You provide sufficient and meaningful comments
Your code is well structured
@@@@@@@@@@@@@@@@@@@@@@@@
#ifndef LINKED_LIST_H
#define LINKED_LIST_H
#include
#include
#include "LinkedListNode.h"
#include
using namespace std;
template
class LinkedList
{
private:
//points to the front of the linked list
LinkedListNode *_front = nullptr;
//keeping track of size in a variable eliminates need to continually
//count LL boxes.
int _size = 0;
protected:
//creates a new LinkedListNode for us
virtual LinkedListNode *createNode(T value)
{
return new LinkedListNode < T > { value };
}
public:
//default constructor
LinkedList()
{
_front = nullptr;
}
//copy constructor
LinkedList(const LinkedList &other)
{
for (int i = 0; i < other.getSize(); i++)
{
addElement(other.getElementAt(i));
}
}
//move constructor
LinkedList(LinkedList &&other)
{
//take other's data
_front = other._front;
_size = other._size;
//reset other's pointers
other._front = nullptr;
}
//initializer list constructor
LinkedList(initializer_list values)
{
for (auto item : values)
{
addElement(item);
}
}
//Always remember to clean up pointers in destructor!
virtual ~LinkedList()
{
LinkedListNode *current = _front;
while (current != nullptr)
{
LinkedListNode *temp = current->getNext();
delete current;
current = temp;
}
}
//will return true if the LL is empty.
virtual bool isEmpty() const
{
return _size == 0;
}
//returns the size of the LL.
virtual int getSize() const
{
return _size;
}
.
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.
How do you update an address according to this code- Currently- I fig.pdf
How do you update an address according to this code?
Currently, I figured out up to (the bolded line) How do I update address, display all entries and
quit.
Add a name (n)
Look up a name (l)
Update address (u)
Delete an entry (d)
Display all entries (a)
Quit (q)
-> n
import java.util.Scanner;
public class AddressBook {
public static void main(String[] args) {
Tablea addressBook = new Tablea();
Scanner scanner = new Scanner(System. in );
String key, value;
// Insert a new entry into the address book
System. out .print("Enter a name: ");
key = scanner.nextLine();
System. out .print("Enter a address: ");
value = scanner.nextLine();
addressBook.insert(key, value);
// Lookup an entry in the address book
System. out .print("Enter a name to look up: ");
key = scanner.nextLine();
String result = addressBook.lookUp(key);
if (result != null) {
System. out .println("address: " + result);
} else {
System. out .println("name not found");
}
// Delete an entry from the address book
System. out .print("Enter a name to delete: ");
key = scanner.nextLine();
boolean deleted = addressBook.delete(key);
if (deleted) {
System. out .println("address deleted");
} else {
System. out .println("name not found");
}
}
}
public class Node {
private String key;
private String value;
private Node next;
Node() {
// add here ..
}
Node(String key, String value) {
// add here ..
this.key = key;
this.value = value;
this.next = null;
}
public String getKey() {
// add here ..
return this.key;
}
public void setKey(String key) {
// add here ..
this.key = key;
}
public String getValue() {
// add here ..
return this.value;
}
public void setValue(String value) {
// add here ..
this.value = value;
}
public Node getNext() {
// add here ..
return this.next;
}
public void setNext(Node next) {
// add here ..
this.next = next;
}
}
import java.util.Scanner;
public class Tablea {
private Node mark;
public Node getMark() {
return this.mark;
}
public void setMark(Node mark) {
this.mark = mark;
}
public boolean insert(String key, String value) {
Node newNode= new Node(key,value);
if (this.mark== null) {
this.mark = newNode;
}else {
newNode.setNext(this.mark.getNext());
this.mark.setNext(newNode);
}
return true;
}
public String lookUp(String key) {
Node current = this.mark;
while(current!=null) {
if (current.getKey().equals(key)) {
return current.getValue();
}
current = current.getNext();
}
return null;
}
public boolean delete(String key) {
Node current = this.mark;
Node prev = null;
while (current!= null) {
if (current.getKey().equals(key)) {
if (prev == null) {
this.mark = current.getNext();
}else {
prev.setNext(current.getNext());
}
return true;
}
prev = current;
current = current.getNext();
}
return false;
}
public boolean update(String key, String newValue) {
Node current = this.mark;
while (current != null) {
if (current.getKey().equals(key)) {
current.setValue(newValue);
return true;
}
current = current.getNext();
}
return false;
}
public boolean markToStart() {
if (this.mark== null) {
retur.
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.
#include iostream #include cstddefusing namespace std;temp.pdf
#include
#include
using namespace std;
template class Node{
private:
T value;
Node *next;
public:
Node();
Node (T data);
T remove();
T getValue();
Node* getNext();
void setValue(T data);
void setNext(Node *nextNode);
};
template
Node::Node(){
T value =NULL;
}
template
Node::Node(T data){
T value=data;
}
template
T Node::getValue(){
return value;
}
template
Node* Node::getNext(){
return next;
}
template
void Node::setValue(T data){
T value=data;
}
template
void Node::setNext(Node *nextNode){
next=nextNode;
}
template class MyLinkedList{
Node start;
//Define a constructor
MyLinkedList()
{
start->setValue(0);
start->setNext(NULL);
}
//Member Functions:
// Adds a new element to the list
void add (T val)
{
/* creating a new node to hold value*/
Nodetemp(val,NULL);
/* If no start node till now, this node is made as start node*/
if (start->next == NULL)
{
start.setNext(temp);
}
/* if start node exists, insert new node at end of list*/
else
{
s = start->next;
/* locating end of list*/
while (s->next != NULL)
{
s= s->next;
}
temp->next = NULL;
/* attaching new node */
s->next = temp;
cout<<\"Element Inserted\"<next->next != NULL)
{
s = s->next;
}
//After locating the end node
ptr=s->next;
s->next=NULL;
//deletting end node
return ptr;
}
}
//Prints elements of the list
void printContent()
{
struct node *ptr = start;
cout<data;
ptr=ptr->next;
}
}
}
//Prints total number of elements
int getSize()
{
struct node *ptr, *s;
int count=0;
if (start == NULL)
{
cout<<\"The List is empty\"<next;
count=1;
while (ptr != NULL)
{
count++;
ptr=ptr->next;
}
return count;
}
//Returns true if there is no element in the list.
bool empty()
{
if(start==NULL)
{
return empty;
}
}
};
Solution
#include
#include
using namespace std;
template class Node{
private:
T value;
Node *next;
public:
Node();
Node (T data);
T remove();
T getValue();
Node* getNext();
void setValue(T data);
void setNext(Node *nextNode);
};
template
Node::Node(){
T value =NULL;
}
template
Node::Node(T data){
T value=data;
}
template
T Node::getValue(){
return value;
}
template
Node* Node::getNext(){
return next;
}
template
void Node::setValue(T data){
T value=data;
}
template
void Node::setNext(Node *nextNode){
next=nextNode;
}
template class MyLinkedList{
Node start;
//Define a constructor
MyLinkedList()
{
start->setValue(0);
start->setNext(NULL);
}
//Member Functions:
// Adds a new element to the list
void add (T val)
{
/* creating a new node to hold value*/
Nodetemp(val,NULL);
/* If no start node till now, this node is made as start node*/
if (start->next == NULL)
{
start.setNext(temp);
}
/* if start node exists, insert new node at end of list*/
else
{
s = start->next;
/* locating end of list*/
while (s->next != NULL)
{
s= s->next;
}
temp->next = NULL;
/* attaching new node */
s->next = temp;
cout<<\"Element Inserted\"<next->next != NULL)
{
s = s->next;
}
//After locating the end node
ptr=s->next;
s->next=NULL;
//deletting end node
return ptr;
}
}
//Prints elements of the list
void printContent()
{
struct .
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.
C code on linked list #include stdio.h #include stdlib.h.pdf
For the ethanol water graph the simple distillation should be a steady smooth
increase, the fractional distillation should look like an s, that sort of takes a straight up jump in
the middle. I am unsure about the other mixture though sorry, hope that helps
Solution
For the ethanol water graph the simple distillation should be a steady smooth
increase, the fractional distillation should look like an s, that sort of takes a straight up jump in
the middle. I am unsure about the other mixture though sorry, hope that helps.
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.
Below is a depiction of a doubly-linked list implementation of the bag.docx
Below is a depiction of a doubly-linked list implementation of the bag ADT with two sentinels -one at the head and one at the tail. Strict DLink { TYPE value; strict DLink * next; struct DLink * prev;}; strict linked List { int size; struct DLink *frontSentinel; struct DLink *backSentinel;}; Write a function that at first reverses the list and then prints the values of the links in the list from front to back. Don\'t redirect to any other function. You must use assertions if necessary.
Solution
/* Program to reverse a doubly linked list */
#include <stdio.h>
#include <stdlib.h>
/* a node of the doubly linked list */
struct node
{
int data;
struct node *n;
struct node *pr;
};
/* Function to reverse a Doubly Linked List */
void reverse(struct node **head_r)
{
struct node *t = NULL;
struct node *cur = *head_r;
/* swap next and prev for all nodes of
doubly linked list */
while (cur != NULL)
{
t = cur->pr;
cur->pr = cur->n;
cur->n = t;
cur = cur->pr;
}
/* Before changing head, check for the cases like empty
list and list with only one node */
if(t != NULL )
*head_r = t->pr;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginging of the Doubly Linked List */
void push(struct node** head_r, int new_data)
{
/* allocate node */
struct node* n_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
n_node->data = n_data;
/* since we are adding at the begining,
prev is always NULL */
n_node->pr = NULL;
/* link the old list off the new node */
n_node->n = (*head_r);
/* change prev of head node to new node */
if((*head_r) != NULL)
(*head_r)->pr = n_node ;
/* move the head to point to the new node */
(*head_r) = n_node;
}
/* Function to print nodes in a given doubly linked list
This function is same as printList() of singly linked lsit */
void printList(struct node *node)
{
while(node!=NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
}
/* Drier program to test above functions*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Let us create a sorted linked list to test the functions
Created linked list will be 10->8->4->2 */
push(&head, 2);
push(&head, 4);
push(&head, 8);
push(&head, 10);
printf(\"\ Original Linked list \");
printList(head);
/* Reverse doubly linked list */
reverse(&head);
printf(\"\ Reversed Linked list \");
printList(head);
getchar();
}
.
Create a link list. Add some nodes to it, search and delete nodes fro.pdf
Create a link list. Add some nodes to it, search and delete nodes from it.
Solution
#include
#include
using std::cout;
using std::endl;
/*
A linked list is a list constructed using pointers. It is not fixed in
size and could grow and shrink while the program is running.
A typical defination of list nodes contains at least two parts, both the
content or date of each element and the pointer to the next element,
which is shown in the figure below.
+---------+
| Data | -----> holds the data of this element in the list.
+---------+
| pointer | -----> is a pointer to the next element in the list.
+---------+
***Attention***:
The pointer holds the address of the next element, not the address of the
data in the next element, even though they are the same in value sometimes.
And It should be set to NULL while acting as the last node of the list.
Implementation of the single linked list:
+---------+ --->+---------+ --->+---------+
| Data | | | Data | | | Data |
+---------+ | +---------+ | +---------+
| pointer |----- | pointer |----- | pointer |
+---------+ +---------+ +---------+
*/
/* definition of the list node class */
class Node
{
friend class LinkedList;
private:
int _value; /* data, can be any data type, but use integer for easiness */
Node *_pNext; /* pointer to the next node */
public:
/* Constructors with No Arguments */
Node(void)
: _pNext(NULL)
{ }
/* Constructors with a given value */
Node(int val)
: _value(val), _pNext(NULL)
{ }
/* Constructors with a given value and a link of the next node */
Node(int val, Node* next)
: _value(val), _pNext(next)
{}
/* Getters */
int getValue(void)
{ return _value; }
Node* getNext(void)
{ return _pNext; }
};
/* definition of the linked list class */
class LinkedList
{
private:
/* pointer of head node */
Node *_pHead;
/* pointer of tail node */
Node *_pTail;
public:
/* Constructors with No Arguments */
LinkedList(void);
/* Constructors with a given value of a list node */
LinkedList(int val);
/* Destructor */
~LinkedList(void);
/* Function to add a node at the head of a linked list */
void headInsert(int val);
/* Function to add a node in the middle of a linked list */
void insertAfter(Node* afterMe, int val);
/* Function to append a node to the end of a linked list */
void tailAppend(int val);
/* Function to get the node in the specific position */
Node* getNode(int pos);
/* Traversing the list and printing the value of each node */
void traverse_and_print();
};
LinkedList::LinkedList()
{
/* Initialize the head and tail node */
_pHead = _pTail = NULL;
}
LinkedList::LinkedList(int val)
{
/* Create a new node, acting as both the head and tail node */
_pHead = new Node(val);
_pTail = _pHead;
}
LinkedList::~LinkedList()
{
/*
* Leave it empty temporarily.
* It will be described in detail in the example \"How to delete a linkedlist\".
*/
}
void LinkedList::headInsert(int val)
{
/* The list is empty? */
if (_pHead == NULL) {
/* the same to create a new list with a given value */
_pTail = _pHead = new Node(val);
.
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.
write recursive function that calculates and returns the length of a.pdf
write recursive function that calculates and returns the length of a linked list
Solution
// C code to determine length of linked list recursively
#include
#include
// node of linked list
struct node
{
int key;
struct node* next;
};
// push key to linked list
void push(struct node** root, int new_element)
{
struct node* nodeNew = (struct node*) malloc(sizeof(struct node));
nodeNew->key = new_element;
nodeNew->next = (*root);
(*root) = nodeNew;
}
// recursively get length of linked list
int lengthLinkedlist(struct node* root)
{
// base case
if (root == NULL)
return 0;
// recursively call the function
return 1 + lengthLinkedlist(root->next);
}
int main()
{
struct node* root = NULL;
push(&root, 11);
push(&root, 34);
push(&root, 4);
push(&root, 23);
push(&root, 14);
push(&root, 88);
push(&root, 56);
push(&root, 84);
push(&root, 6);
/* Check the count function */
printf(\"Length of Linked list: %d\ \", lengthLinkedlist(root));
return 0;
}
/*
output:
Length of Linked list: 9
*/.
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.
Case studyInitial history27-year-old female comlaining of sympto.pdf
Case study
Initial history
27-year-old female comlaining of symptoms of urgency to urinate, frequent urination, and
urethral burning during urination persisting for 48 hours
Patient awoke from sleep with urgency and suprapubic discomfort 2 nights ago
Urine now has strong odor and cloudy appearance
Additional history
Recurring urinary tract infections since she married at age 22 years
Prior UTI associated with similar symptoms
Denies history of febrile or hemorrhagic UTI
Three episodes over past 2 years
May be associated with strenuous physical exertion and sexual intercourse
No other medical history
She reports an allergy to penicillin which causes a “rush” and “trouble breathing”
Physical examination
Well-nourished female experiencing mild discomfort
Tenderness on palpation of abdominal pelvic area; physical examination otherwise unremarkable
T=98.6°C orally; BP=114/64 mmHg; P=68; RR=12
Laboratory results
Dipstick urinalysis:
Color: Dark yellow, specific gravity=1.030, pH=6.5
Protein: Negative, glucose: negative, Ketone: negative, bilirubin: negative
Trace occult blood
Leukocytes:large amount
Nitrates: positive, urobilinogen:negative
Microscopic examination:
WBC: too numerous to count (TNTC)/HPF (high power field)
Bacteria: TNTC
RBC: 3 or 4/HPF
Casts: negative
Urine culture:
E.Coli:> 104 CFU/mL
Organism sensitive to ampicillin, nitrofutantoin, trimethoprim-sulfamethoxazole (TMP-SMX),
ciprofloxacin, cephalexin
What is the history of this patient?
Solution
The organism is sensitive to antibiotics like ampicillin, nitrofutantoin, trimethoprim-
sulfamethoxazole (TMP-SMX), ciprofloxacin, cephalexin. The urine culture shows high number
of E.coli present in the urine with 3-4 RBC. Also the WBC is too numerous to count . An initial
increase in WBC is indicative of an infection and with such high amount of bacteria in urine , is
indicative of a bacterial infection , specifically caused by E.Coli. Although, the urine analysis do
not show anymore anomalies, except presence of nitrates. Nitrates in urine occur are also
indicative of an urinary tract infection.
The initial patient history is also suggestive of an Urinary tract infection, due to cloudy
appearance of urine. The indication is of a Cystitis, which is generally caused by an E.coli in the
urine . One of the reasons of the recurring urinary infections might be due to sexual intercourse
or strenuous activities..
I appreciate any help with the following questions.(1) What typ.pdf
Hypothesize a set of at least three arguments as to why innovation is important to the
competitiveness of nations, companies, leaders, and individual employees.
Solution
One of the main reasons why innovation is important is the fact that it creates development on
nations and business, it opens opportunities to new projects that on the past have not being taken
in consideration.
Innovation sometimes is the only way an organization has to stay alive on business because of
the hard and globalize competition, it creates the unique value for your brand our country, and it
is an added value. For example recently Nicaragua offer new investors that they will not need to
pay taxes for 10 years as long as they start a business there and offer job to the people in
Nicaragua.
Innovation makes the organization better every day, it is a constant improvement on the
processes on a positive manner, and this increases creativity. You don’t need to create something
new that no one has ever think you can take a product and add it qualities that your competition
does not have and making your target market to prefer your brand..
More Related Content
Similar to could you implement this function please, im having issues with it..pdf
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.
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
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.
C++ Please write the whole code that is needed for this assignment- wr.docx
Brianca plans to save $5,000, $1,000, and $42,000 a year over the next three years, respectively. How much would you need to deposit in one lump sum today to have the same amount as Brianca three years from now if you both earn 10.9 percent, compounded annually? $36,115 $35,192 $43,282 $41,635 $35,372
.
For this micro assignment, you must implement two Linked List functi.docx
For this micro assignment, you must implement two Linked List functions. We will use the following
example Linked List:
2 0 -1 5 7
getElementAt(index)
This function should return the element (i.e. value) of the Nth item inside the linked list. For example,
on the Linked List above, getElementAt(0) should return 2; getElementAt(3) should return 5.
addElementAt(value, location)
This function should insert a new value at the given location. Note that the location supplied must be
within bounds of the LinkedList. For example, we cannot call addElementAt(4, 11) on the above Linked
List because 11 is beyond the size of the Linked List.
Here are some examples. If we call addElementAt(0, 1), the above Linked List would now look like:
1 2 0 -1 5 7
If we again call addElementAt(2, 123), we would get:
1 2 123 0 -1 5 7
Grading
Your submission will be graded based on the following:
1. [7] Your solution does not cause any runtime issues and your file passes all test cases
2. [3] Your code contains good style. For example,
You provide meaningful variable names
You provide sufficient and meaningful comments
Your code is well structured
@@@@@@@@@@@@@@@@@@@@@@@@
#ifndef LINKED_LIST_H
#define LINKED_LIST_H
#include
#include
#include "LinkedListNode.h"
#include
using namespace std;
template
class LinkedList
{
private:
//points to the front of the linked list
LinkedListNode *_front = nullptr;
//keeping track of size in a variable eliminates need to continually
//count LL boxes.
int _size = 0;
protected:
//creates a new LinkedListNode for us
virtual LinkedListNode *createNode(T value)
{
return new LinkedListNode < T > { value };
}
public:
//default constructor
LinkedList()
{
_front = nullptr;
}
//copy constructor
LinkedList(const LinkedList &other)
{
for (int i = 0; i < other.getSize(); i++)
{
addElement(other.getElementAt(i));
}
}
//move constructor
LinkedList(LinkedList &&other)
{
//take other's data
_front = other._front;
_size = other._size;
//reset other's pointers
other._front = nullptr;
}
//initializer list constructor
LinkedList(initializer_list values)
{
for (auto item : values)
{
addElement(item);
}
}
//Always remember to clean up pointers in destructor!
virtual ~LinkedList()
{
LinkedListNode *current = _front;
while (current != nullptr)
{
LinkedListNode *temp = current->getNext();
delete current;
current = temp;
}
}
//will return true if the LL is empty.
virtual bool isEmpty() const
{
return _size == 0;
}
//returns the size of the LL.
virtual int getSize() const
{
return _size;
}
.
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.
How do you update an address according to this code- Currently- I fig.pdf
How do you update an address according to this code?
Currently, I figured out up to (the bolded line) How do I update address, display all entries and
quit.
Add a name (n)
Look up a name (l)
Update address (u)
Delete an entry (d)
Display all entries (a)
Quit (q)
-> n
import java.util.Scanner;
public class AddressBook {
public static void main(String[] args) {
Tablea addressBook = new Tablea();
Scanner scanner = new Scanner(System. in );
String key, value;
// Insert a new entry into the address book
System. out .print("Enter a name: ");
key = scanner.nextLine();
System. out .print("Enter a address: ");
value = scanner.nextLine();
addressBook.insert(key, value);
// Lookup an entry in the address book
System. out .print("Enter a name to look up: ");
key = scanner.nextLine();
String result = addressBook.lookUp(key);
if (result != null) {
System. out .println("address: " + result);
} else {
System. out .println("name not found");
}
// Delete an entry from the address book
System. out .print("Enter a name to delete: ");
key = scanner.nextLine();
boolean deleted = addressBook.delete(key);
if (deleted) {
System. out .println("address deleted");
} else {
System. out .println("name not found");
}
}
}
public class Node {
private String key;
private String value;
private Node next;
Node() {
// add here ..
}
Node(String key, String value) {
// add here ..
this.key = key;
this.value = value;
this.next = null;
}
public String getKey() {
// add here ..
return this.key;
}
public void setKey(String key) {
// add here ..
this.key = key;
}
public String getValue() {
// add here ..
return this.value;
}
public void setValue(String value) {
// add here ..
this.value = value;
}
public Node getNext() {
// add here ..
return this.next;
}
public void setNext(Node next) {
// add here ..
this.next = next;
}
}
import java.util.Scanner;
public class Tablea {
private Node mark;
public Node getMark() {
return this.mark;
}
public void setMark(Node mark) {
this.mark = mark;
}
public boolean insert(String key, String value) {
Node newNode= new Node(key,value);
if (this.mark== null) {
this.mark = newNode;
}else {
newNode.setNext(this.mark.getNext());
this.mark.setNext(newNode);
}
return true;
}
public String lookUp(String key) {
Node current = this.mark;
while(current!=null) {
if (current.getKey().equals(key)) {
return current.getValue();
}
current = current.getNext();
}
return null;
}
public boolean delete(String key) {
Node current = this.mark;
Node prev = null;
while (current!= null) {
if (current.getKey().equals(key)) {
if (prev == null) {
this.mark = current.getNext();
}else {
prev.setNext(current.getNext());
}
return true;
}
prev = current;
current = current.getNext();
}
return false;
}
public boolean update(String key, String newValue) {
Node current = this.mark;
while (current != null) {
if (current.getKey().equals(key)) {
current.setValue(newValue);
return true;
}
current = current.getNext();
}
return false;
}
public boolean markToStart() {
if (this.mark== null) {
retur.
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.
#include iostream #include cstddefusing namespace std;temp.pdf
#include
#include
using namespace std;
template class Node{
private:
T value;
Node *next;
public:
Node();
Node (T data);
T remove();
T getValue();
Node* getNext();
void setValue(T data);
void setNext(Node *nextNode);
};
template
Node::Node(){
T value =NULL;
}
template
Node::Node(T data){
T value=data;
}
template
T Node::getValue(){
return value;
}
template
Node* Node::getNext(){
return next;
}
template
void Node::setValue(T data){
T value=data;
}
template
void Node::setNext(Node *nextNode){
next=nextNode;
}
template class MyLinkedList{
Node start;
//Define a constructor
MyLinkedList()
{
start->setValue(0);
start->setNext(NULL);
}
//Member Functions:
// Adds a new element to the list
void add (T val)
{
/* creating a new node to hold value*/
Nodetemp(val,NULL);
/* If no start node till now, this node is made as start node*/
if (start->next == NULL)
{
start.setNext(temp);
}
/* if start node exists, insert new node at end of list*/
else
{
s = start->next;
/* locating end of list*/
while (s->next != NULL)
{
s= s->next;
}
temp->next = NULL;
/* attaching new node */
s->next = temp;
cout<<\"Element Inserted\"<next->next != NULL)
{
s = s->next;
}
//After locating the end node
ptr=s->next;
s->next=NULL;
//deletting end node
return ptr;
}
}
//Prints elements of the list
void printContent()
{
struct node *ptr = start;
cout<data;
ptr=ptr->next;
}
}
}
//Prints total number of elements
int getSize()
{
struct node *ptr, *s;
int count=0;
if (start == NULL)
{
cout<<\"The List is empty\"<next;
count=1;
while (ptr != NULL)
{
count++;
ptr=ptr->next;
}
return count;
}
//Returns true if there is no element in the list.
bool empty()
{
if(start==NULL)
{
return empty;
}
}
};
Solution
#include
#include
using namespace std;
template class Node{
private:
T value;
Node *next;
public:
Node();
Node (T data);
T remove();
T getValue();
Node* getNext();
void setValue(T data);
void setNext(Node *nextNode);
};
template
Node::Node(){
T value =NULL;
}
template
Node::Node(T data){
T value=data;
}
template
T Node::getValue(){
return value;
}
template
Node* Node::getNext(){
return next;
}
template
void Node::setValue(T data){
T value=data;
}
template
void Node::setNext(Node *nextNode){
next=nextNode;
}
template class MyLinkedList{
Node start;
//Define a constructor
MyLinkedList()
{
start->setValue(0);
start->setNext(NULL);
}
//Member Functions:
// Adds a new element to the list
void add (T val)
{
/* creating a new node to hold value*/
Nodetemp(val,NULL);
/* If no start node till now, this node is made as start node*/
if (start->next == NULL)
{
start.setNext(temp);
}
/* if start node exists, insert new node at end of list*/
else
{
s = start->next;
/* locating end of list*/
while (s->next != NULL)
{
s= s->next;
}
temp->next = NULL;
/* attaching new node */
s->next = temp;
cout<<\"Element Inserted\"<next->next != NULL)
{
s = s->next;
}
//After locating the end node
ptr=s->next;
s->next=NULL;
//deletting end node
return ptr;
}
}
//Prints elements of the list
void printContent()
{
struct .
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.
C code on linked list #include stdio.h #include stdlib.h.pdf
For the ethanol water graph the simple distillation should be a steady smooth
increase, the fractional distillation should look like an s, that sort of takes a straight up jump in
the middle. I am unsure about the other mixture though sorry, hope that helps
Solution
For the ethanol water graph the simple distillation should be a steady smooth
increase, the fractional distillation should look like an s, that sort of takes a straight up jump in
the middle. I am unsure about the other mixture though sorry, hope that helps.
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.
Below is a depiction of a doubly-linked list implementation of the bag.docx
Below is a depiction of a doubly-linked list implementation of the bag ADT with two sentinels -one at the head and one at the tail. Strict DLink { TYPE value; strict DLink * next; struct DLink * prev;}; strict linked List { int size; struct DLink *frontSentinel; struct DLink *backSentinel;}; Write a function that at first reverses the list and then prints the values of the links in the list from front to back. Don\'t redirect to any other function. You must use assertions if necessary.
Solution
/* Program to reverse a doubly linked list */
#include <stdio.h>
#include <stdlib.h>
/* a node of the doubly linked list */
struct node
{
int data;
struct node *n;
struct node *pr;
};
/* Function to reverse a Doubly Linked List */
void reverse(struct node **head_r)
{
struct node *t = NULL;
struct node *cur = *head_r;
/* swap next and prev for all nodes of
doubly linked list */
while (cur != NULL)
{
t = cur->pr;
cur->pr = cur->n;
cur->n = t;
cur = cur->pr;
}
/* Before changing head, check for the cases like empty
list and list with only one node */
if(t != NULL )
*head_r = t->pr;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginging of the Doubly Linked List */
void push(struct node** head_r, int new_data)
{
/* allocate node */
struct node* n_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
n_node->data = n_data;
/* since we are adding at the begining,
prev is always NULL */
n_node->pr = NULL;
/* link the old list off the new node */
n_node->n = (*head_r);
/* change prev of head node to new node */
if((*head_r) != NULL)
(*head_r)->pr = n_node ;
/* move the head to point to the new node */
(*head_r) = n_node;
}
/* Function to print nodes in a given doubly linked list
This function is same as printList() of singly linked lsit */
void printList(struct node *node)
{
while(node!=NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
}
/* Drier program to test above functions*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Let us create a sorted linked list to test the functions
Created linked list will be 10->8->4->2 */
push(&head, 2);
push(&head, 4);
push(&head, 8);
push(&head, 10);
printf(\"\ Original Linked list \");
printList(head);
/* Reverse doubly linked list */
reverse(&head);
printf(\"\ Reversed Linked list \");
printList(head);
getchar();
}
.
Create a link list. Add some nodes to it, search and delete nodes fro.pdf
Create a link list. Add some nodes to it, search and delete nodes from it.
Solution
#include
#include
using std::cout;
using std::endl;
/*
A linked list is a list constructed using pointers. It is not fixed in
size and could grow and shrink while the program is running.
A typical defination of list nodes contains at least two parts, both the
content or date of each element and the pointer to the next element,
which is shown in the figure below.
+---------+
| Data | -----> holds the data of this element in the list.
+---------+
| pointer | -----> is a pointer to the next element in the list.
+---------+
***Attention***:
The pointer holds the address of the next element, not the address of the
data in the next element, even though they are the same in value sometimes.
And It should be set to NULL while acting as the last node of the list.
Implementation of the single linked list:
+---------+ --->+---------+ --->+---------+
| Data | | | Data | | | Data |
+---------+ | +---------+ | +---------+
| pointer |----- | pointer |----- | pointer |
+---------+ +---------+ +---------+
*/
/* definition of the list node class */
class Node
{
friend class LinkedList;
private:
int _value; /* data, can be any data type, but use integer for easiness */
Node *_pNext; /* pointer to the next node */
public:
/* Constructors with No Arguments */
Node(void)
: _pNext(NULL)
{ }
/* Constructors with a given value */
Node(int val)
: _value(val), _pNext(NULL)
{ }
/* Constructors with a given value and a link of the next node */
Node(int val, Node* next)
: _value(val), _pNext(next)
{}
/* Getters */
int getValue(void)
{ return _value; }
Node* getNext(void)
{ return _pNext; }
};
/* definition of the linked list class */
class LinkedList
{
private:
/* pointer of head node */
Node *_pHead;
/* pointer of tail node */
Node *_pTail;
public:
/* Constructors with No Arguments */
LinkedList(void);
/* Constructors with a given value of a list node */
LinkedList(int val);
/* Destructor */
~LinkedList(void);
/* Function to add a node at the head of a linked list */
void headInsert(int val);
/* Function to add a node in the middle of a linked list */
void insertAfter(Node* afterMe, int val);
/* Function to append a node to the end of a linked list */
void tailAppend(int val);
/* Function to get the node in the specific position */
Node* getNode(int pos);
/* Traversing the list and printing the value of each node */
void traverse_and_print();
};
LinkedList::LinkedList()
{
/* Initialize the head and tail node */
_pHead = _pTail = NULL;
}
LinkedList::LinkedList(int val)
{
/* Create a new node, acting as both the head and tail node */
_pHead = new Node(val);
_pTail = _pHead;
}
LinkedList::~LinkedList()
{
/*
* Leave it empty temporarily.
* It will be described in detail in the example \"How to delete a linkedlist\".
*/
}
void LinkedList::headInsert(int val)
{
/* The list is empty? */
if (_pHead == NULL) {
/* the same to create a new list with a given value */
_pTail = _pHead = new Node(val);
.
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.
write recursive function that calculates and returns the length of a.pdf
write recursive function that calculates and returns the length of a linked list
Solution
// C code to determine length of linked list recursively
#include
#include
// node of linked list
struct node
{
int key;
struct node* next;
};
// push key to linked list
void push(struct node** root, int new_element)
{
struct node* nodeNew = (struct node*) malloc(sizeof(struct node));
nodeNew->key = new_element;
nodeNew->next = (*root);
(*root) = nodeNew;
}
// recursively get length of linked list
int lengthLinkedlist(struct node* root)
{
// base case
if (root == NULL)
return 0;
// recursively call the function
return 1 + lengthLinkedlist(root->next);
}
int main()
{
struct node* root = NULL;
push(&root, 11);
push(&root, 34);
push(&root, 4);
push(&root, 23);
push(&root, 14);
push(&root, 88);
push(&root, 56);
push(&root, 84);
push(&root, 6);
/* Check the count function */
printf(\"Length of Linked list: %d\ \", lengthLinkedlist(root));
return 0;
}
/*
output:
Length of Linked list: 9
*/.
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.
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More from feroz544
Case studyInitial history27-year-old female comlaining of sympto.pdf
Case study
Initial history
27-year-old female comlaining of symptoms of urgency to urinate, frequent urination, and
urethral burning during urination persisting for 48 hours
Patient awoke from sleep with urgency and suprapubic discomfort 2 nights ago
Urine now has strong odor and cloudy appearance
Additional history
Recurring urinary tract infections since she married at age 22 years
Prior UTI associated with similar symptoms
Denies history of febrile or hemorrhagic UTI
Three episodes over past 2 years
May be associated with strenuous physical exertion and sexual intercourse
No other medical history
She reports an allergy to penicillin which causes a “rush” and “trouble breathing”
Physical examination
Well-nourished female experiencing mild discomfort
Tenderness on palpation of abdominal pelvic area; physical examination otherwise unremarkable
T=98.6°C orally; BP=114/64 mmHg; P=68; RR=12
Laboratory results
Dipstick urinalysis:
Color: Dark yellow, specific gravity=1.030, pH=6.5
Protein: Negative, glucose: negative, Ketone: negative, bilirubin: negative
Trace occult blood
Leukocytes:large amount
Nitrates: positive, urobilinogen:negative
Microscopic examination:
WBC: too numerous to count (TNTC)/HPF (high power field)
Bacteria: TNTC
RBC: 3 or 4/HPF
Casts: negative
Urine culture:
E.Coli:> 104 CFU/mL
Organism sensitive to ampicillin, nitrofutantoin, trimethoprim-sulfamethoxazole (TMP-SMX),
ciprofloxacin, cephalexin
What is the history of this patient?
Solution
The organism is sensitive to antibiotics like ampicillin, nitrofutantoin, trimethoprim-
sulfamethoxazole (TMP-SMX), ciprofloxacin, cephalexin. The urine culture shows high number
of E.coli present in the urine with 3-4 RBC. Also the WBC is too numerous to count . An initial
increase in WBC is indicative of an infection and with such high amount of bacteria in urine , is
indicative of a bacterial infection , specifically caused by E.Coli. Although, the urine analysis do
not show anymore anomalies, except presence of nitrates. Nitrates in urine occur are also
indicative of an urinary tract infection.
The initial patient history is also suggestive of an Urinary tract infection, due to cloudy
appearance of urine. The indication is of a Cystitis, which is generally caused by an E.coli in the
urine . One of the reasons of the recurring urinary infections might be due to sexual intercourse
or strenuous activities..
I appreciate any help with the following questions.(1) What typ.pdf
Hypothesize a set of at least three arguments as to why innovation is important to the
competitiveness of nations, companies, leaders, and individual employees.
Solution
One of the main reasons why innovation is important is the fact that it creates development on
nations and business, it opens opportunities to new projects that on the past have not being taken
in consideration.
Innovation sometimes is the only way an organization has to stay alive on business because of
the hard and globalize competition, it creates the unique value for your brand our country, and it
is an added value. For example recently Nicaragua offer new investors that they will not need to
pay taxes for 10 years as long as they start a business there and offer job to the people in
Nicaragua.
Innovation makes the organization better every day, it is a constant improvement on the
processes on a positive manner, and this increases creativity. You don’t need to create something
new that no one has ever think you can take a product and add it qualities that your competition
does not have and making your target market to prefer your brand..
How do the IPv6 autoconfiguration and numbering features work What .pdf
How can you set your organization apart from others in the marketplace, thus attracting the best
candidates?(the class is HR management)
Solution
To attract best candidates, you can have robust hiring / recruiting practices in the HR function.
Some points to keep in mind while drafting the policies:
1: Demonstrate the big critical problem to them that they’re going to crack
2: Become personal fast: Present hopefuls to the team initially in the process
3: Build a long-term connection with the society
4: Involve the community in answering a problem
5: Take your company philosophy on a campus excursion
6: Be careful about staging too many benefits
7: Be genuine; don’t fake it
8: Offer training and opportunities
9: Ask prospective employees what they want
10: Lower the ‘entry level’ expectations and offer training for those positions
11: Lure with “sexy” tech and the assurance of training
12: Make sure your recruiters are tech confident
13: Be enthusiastic to hire candidates with 80 percent of the prerequisites
14: Treat your employees well so they can be your word-of-mouth engaging influence
15: Write a job description with genuine expectations
16: Be honest. No one enjoys a bait and switch
17: Communicate during the hiring activity
18: Vet your recruiters’ methods to discover aspirants
19: Be disability welcoming
20: Propose market value even if they’ll consent lesser salary
21: Improve retention by decreasing employing requirements and increase your engagement
brand
22: Hire the best talent to get the entire group
23: Hang out in open-source societies
24: Keenly manage people during downsizings
25: Encourage ‘pay for performance’ instead of hourly wages
26: Incentives, perks, benefits
27: Sell your organization as a intelligent asset
28: Endorse your company’s innovative culture
29: Encourage employees to continue past employer contacts so you can obtain from rivals
30: Forget keyword matches and pay attention on candidates for the long tenure.
Government Audits & Fraud Reporting For discussion During an audit .pdf
Global businesses that have extended their public relations strategies worldwide should:
Multiple Choice Ignore local customs and emerging issues in the locations where they operate
Assume that there is no anti-American sentiment in the countries where they have operations.
Ensure that communications with local media take place in the area\'s native language Never
decentralize their global public relations programs to ensure the company has \"one voice.
Solution
Answer is C ensure that........... The reason is native language gives brand local touch and helps
people to understand what they say. This will increase demand. Other options are totally wrong.
Given L1 and Prove a llb are supplementary. 3 2 Proof It is given .pdf
Give some examples of absolute and qualified privilege with regard to invasion of privacy and
defamation as they may occur in the workplace. Explain your answer and please conduct
research to support your findings, citing at least one scholarly resource in APA format.
Solution
Absolute privilege is a complete defence to an action for defamation in English law. If the
defence of absolute privilege applies it is irrelevant that a defendant has acted with malice, knew
information was false or acted solely to damage the reputation of the plaintiff.
The defense of qualified privilege permits persons in positions of authority or trust to make
statements or relay or report statements that would be considered slander and libel if made by
anyone else..
Explain viewing the nonprofit organization as an economic entity and.pdf
Explain viewing the nonprofit organization as an economic entity and, as such, subject to
economic factors
Solution
The objective of non profit organization is to serve the society with respect to social services,
health care, education, employment and culture with no profit motive. The NPO’s are majorly
expected to work for poor people with the intention of upliftment of a nation.
But if we look at closely, the NPOs are formed to produce a good or service with scarce
resources. The NPOs as a firm needs to establish some level of efficiency in managing the
resources committed to maximize social benefits and it needs to raise its own resources, manage
them effectively, and add value to a service which justifies the investment, efforts and support
from government in terms of tax exemption..
Explain the concept and cause of freezing point depression. Elaborat.pdf
Explain the concept and cause of freezing point depression. Elaborate on how it is related to
colligative properties and melting points.
Solution
Freezing point depression:
It is the process of adding solute substance in solvent to decrease freezing point. Sodium
chloride-water solution decreases their temperature below the freezing point. This phenomenon
is practically used in the radiator fluid. Ethylene glycol-water mixture is useful for a radiator in
cold season.
Formula For Depression Of Freezing Point:
TF = KF · b · i
Where,
TF is freezing point depression (TF (pure solvent) - T F(solution))
KF is the cryoscopic constant.
b is the molality.
i is the vant’t Hoff factor.
The freezing point depression can also be explained in terms of vapor pressure. Adding solute to
a solvent will essentially dilute the solvent molecules, and according to Raoult’s law, this leads
to a decrease in vapor pressure. Considering the fact that the vapor pressure of the solid and
liquid forms must be the same at freezing point, because otherwise the system would not be at
equilibrium, the lowering of the vapor pressure leads to the lowering of the temperature at which
the vapor pressures of the liquid and frozen forms of the solution will be equal.
What causes freezing point depression?
When a solute and solvent are mixed, there is increase in entropy and the liquid finds it
thermodynamically more preferable to stay in the liquid state due to more randomness. Hence
the added solute increases the ‘desire’ for the solvent to stay in the liquid state thus decreasing
the freezing point.
What is the reason why freezing point depression is a colligative property
Freezing point depression is a colligative property observed in solutions that results from the
introduction of solute molecules to a solvent. The freezing points of solutions are all lower than
that of the pure solvent and is directly proportional to the molality of the solute.(so its a
colligative property because it depends on number of moles of solute present)
Colligative properties are properties of solutions that depend on the number of molecules in a
given volume of solvent and not on the properties (e.g. size or mass) of the molecules. Boiling
point and melting point (freezing point) depend only on the the number of particles dissolved in
the solution, not the identity of the particles..
Company management has asked that you compare the OSSTMM and the PTE.pdf
Company management has asked that you compare the OSSTMM and the PTES in order to
determine which methodology to select for internal testing. Compare these two methodologies
and explain in detail about both.
Solution
Penetration testing has several methodologies-OSSTMM and PTES is one of them.
Basic comparison between OSSTMM and PTES are as follows -
Penetration testing methodoogy
Penetration Testing Execution Standard (PTES) :There are 7 phases which is used to define
PTES for penetration testing.
1)Pre-engagement Interactions
2)Intelligence Gathering
3)Threat Modeling
4)Vulnerability Analysis
5)Exploitation
6)Post Exploitation
7)Reporting
On the other hand,OSSTMM is used for performing penetration testing and to obtain security
metrics.The OSSTMM provides transparency to those who have inadequate security
configurations and policies.The OSSTMM includes the entire risk assessment process starting
from requirement analysis to report creation.
The six areas covered by OSSTMM are -
1)Information security
2)Process security
3)Internet technology security
4)Communications security
5)Wireless security
6)Physical securityOSSTMMPTESMore TheoreticalTechnology orientedSecurity assessment
methodology
Penetration testing methodoogyMetrics basedExtended analysis of all stages.
As people engage in more international travel and become more famili.pdf
As people engage in more international travel and become more familiar with other countries,
will cultural difference decline as a roadblock to international understanding or will the continue
to be a major barrier? Defend your answer. Why are multinational enterprises getting involved in
corporate social responsibility and sustainable business practices? Are they displaying a sense of
social responsibility or it is merely a matter of good business, or both?
Solution
Multinational and cross-cultural teams are likewise becoming ever more common, meaning
businesses can benefit from an increasingly diverse knowledge base and new, insightful
approaches to business problems. However, along with the benefits of insight and expertise,
global organizations also face potential stumbling blocks when it comes to culture and
international business.
While there are a number of ways to define culture, put simply it is a set of common and
accepted norms shared by a society. But in an international business context, what is common
and accepted for a professional from one country, could be very different for a colleague from
overseas. Recognizing and understanding how culture affectsinternational business in three core
areas: communication, etiquette, and organizational hierarchy can help you to avoid
misunderstandings with colleagues and clients from abroad and excel in a globalized business
environment.
1. Communication
Effective communication is essential to the success of any business venture, but it is particularly
critical when there is a real risk of your message getting “lost in translation.” In many
international companies, English is the de facto language of business. But more than just the
language you speak, it’s how you convey your message that’s important. For instance, while the
Finns may value directness and brevity, professionals from India can be more indirect and
nuanced in their communication. Moreover, while fluent English might give you a professional
boost globally, understanding the importance of subtle non-verbal communication between
cultures can be equally crucial in international business.
What might be commonplace in your culture — be it a firm handshake, making direct eye
contact, or kiss on the cheek — could be unusual or even offensive to a foreign colleague or
client. Where possible, do your research in advance of professional interactions with individuals
from a different culture. Remember to be perceptive to body language, and when in doubt, ask.
While navigating cross-cultural communication can be a challenge, approaching cultural
differences with sensitivity, openness, and curiosity can help to put everyone at ease.
2. Workplace etiquette
Different approaches to professional communication are just one of the innumerable differences
in workplace norms from around the world. CT Business Travel has put together a useful
infographic for a quick reference of cultural differences in business etiquette globally.
For instance, the.
Consider a l-D elastic bar problem defined on [0, 4]. The domain .pdf
Consider a l-D elastic bar problem defined on [0, 4]. The domain is devided into 4 linear 2-node
elements as follows. Write down element shape functions N (x) for each element (e = 1, 2, 3),
Give a sketch of the shape functions N_i(x) = (f = 1, 2, 3, 4, 5) on the entire domain [0, 4].
Given the nodal displacement matrix d = [4 8 6 2]^T, find the displacement solutions at x = 1.5
and x = 3.6. Given the nodal displacement matrix d = [4 8 6 2]^T, find the strain (defined as
du/dx) solutions at x = 0.4 and x = 2.3.
Solution
Using the nodal-based smoothing operation, the strains to be used in Equation (23) is assumedto
be the
smoothed
strain for node
k
dened by
e
k
e
(
x
k
)
=
k
e
(
x
)
W
(
x
x
k
)
d
where
W
=
W
W
W
is a diagonal matrix of smoothing function
W
. For simplicity, the smoothingfunction is taken as
W
(
x
x
k
)
=
1
/
A
k
,
x
k
0
,
x
/
k
where
A
k
=
k
d
is the area of smoothing domain for node
k
.Substituting into Equation and integrating by parts, the smoothed strain canbe calculated using
e
k
=
1
A
k
k
e
(
x
)
d
=
1
A
k
k
L
n
u
(
x
)
d
=
e
k
(
u
)
where
k
is the boundary of the smoothing domain for node
k
, and
L
n
is the matrix of the outwardnormal vector on
k
. Equation states the fact that the assumed strain
e
k
is a function of theassumed displacement
u
.Substituting Equation into Equation , the smoothed strain can be expressed in thefollowing
matrix form of nodal displacements:
e
k
=
i
N
in
B
i
(
x
k
)
d
i
where
N
in
is the number of nodes in the inuence domain of node
k
(including node
k
). Whenlinear shape functions are used, it is the number of nodes that is directly connected to
node
k
inthe triangular mesh (see Figure 1). In Equation , the
B
i
(
x
k
)
is termed as the
smoothed
strainmatrix that is calculated using
B
i
(
x
k
)
=
b
ix
(
x
k
)
00
b
iy
(
x
k
)
b
iy
(
x
k
)
b
ix
(
x
k
)
Using the Gauss integration along each segment of boundary
k
, we have
b
il
=
1
A
k N
s
m
=
1
N
g
n
=
1
w
n
i
(
x
mn
)
n
l
(
x
m
)
(
l
=
x
,
y
)
(30)where
N
s
is the number of segments of the boundary
k
,
N
g
is the number of Gauss points usedin each segment,
w
n
is the corresponding weight number of Gauss integration scheme, and
n
l
isthe unit outward normal corresponding to each segment on the smoothing domain boundary. In
theLC-PIM using linear shape functions,
n
g
=
1 is used. The entries in sub-matrices of the stiffnessmatrix
K
in Equation are then expressed as
K
ij
=
N
k
=
1
K
ij
(
k
where the summation means an assembly process as we practice in the FEM, and
K
ij
(
k
)
is thestiffness matrix associated with node
k
that is computed using
K
ij
(
k
)
=
k
B
T
i
D
B
j
d
=
B
T
i
D
B
j
A
k
The entries (in sub-vectors of nodal forces) of the force vector
f
in Equation can be simplyexpressed as
f
i
=
k
N
in
f
i
(
k
)
The above integration is also performed by a summation of integrals over smoothing
domains;hence,
f
i
is an assembly of nodal force vectors at the surrounding nodes of node
k
:
f
i
(
k
)
=
t
(
k
)
U
i
ˆ
t
d
+
(
k
)
U
i
b
d
Note again th.
Click to add title 2. Explain the nature of the social structurecast.pdf
Click to add title 2. Explain the nature of the social structure/caste system, creolization and new
cultures, and the black family system that developed under slavery in the Americas Explain the
ways in which slaves were treated and the nature of the laws that were developed to regulate the
treatment of slaves
Solution
Caste system is one major demon the people used to face during the 18, 19 centuries where they
used to treat the lower caste people as their slaves where the black family system is one form of
slavery which used to prevail in America.
Slavery not only made family life of those people miserable but also impossible they are not
treated as humans but as commodities. They are never given any legal rights. If the people of a
family in ageneration work for one rich family the next geneartion and generations are
considered as slaves. Slaves could never marry legally, should not have any properties, women
were exploited, children were also slaves. They are used as unpaid labour to generate wealth.
The Act of 1807, 13th ammendment of US constitution are major laws which are made to
abolish slavery.
Based on the following data, what is the working capital Accounts p.pdf
Based on the following data, what is the working capital? Accounts payable $ 30,000 Accounts
receivable 60,000 Accrued liabilities 5,000 Cash 40,000 Fixed assets 670,000 Intangible assets
50,000 Inventory 69,000 Long-term investments 80,000 Long-term liabilities 100,000 Prepaid
expenses 1,000 Temporary investments 30,000 a. $130,000 b. $95,000 c. $30,000 d. $165,000
Solution
Total current assets=Accounts receivable+Cash+Inventory+Prepaid expenses+Temporary
investments
=(60000+40000+69000+1000+30000)=$200,000
Total current liabilities=Accounts payable+Accrued liabilities
=(30000+5000)=$35000
Working capital=Total current assets-Total current liabilities
=(200,000-35000)
=$165000..
who are the people that steal cashSolutionPeople that steal c.pdf
Which of the following wireless standards uses direct sequence spread spectrum (DSSS) by
default?
A. Bluetooth
B. 802.11a
C. 802.11b
D. 802.11ac
Solution
C. 802.11b uses direct sequence spread spectrum (DSSS) technology. DSSS is a transmission
method that transmits the data along with a chipping bit to increase the signal’s resistance to
interference. Answer A is incorrect because Bluetooth uses frequency-hopping spread spectrum.
Answer B is incorrect because 802.11a uses orthogonal frequency-division multiplexing. Answer
D is incorrect because 802.11ac uses MIMO-OFDM.
Why is nucleotide synthesis an important pathway for medical interve.pdf
Why do you need to use aprotic solvents (THF, diethyl ether) in a Grignard Reaction?
Solution
RMgX is general representation of grignard reagent it exist as R- and MgX+
R- is very strong base.
if we use any protic solvents like H2O and ROH
R- abstracts proton from solvent becomes neutral
R- + H2O -------------> RH
so there will be reaction between grignard reagent and protoc solvent.
so we must use aprotic solvent to avoid above mentioned problem..
Why do some argue that the Fed made the Great Depression worse S.pdf
who are the people that steal cash?
Solution
People that steal cash are as follows:
Thief - Someone who steals cash from person\'s bank account, house or anywhere else.
Burglar - Someone who enters a building illegally in order to steal things
Pickpocket - Someone wh steals money from people’s pockets and bags
Mugger - Someone who attacks people in public places and steals their money
Robber - Someone who take people money by giving threats (example - bank robbery).
Which of the following is a characteristic of TQMI. A focus on th.pdf
Which has more power, business or society? In a major confrontation, which will win? Explain
with an example.
Solution
I think society has more power than business. Therefore, in a major confrontation, the society
will win. Businesses cannot survive in a conflict with the society. Businesses survive as long as
they fulfill the needs of the society. Therefore, society is primary and businesses are secondary.
Society comes with people, systems, traditions, emotions, ethics, etc. Businesses must adapt to
the realities of the society and businesses cannot survive without being in a very cohesive and
integrated relationship with the society. It is the resources, people, etc. of the society which
makes the existence of businesses possible.
For example, would Walmart be able to survive without hiring workers from the society, or
sourcing materials from different stakeholders of the society, or using infrastructure, the judicial
system, political system, payment system, education, system, or selling its goods and services to
the society? Walmart will collapse in a day without active support from various stakeholders of
the society. In a major confrontation, Walmart will cease to exist and the society will win..
Which of the following wireless standards uses direct sequence sprea.pdf
Which of the following statements is true for ideal gases, but is not always true for real gases?
Choose all that apply There are no attractive forces between molecules. Pressure is caused by
molecule-wall collisions Replacing one gas by another under the same conditions, has no effect
on pressure. Molecules are in constant random motion. Submit Answer Retry Entire Group 2
more group attempts remaining
Solution
1. It is true that in real gases the attractive forces cannot be neglected.
2.In the real gas, the compressibility factor is not equal to 1. This implies that the molecule-wall
collisions are in fact carry a prominent significance in projecting Pressure.
3. In ideal gases, the particle size is assumed to be constant and the interatomic spacing is
considered to be almost negligible but it\'s not true in case of real gases..
A 0.28 mol sample of a weak acid with an unknown pKa was combined wi.pdf
A 0.28 mol sample of a weak acid with an unknown pKa was combined with 12.0 mL of 3.10M
KOH and the resulting solution was diluted to 1.500 L. The measured pH of the solution was
3.95.
Solution
moles of KOH=3.1*0.012
=0.0372
so,
pH=pKa+log(salt/acid)
or 3.95=x+log(0.0372/(0.28-0.0372))
or x=4.76471.
What was the first primate to have Y-5 molar pattern GibbonsSol.pdf
What special problems does ECB? confront? A. The ECB must function with a highly
centralized structure. B. The ECB is required to employ separate interest rate policies for each of
the member nations. C. The ECB must coordinate the use of multiple currencies. D. The
decentralized organization of the ECB with the governors of the national central banks holding a
majority of the? votes, instead of the executive? board, may make it harder to achieve a
consensus during a crisis.
Solution
D) The decentralized organization of the ECB with the governors of the national central banks
holding a majority of the votes, instead of the executive board, may make it harder to achieve a
consensus during a crisis.
(The ECB is the central bank for the euro while managing monetary policy of the Eurozone,
consisting of 19 EU member states).
What special problems does ECB confront A. The ECB must function w.pdf
What is Walmarts vision and execution of marketing strategy that set its direction?
Solution
Walmart\'s Vision statement:
\"To be the best retailer in the hearts and minds of consumers and employees\"
to achieve this vision Walmart follows various marketing strategies. Some them are as below:
In Intensive growth strategies Walmart uses three strategies i.e market penetration, market
development, product development
For market penetration Walmart offer discounts, promotions and special packages. For market
development Walmart build new stores in overseas market to gain consumers there, product
development investment are not incurred by Walmart but it\'s focus is on offering new products
to the market.
In Generic strategy which is based on Porter\'s model Walmart focus on maintaining low prices
for this they do cost reduction in Operations, like using Information technology to maximize
operational efficiency or minimizing wages.
Walmart follows it\'s vision by keeping it\'s cost and prices low which is based on cost leadership
generic strategy. For cost leadership Walmart expands and achieve better economies of scale..
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could you implement this function please, im having issues with it..pdf
- 1. 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&);
- 2. 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;
- 3. }; 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;
- 4. newNode = new ListNode; newNode->setDatum(item); newNode->setNext(NULL); if (head == NULL) { head = newNode; } else { while (cur->getNext() != NULL) { cur = cur->getNext(); } cur->setNext(newNode); } } void LinkedList::makeList(const ListNode::value_type num[], const size_t& count) { int i = 0; while ( i < count) { insertItem(num[i++]); } } void LinkedList::deleteList() { ListNode *tmp = head; while (tmp != NULL) { head = head->getNext(); delete tmp; tmp = head; } } std::ostream& operator<<(std::ostream& os, const LinkedList &srcList) { //Set a current-pointer to the "head".
- 5. ListNode* cursor = srcList.head; //While current-pointer is not NULL while (cursor != NULL) { //Print the data member ("datum") of the current node os << "->[" << cursor->getDatum() << "]"; //Set the current-pointer to the "next" node in the list. cursor = cursor->getNext(); } //Print out a basic termination symbol std::cout << "--X" << std::endl; return os; } --------------------------------------- //main.cpp #include #include "linkedlist.h" using namespace std; int main() { LinkedList list1; //Test of adding items out of order list1.insertItem(5); list1.insertItem(20); list1.insertItem(10); cout << "After adding 5,20,10 to list ,List contains" << endl; cout << list1 << endl; //Test of deleting entire list list1.deleteList(); cout << "After deleteing list,List contains" << endl; cout << list1 << endl; //Add items again in same order as before cout << "After adding 5,20,10 to list ,List contains" << endl; list1.insertItem(5); list1.insertItem(20); list1.insertItem(10); cout << list1 << endl;
- 6. //Now replace list with a new one in a specific order int pow2[] = { 1, 2, 4, 8, 16, 32, 16, 8, 4, 2, 1 }; list1.makeList(pow2, sizeof(pow2) / sizeof(int)); cout << "After calling makeList function" << endl; cout << list1 << endl; //Returning a non-zero number, if not 3, then we know it seg-faulted return 3; } ----------------------------------------------------------------------------------------------------------- After adding 5,20,10 to list ,List contains ->[5]->[20]->[10]--X After deleteing list,List contains --X After adding 5,20,10 to list ,List contains ->[5]->[20]->[10]--X After calling makeList function ->[5]->[20]->[10]->[1]->[2]->[4]->[8]->[16]->[32]->[16]->[8]->[4]->[2]->[1]--X