Use C++class Node{public Node ( int = 0 ); constru.pdf
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Use C++ class Node{ public: Node ( int = 0 ); // constructor with default value for // info field int info; // data Node * next; // pointer to next node in the list }; and put the constructor in Node.cpp // Constructor Node::Node ( int data ) { info = data; next = nullptr; } Write the following using the Node class. A function that copies a linked list pointed to by ptr and returns a pointer to the copied list. Please note: the copied list should be independent of the original list; the original list can be destroyed and the copy should still remain. const Node * copy (Node * ptr); Solution Need to perform \'Deep copy\' here : const Node * copy (Node * ptr) { //Create new list to hold copied items Node* newNode = new Node; //Return if empty list if (ptr->next == 0) { return newNode; } while(ptr !=NULL) { newNode->info = ptr->info; newNode->next = ptr->next; ptr = ptr ->next; } return newNode; }.
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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.
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.
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.
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.
Need Help!! C++ #include-iostream- #include-linkedlist-h- using namesp.pdf
Need Help!! C++
#include<iostream>
#include"linkedlist.h"
using namespace std;
/**
* @brief Destructor to destroy all nodes and release memory
*/
LinkedList::~LinkedList() {
//TODO: Add code here. Make sure memory is released properly.
}
/**
* @brief Purpose: Checks if the list is empty
* @return true if the list is empty, false otherwise
*/
bool LinkedList::isEmpty() const {
// TODO: Add code here
return count == 0;
}
/**
* @brief Get the number of nodes in the list
* @return int The number of nodes in the list
*/
int LinkedList::length() const{
//TODO: Add code here
}
/**
* @brief Convert the list to a string
*
*/
string LinkedList::toString() {
string str = "[";
Node *ptr = front;
if (ptr != nullptr) {
// Head node is not preceded by separator
str += to_string(ptr->val);
ptr = ptr->next;
}
while (ptr != nullptr) {
str += ", " + to_string(ptr->val);
ptr = ptr->next;
}
str += "]";
return str;
}
/**
* @brief Displays the contents of the list
*/
void LinkedList::displayAll() {
cout << toString() << endl;
}
//TODO: Add comments
void LinkedList::addRear(T val) {
// TODO: Add code here
// consider the two cases of whether the list was empty
}
//TODO: Add comments
void LinkedList::addFront(T val) {
// TODO: Add code here
// consider the two cases of whether the list was empty
}
//TODO: Add comments
bool LinkedList::deleteFront(T &OldNum) {
// TODO: Add code here
// consider if the list was empty and return false if the list is empty
// consider the special case of deleting the only node in the list
}
//TODO: Add comments
bool LinkedList::deleteRear(T &OldNum) {
// TODO: Add code here
// consider if the list was empty and return false if the list is empty
// consider the special case of deleting the only node in the list
}
/* --- harder ones for test 2 and 3 -- */
/**
* @brief Delete a node at a given position from the list. The
* node at position pos is deleted and the value of the deleted node is returned in val.
* The valid range of pos is 1 to count. pos = 1 is the first node, and pos = count is the last node.
* @param pos: position of the node to be deleted
* @param val: it is set to the value of the node to be deleted
* @return true: if the node was deleted successfully
* @return false: if the node was not deleted successfully because the position was out of range
*/
bool LinkedList::deleteAt(int pos, T &val) {
//TODO: Add code here
// check if the pos is valid first, then move the ptr to the rigth positon
// consider the special case of deleting the first node and the last node
// Do not forget to set value.
}
/**
* @brief Insert a value at a specified position in the list. The valid pos is in the range of 1 to
count+1.
* The value will be inserted before the node at the specified position. if pos = 1, the value will be
inserted
* at the front of the list. if pos = count+1, the value will be inserted at the rear of the list.
* @param pos: position to insert the value at.
* @param val: value to insert.
* @return true: if the value was inserte.
In C++ I need help with this method that Im trying to write fillLi.pdf
In C++ I need help with this method that I'm trying to write: fillListFromFile( ) : void -- this
method will add elements to the list from a file called linkedListData1.txt HINT: this is where
you will keep track of the number of elements in the list
I have all of the other methods, but I'm stuck on this one and I've added it as the last method in
the application. The code I have so far is below.
.cpp file:
.h file:
main:
#include
#include
#include
#include "Node.h"
using namespace std;
void displayList(Node*);
void insertFront(Node** head, int value);
void insertBack(Node** ptrToHead, int value);
void insertAfter(Node* previous, int value);
void deleteAtN(Node** ptrToHead, int n);
void deleteNode(Node** ptrToHead, int searchVal);
void clearList(Node**);
bool searchVal(Node** head, int value);
int countFrequencyOf(Node** ptrTohead, int value);
int size(Node** ptrTohead);
void displayFromFrontToN(Node** ptrToHead, int n);
void displayFromNToEnd(Node** ptrToHead, int n);
void insertInOrder(Node** head, Node* ptrToHead);
void displayInRows(Node** ptrToHead);
int main()
{
Node* node1 = new Node();
node1->setValue(10);
Node* node2 = new Node();
node2->setValue(20);
Node* node3 = new Node();
node3->setValue(30);
node1->next = node2;
node2->next = node3;
Node* head = node1;
displayList(head);
insertFront(&head, 5);
displayList(head);
insertBack(&head, 50);
displayList(head);
insertBack(&head, 60);
displayList(head);
insertAfter(node2, 25);
displayList(head);
deleteAtN(&head, 2);
displayList(head);
deleteNode(&head, 50);
displayList(head);
clearList(&head);
displayList(head);
insertFront(&head, 5);
insertFront(&head, 10);
insertFront(&head, 20);
insertBack(&head, 50);
searchVal(&head, 50);
displayList(head);
countFrequencyOf(&head, 25);
displayList(head);
insertBack(&head, 25);
insertFront(&head, 25);
countFrequencyOf(&head, 25);
displayList(head);
size(&head);
cout << "add one item to list" << endl;
insertBack(&head, 25);
size(&head);
displayList(head);
cout << "Display From Front To N" << endl;
displayFromFrontToN(&head, 4);
cout << "Display From N To End" << endl;
displayFromNToEnd(&head, 3);
cout << "Display In Rows" << endl;
displayInRows(&head);
insertFront(&head, 12);
insertFront(&head, 17);
insertFront(&head, 22);
insertBack(&head, 58);
cout << "Display In Rows" << endl;
displayInRows(&head);//call displayInRows()
return 0;
}//end main()
void displayList(Node* ptr) {
if (ptr == nullptr)
cout << "empty list" << endl;
else {
while (ptr != nullptr) {
cout << ptr->getValue() << endl;
ptr = ptr->next;
}
}
cout << "---------------------------" << endl;
}//end displayList()
void insertFront(Node** head, int value) {
Node* newNode = new Node();
newNode->setValue(value);
//cout<< newNode->getValue() << endl;
newNode->next = *head;
*head = newNode;
}
void insertBack(Node** ptrToHead, int value) {
Node* newNode = new Node();
newNode->setValue(value);
newNode->next = nullptr;
if (*ptrToHead == nullptr)
{
*ptrToHead = newNode;
}//end if
else
{
Node* last .
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.
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.
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.
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.
Need Help!! C++ #include-iostream- #include-linkedlist-h- using namesp.pdf
Need Help!! C++
#include<iostream>
#include"linkedlist.h"
using namespace std;
/**
* @brief Destructor to destroy all nodes and release memory
*/
LinkedList::~LinkedList() {
//TODO: Add code here. Make sure memory is released properly.
}
/**
* @brief Purpose: Checks if the list is empty
* @return true if the list is empty, false otherwise
*/
bool LinkedList::isEmpty() const {
// TODO: Add code here
return count == 0;
}
/**
* @brief Get the number of nodes in the list
* @return int The number of nodes in the list
*/
int LinkedList::length() const{
//TODO: Add code here
}
/**
* @brief Convert the list to a string
*
*/
string LinkedList::toString() {
string str = "[";
Node *ptr = front;
if (ptr != nullptr) {
// Head node is not preceded by separator
str += to_string(ptr->val);
ptr = ptr->next;
}
while (ptr != nullptr) {
str += ", " + to_string(ptr->val);
ptr = ptr->next;
}
str += "]";
return str;
}
/**
* @brief Displays the contents of the list
*/
void LinkedList::displayAll() {
cout << toString() << endl;
}
//TODO: Add comments
void LinkedList::addRear(T val) {
// TODO: Add code here
// consider the two cases of whether the list was empty
}
//TODO: Add comments
void LinkedList::addFront(T val) {
// TODO: Add code here
// consider the two cases of whether the list was empty
}
//TODO: Add comments
bool LinkedList::deleteFront(T &OldNum) {
// TODO: Add code here
// consider if the list was empty and return false if the list is empty
// consider the special case of deleting the only node in the list
}
//TODO: Add comments
bool LinkedList::deleteRear(T &OldNum) {
// TODO: Add code here
// consider if the list was empty and return false if the list is empty
// consider the special case of deleting the only node in the list
}
/* --- harder ones for test 2 and 3 -- */
/**
* @brief Delete a node at a given position from the list. The
* node at position pos is deleted and the value of the deleted node is returned in val.
* The valid range of pos is 1 to count. pos = 1 is the first node, and pos = count is the last node.
* @param pos: position of the node to be deleted
* @param val: it is set to the value of the node to be deleted
* @return true: if the node was deleted successfully
* @return false: if the node was not deleted successfully because the position was out of range
*/
bool LinkedList::deleteAt(int pos, T &val) {
//TODO: Add code here
// check if the pos is valid first, then move the ptr to the rigth positon
// consider the special case of deleting the first node and the last node
// Do not forget to set value.
}
/**
* @brief Insert a value at a specified position in the list. The valid pos is in the range of 1 to
count+1.
* The value will be inserted before the node at the specified position. if pos = 1, the value will be
inserted
* at the front of the list. if pos = count+1, the value will be inserted at the rear of the list.
* @param pos: position to insert the value at.
* @param val: value to insert.
* @return true: if the value was inserte.
In C++ I need help with this method that Im trying to write fillLi.pdf
In C++ I need help with this method that I'm trying to write: fillListFromFile( ) : void -- this
method will add elements to the list from a file called linkedListData1.txt HINT: this is where
you will keep track of the number of elements in the list
I have all of the other methods, but I'm stuck on this one and I've added it as the last method in
the application. The code I have so far is below.
.cpp file:
.h file:
main:
#include
#include
#include
#include "Node.h"
using namespace std;
void displayList(Node*);
void insertFront(Node** head, int value);
void insertBack(Node** ptrToHead, int value);
void insertAfter(Node* previous, int value);
void deleteAtN(Node** ptrToHead, int n);
void deleteNode(Node** ptrToHead, int searchVal);
void clearList(Node**);
bool searchVal(Node** head, int value);
int countFrequencyOf(Node** ptrTohead, int value);
int size(Node** ptrTohead);
void displayFromFrontToN(Node** ptrToHead, int n);
void displayFromNToEnd(Node** ptrToHead, int n);
void insertInOrder(Node** head, Node* ptrToHead);
void displayInRows(Node** ptrToHead);
int main()
{
Node* node1 = new Node();
node1->setValue(10);
Node* node2 = new Node();
node2->setValue(20);
Node* node3 = new Node();
node3->setValue(30);
node1->next = node2;
node2->next = node3;
Node* head = node1;
displayList(head);
insertFront(&head, 5);
displayList(head);
insertBack(&head, 50);
displayList(head);
insertBack(&head, 60);
displayList(head);
insertAfter(node2, 25);
displayList(head);
deleteAtN(&head, 2);
displayList(head);
deleteNode(&head, 50);
displayList(head);
clearList(&head);
displayList(head);
insertFront(&head, 5);
insertFront(&head, 10);
insertFront(&head, 20);
insertBack(&head, 50);
searchVal(&head, 50);
displayList(head);
countFrequencyOf(&head, 25);
displayList(head);
insertBack(&head, 25);
insertFront(&head, 25);
countFrequencyOf(&head, 25);
displayList(head);
size(&head);
cout << "add one item to list" << endl;
insertBack(&head, 25);
size(&head);
displayList(head);
cout << "Display From Front To N" << endl;
displayFromFrontToN(&head, 4);
cout << "Display From N To End" << endl;
displayFromNToEnd(&head, 3);
cout << "Display In Rows" << endl;
displayInRows(&head);
insertFront(&head, 12);
insertFront(&head, 17);
insertFront(&head, 22);
insertBack(&head, 58);
cout << "Display In Rows" << endl;
displayInRows(&head);//call displayInRows()
return 0;
}//end main()
void displayList(Node* ptr) {
if (ptr == nullptr)
cout << "empty list" << endl;
else {
while (ptr != nullptr) {
cout << ptr->getValue() << endl;
ptr = ptr->next;
}
}
cout << "---------------------------" << endl;
}//end displayList()
void insertFront(Node** head, int value) {
Node* newNode = new Node();
newNode->setValue(value);
//cout<< newNode->getValue() << endl;
newNode->next = *head;
*head = newNode;
}
void insertBack(Node** ptrToHead, int value) {
Node* newNode = new Node();
newNode->setValue(value);
newNode->next = nullptr;
if (*ptrToHead == nullptr)
{
*ptrToHead = newNode;
}//end if
else
{
Node* last .
How do you stop infinite loop Because I believe that it is making a.pdf
How do you stop infinite loop? Because I believe that it is making an infinite circular list.
c++ code:
Here is the list class:
#ifndef LIN_J_LIST
#define LIN_J_LIST
typedef unsigned int uint;
#include
#include
using namespace std;
/**
* a simplified generic singly linked list class to illustrate C++ concepts
* @author Jerry Lin
* @version 2/17/17
*/
template< typename Object >
class List
{
private:
/**
* A class to store data and provide a link to the next node in the list
*/
class Node
{
public:
/**
* The constructor
* @param data the data to be stored in this node
*/
explicit Node( const Object & data )
: data{ data }, next{ nullptr } {}
Object data;
Node * next;
};
public:
/**
* The constructor for an empty list
*/
List()
: size{ 0 }, first{ nullptr }, last{ nullptr } {}
/**
* the copy constructor-creates and copy the list
*/
List( List && rhs ) = delete;
List( const List & rhs )
{
count = 0;
size = 0;
if(rhs.size != 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first;
size++;
while(current->next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op.
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode->next;
size++;
}
}
// you document and implement this method
}
/**
* the operator= method-copies the list
*/
List & operator=( List && rhs) = delete;
List & operator=( const List & rhs )
{
count = 0;
size = 0;
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
if(rhs.size!= 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first; //create a temporary to store
size++;
while(current -> next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode -> next;
size++;
}
}
return *this;
// you document and implement this method
}
/**
* accessor
* @return count
*/
int get_count() const
{
return count;
}
/**
* The destructor that gets rid of everything that\'s in the list and
* resets it to empty. If the list is already empty, do nothing.
*/
~List()
{
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
}
/**
* Put a new element onto the beginning of the list
* @param item the data the new element will contain
*/
void push_front( const Object& item )
{
Node * new_node = new Node( item );//basic op.
if(is_empty())
{
last = new_node;
}
else
{
new_node->next = first;
}
first = new_node;
size++;
/* you complete the rest */
}
/**
* Remove the element that\'s at the front of the list. Causes an
* assertion error if the list is empty.
*/
void pop_front()
{
assert( !is_empty() );
Node * temp = first;
if( first == last )
{
first = last = nullptr;
}
else
{
first = first->next;
}
delete temp;
size--;
}
/**
* Accessor to return the da.
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\");
}
.
I need help completing this C++ code with these requirements.instr.pdf
I need help completing this C++ code with these requirements.
instructions: IN C++ LANGUAGE PLEASE
Update the comments for each prototype by filling in the pre and post conditions.
Remember that pre conditions indicate what conditions must be met BEFORE the user calls the
function. The function will not work properly unless the pre conditions are met.
Post conditions indicate what will be the result of the function call. In other words, what
conditions will exist after the function is called.
See class notes from Thursday on pre and post conditions for some examples and further
explanation.
Code begins here:
#include //for NULL
class List
{
private:
struct Node
{
int data;
Node* next;
Node(int data): data(data), next(NULL){}
};
typedef struct Node* Nodeptr;
Nodeptr first;
Nodeptr last;
int size;
public:
/**Constructors and Destructors*/
List();
//Default constructor; initializes and empty list
//Postcondition:
~List();
//Destructor. Frees memory allocated to the list
//Postcondition:
/**Accessors*/
int getFirst();
//Returns the first element in the list
//Precondition:
int getLast();
//Returns the last element in the list
//Precondition:
bool isEmpty();
//Determines whether a list is empty.
int getSize();
//Returns the size of the list
/**Manipulation Procedures*/
void removeLast();
//Removes the value of the last element in the list
//Precondition:
//Postcondition:
void removeFirst();
//Removes the value of the first element in the list
//Precondition:
//Postcondition:
void insertLast(int data);
//Inserts a new element at the end of the list
//If the list is empty, the new element becomes both first and last
//Postcondition:
void insertFirst(int data);
//Inserts a new element at the start of the list
//If the list is empty, the new element becomes both first and last
//Postcondition:
/**Additional List Operations*/
void printList();
//Prints to the console the value of each element in the list sequentially
//and separated by a blank space
//Prints nothing if the list is empty
};
Solution
PROGRAM CODE:
#include
#include //for NULL
using namespace std;
class List
{
private:
struct Node
{
int data;
Node* next;
Node(int data): data(data), next(NULL){}
};
typedef struct Node* Nodeptr;
Nodeptr first;
Nodeptr last;
int size;
public:
/**Constructors and Destructors*/
List()
{
first = (Node*) malloc(sizeof(Node));
last = (Node*) malloc(sizeof(Node));
size = 0;
}
//Default constructor; initializes and empty list
//Postcondition:
~List()
{
delete first;
delete last;
}
//Destructor. Frees memory allocated to the list
//Postcondition:
/**Accessors*/
int getFirst()
{
return first->data;
}
//Returns the first element in the list
//Precondition:
int getLast()
{
return last->data;
}
//Returns the last element in the list
//Precondition:
bool isEmpty()
{
if(first == NULL)
return true;
else return false;
}
//Determines whether a list is empty.
int getSize()
{
return size;
}
//Returns the size of the list
/**Manipulation Procedures*/
void removeLast()
{
last = .
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.
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
.
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
Answer:
Note: LinkedList.cpp is written and driver program main.cpp is also given.
//ListInterface.h
#ifndef _LIST_INTERFACE
#define _LIST_INTERFACE
template
class ListInterface
{
public:
virtual bool isEmpty() const = 0;
virtual int getLength() const = 0;
virtual bool insert(int newPosition, const ItemType& newEntry) = 0;
virtual bool remove(int position) = 0;
virtual void clear() = 0;
virtual ItemType getEntry(int position) const = 0;
virtual void replace(int position, const ItemType& newEntry) = 0;
}; // end ListInterface
#endif
//Node.h
#ifndef NODE_
#define NODE_
template
class Node
{
private:
ItemType item; // A data item
Node* next; // Pointer to next node
public:
Node();
Node(const ItemType& anItem);
Node(const ItemType& anItem, Node* nextNodePtr);
void setItem(const ItemType& anItem);
void setNext(Node* nextNodePtr);
ItemType getItem() const ;
Node* getNext() const ;
}; // end Node
#endif
//Node.cpp
#include \"Node.h\"
template
Node::Node() : next(nullptr)
{
} // end default constructor
template
Node::Node(const ItemType& anItem) : item(anItem), next(nullptr)
{
} // end constructor
template
Node::Node(const ItemType& anItem, Node* nextNodePtr) :
item(anItem), next(nextNodePtr)
{
} // end constructor
template
void Node::setItem(const ItemType& anItem)
{
item = anItem;
} // end setItem
template
void Node::setNext(Node* nextNodePtr)
{
next = nextNodePtr;
} // end setNext
template
ItemType Node::getItem() const
{
return item;
} // end getItem
template
Node* Node::getNext() const
{
return next;
} // end getNext
//PrecondViolatedExcept.h
#ifndef PRECOND_VIOLATED_EXCEPT_
#define PRECOND_VIOLATED_EXCEPT_
#include
#include
class PrecondViolatedExcept : public std::logic_error
{
public:
PrecondViolatedExcept(const std::string& message = \"\");
}; // end PrecondViolatedExcept
#endif
//PrecondViolatedExcept.cpp
#include \"PrecondViolatedExcept.h\"
PrecondViolatedExcept::PrecondViolatedExcept(const std::string& message)
: std::logic_error(\"Precondition Violated Exception: \" + message)
{
} // end constructor
//LinkedList.h
#ifndef LINKED_LIST_
#define LINKED_LIST_
#include \"ListInterface.h\"
#include \"Node.h\"
#include
#include \"PrecondViolatedExcept.h\"
template
class LinkedList : public ListInterface
{
private:
Node* headPtr; // Pointer to first node in the chain;
// (contains the first entry in the list)
int itemCount; // Current count of list items
// Locates a specified node in this linked list.
Node* getNodeAt(int position) const;
public:
LinkedList();
LinkedList(const LinkedList& aList);
virtual ~LinkedList();
bool isEmpty() const;
int getLength() const;
bool insert(int newPosition, const ItemType& newEntry);
bool remove(int position);
void clear();
ItemType getEntry(int position) const;
void replace(int position, const ItemType& newEntry);
}; // end LinkedList
#endif
//LinkedList.cpp
#include \"LinkedList.h\"
#include
#include
#include
#include
//Default constructor
template
LinkedList::LinkedList() : headPtr(nullptr), itemCount(0)
{
.
C Homework Help
I am Andrew O. I am a C Homework Expert at programminghomeworkhelp.com. I hold a master’s in Programming from, the University of Southampton, UK. I have been helping students with their homework for the past 10 years. I solve homework related to C.
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You can also call on +1 678 648 4277 for any assistance with C Homework.
--INSTRUCTION- --It helps to first create if-then-else structure to fi.pdf
//INSTRUCTION:
//It helps to first create if-then-else structure to fill in later.
//- Make sure PURPOSE and PARAMETER comments are given in detail
// Add sufficient comments to your code body to describe what it does.
// - Make sure all if-then-else are commented describing which case it is
// - Make sure all local variables are described fully with their purposes
//Complete the TODO parts.
// ====================================================
//Complier: g++
//File type: linkedList.cpp implementation file
//=====================================================
#include<iostream>
#include"linkedlist.h"
using namespace std;
//cunstructor that initializes an empty list
LinkedList::LinkedList() {
front = nullptr ;
rear = nullptr ;
count = 0;
}
/**
* @brief Destructor to destroy all nodes and release memory
*/
LinkedList::~LinkedList() {
Node *p, *q;
p = front;
while (p != nullptr ) {
q = p->next;
delete p;
p = q;
}
}
/**
* @brief Purpose: Checks if the list is empty
* @return true if the list is empty, false otherwise
*/
bool LinkedList::isEmpty() const {
// TODO: Add code here
return count == 0;
}
/**
* @brief Get the number of nodes in the list
* @return int The number of nodes in the list
*/
int LinkedList::length() const {
return count;
}
/**
* @brief Convert the list to a string
*
*/
string LinkedList::toString() {
string str = "[";
Node *ptr = front;
if (ptr != nullptr ) {
// Head node is not preceded by separator
str += to_string(ptr->val);
ptr = ptr->next;
}
while (ptr != nullptr ) {
str += ", " + to_string(ptr->val);
ptr = ptr->next;
}
str += "]";
return str;
}
/**
* @brief Displays the contents of the list
*/
void LinkedList::displayAll() {
cout << toString() << endl;
}
//Add a node to the end of the list
void LinkedList::addRear(T val) {
Node *p = new Node;
p->val = val;
p->next = nullptr ;
if (isEmpty()) {
front = p;
}
else {
rear->next = p;
}
rear = p;
count++;
// consider the two cases of whether the list was empty
}
//Add a node to the front of the list
void LinkedList::addFront(T val) {
Node *p = new Node;
p->val = val;
p->next = front;
if (isEmpty()) {
rear = p;
}
front = p;
count++;
// consider the two cases of whether the list was empty
}
//Delete the first node of the list
bool LinkedList::deleteFront(T &OldNum) {
if (isEmpty()) {
return false ;
}
Node *p = front;
OldNum = p->val;
front = front->next;
if (front == nullptr ) {
rear = nullptr ;
}
delete p;
count--;
return true ;
// consider if the list was empty and return false if the list is empty
// consider the special case of deleting the only node in the list
}
//Delete the last node of the list
bool LinkedList::deleteRear(T &OldNum) {
if (isEmpty()) {
return false ;
}
Node *p = rear;
OldNum = p->val;
if (front == rear) {
front = rear = nullptr ;
}
else {
Node *q = front;
while (q->next != rear) {
q = q->next;
}
rear = q;
q->next = nullptr ;
}
delete p;
count--;
return true ;
// consider if the list was empty and return false if the list is empty
// consider the specia.
COMP2710: Software Construction - Linked list exercises
Goals:
• To learn how to use structures
• To learn how to use linked data structures
• To design and implement functions
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.
Program to insert in a sorted list #includestdio.h#include.pdf
* Program to insert in a sorted list */
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* function to insert a new_node in a list. Note that this
function expects a pointer to head_ref as this can modify the
head of the input linked list (similar to push())*/
void sortedInsert(struct node** head_ref, struct node* new_node)
{
struct node* current;
/* Special case for the head end */
if (*head_ref == NULL || (*head_ref)->data >= new_node->data)
{
new_node->next = *head_ref;
*head_ref = new_node;
}
else
{
/* Locate the node before the point of insertion */
current = *head_ref;
while (current->next!=NULL &&
current->next->data < new_node->data)
{
current = current->next;
}
new_node->next = current->next;
current->next = new_node;
}
}
/* BELOW FUNCTIONS ARE JUST UTILITY TO TEST sortedInsert */
/* A utility function to create a new node */
struct node *newNode(int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
new_node->next = NULL;
return new_node;
}
/* Function to print linked list */
void printList(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
printf(\"%d \", temp->data);
temp = temp->next;
}
}
/* Drier program to test count function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
struct node *new_node = newNode(5);
sortedInsert(&head, new_node);
new_node = newNode(10);
sortedInsert(&head, new_node);
new_node = newNode(7);
sortedInsert(&head, new_node);
new_node = newNode(3);
sortedInsert(&head, new_node);
new_node = newNode(1);
sortedInsert(&head, new_node);
new_node = newNode(9);
sortedInsert(&head, new_node);
printf(\"\ Created Linked List\ \");
printList(head);
return 0;
}
public class Listnode {
//*** fields ***
private Object data;
private Listnode next;
//*** methods ***
// 2 constructors
public Listnode(Object d) {
this(d, null);
}
public Listnode(Object d, Listnode n) {
data = d;
next = n;
}
// access to fields
public Object getData() {
return data;
}
public Listnode getNext() {
return next;
}
// modify fields
public void setData(Object ob) {
data = ob;
}
public void setNext(Listnode n) {
next = n;
}
}
Thsi below program print out all the varibles in SLL
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* 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) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int .
C++ problemPart 1 Recursive Print (40 pts)Please write the recu.pdf
C++ problem
Part 1: Recursive Print (40 pts)
Please write the recursive List reverse print function, whose iterative version we wrote in class.
Below are the function signatures for the functions you are going to need:
public:
/**Additional Operations*/
void print_reverse();
//Wrapper function that calls the reverse helper function to print a list in reverse
//prints nothing if the List is empty
private:
void reverse(Nodeptr node);
//Helper function for the public printReverse() function.
//Recursively prints the data in a List in reverse.
Why do we need the private helper function here?
Since we are going to be reversing our list node by node, in a recursive fashion, we want to pass
a one node at a time to our reverse function.
However, since our nodes are private, we cannot access them if we call the function inside of
main.
Add these function signatures to your List.h file along with your other function prototypes inside
the class definition.
Make sure that you place the reverse function inside the private portion of your List class
definition and the print_reverse function prototype to the public portion of your List class
definition.
Now, implement these two functions inside of List.h, under your section for additional
operations.
Important: Test each function carefully inside of your ListTest.cpp to make sure that it is
working properly.
Part 2: Adding an Index to Your List Nodes (20 pts)
Next, you will add the following functions to your List.h
/**Accessor Functions*/
int get_index();
//Indicates the index of the Node where the iterator is currently pointing
//Nodes are numbered from 1 to length of the list
//Pre: length != 0
//Pre: !off_end()
...
int List::get_index()
{
//Implement the function here
}
/**Manipulation Procedures*/
void scroll_to_index(int index);
//Moves the iterator to the node whose index is specified by the user
//Pre: length != 0
...
void scroll_to_index(int index)
{
//Implement function here
}
Part 3: Implementing Search as Part of Your List (40 pts)
Now, we are going to add two search functions to our List so that we can search for elements in
our List.
The first of these functions is going to be a simple linear search function.
You will need to add the following function prototype and function definition to your List.h:
/**Additional Operations*/
int linear_search(listitem item);
//Searchs the list, element by element, from the start of the List to the end of the List
//Returns the index of the element, if it is found in the List
//Returns -1 if the element is not in the List
//Pre: length != 0
...
int List::linear_search(listitem item)
{
//Implement the function here
}
You are also going to add a function to perform recursive binary search on your List.
You will need to add the following function prototype and function definition to your List.h:
int binary_search(int low, int high, listitem item);
//Recursively searchs the list by dividing the search space in half
//Returns the index of the element, if it is fo.
Write java program using linked list to get integer from user and.docx
Write java program using linked list to get integer from user and
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.
Inspect the class declaration for a doubly-linked list node in Node-h-.pdf
Inspect the class declaration for a doubly-linked list node in Node.h. Access Node.h by clicking
on the orange arrow next to main.cpp at the top of the coding window. The Node class has three
member variables: a double data value, a pointer to the next node, and a pointer to the previous
node. Each member variable is protected. So code outside of the class must use the provided
getter and setter member functions to get or set a member variable. Node.h is read only, since no
changes are required. Step 2: Implement the Insert() member function A class for a sorted,
doubly-linked list is declared in SortedNumberList.h. Implement the SortedNumberList class's
Insert() member function. The function must create a new node with the parameter value, then
insert the node into the proper sorted position in the linked list. Ex: Suppose a
SortedNumberList's current list is 23 47.25 86, then Insert(33.5) is called. A new node with data
value 33.5 is created and inserted between 23 and 47.25, thus preserving the list's sorted order
and yielding: 23 35.5 47.25 86 Step 3: Test in develop mode Code in main() takes a space-
separated list of numbers and inserts each into a SortedNumberList. The list is displayed after
each insertion. Ex: If input is 77 15 -42 63.5 then output is: List after inserting 77: 77 List after
inserting 15: 15 77 List after inserting -42: -42 15 77 List after inserting 63.5: -42 15 63.5 77 Try
various program inputs, ensuring that each outputs a sorted list. Step 4: Implement the Remove()
member function Implement the SortedNumberList class's Remove() member function. The
function takes a parameter for the number to be removed from the list. If the number does not
exist in the list, the list is not changed and false is returned. Otherwise, the first instance of the
number is removed from the list and true is returned. Uncomment the commented-out part in
main() that reads a second input line and removes numbers from the list. Test in develop mode to
ensure that insertion and removal both work properly, then submit code for grading. Ex: If input
is 84 72 19 61 19 84 then output is: List after inserting 84: 84 List after inserting 72: 72 84 List
after inserting 19: 19 72 84 List after inserting 61: 19 61 72 84 List after removing 19: 61 72 84
List after removing 84: 61 72
main.cpp
#include
#include
#include
#include "Node.h"
#include "SortedNumberList.h"
using namespace std; void PrintList(SortedNumberList& list); vector SpaceSplit(string source);
int main(int argc, char *argv[]) { // Read the line of input numbers string inputLine; getline(cin,
inputLine); // Split on space character vector terms = SpaceSplit(inputLine); // Insert each value
and show the sorted list's contents after each insertion SortedNumberList list; for (auto term :
terms) { double number = stod(term); cout << "List after inserting " << number << ": " << endl;
list.Insert(number); PrintList(list); } /* // Read the input line with numbers to remove getline(cin,
inputLi.
C++ please put everthing after you answer it- thanks Complete the stub.docx
C++ please put everthing after you answer it, thanks
Complete the stub file for the DoublyLinkedList implementation.
Please note: Any changes to the NodeType struct, not using the NodeType struct, or changing from template Type to using a specific data type will all be cause for deductions.
You are required to implement the methods:
insertFirst
insertLast
insertNode - (insert an item at a given index)
deleteNode - (remove a specific item)
The stub file contains the basic class format to implement the virtual methods from ListType. You should only need to complete the code for these four functions. Your code must maintain (or add) the node connections both forward and backward. In the double link list class these are called next and prev.
Your code must also maintain the count variable.
\} template Type ListType : : back ( ) const \{ assert(tail != NULL); //confirm tail exists return tail->data; //return the data of the tail node \} template void ListType >: :copyList (const ListType > \& otherList) \{ NodeType *newNode; //pointer to create a node NodeType current; //pointer to traverse the list if (head ! = NULL) //if the list is nonempty, make it empty destroyList(); if (otherList. head == NULL) //otherList is empty \{ head = NULL; tail = NULL; count = 0 ; \} else \{ current = otherList. head; // current points to the count = otherList . count; // list to be copied //copy the head node head = new NodeType < Type > ; / / create the node head->data = current > data; // copy the data head->next = NULL; // set the link field of tail = head; // the node to NULL // the next node //copy the remaining list while (current != NULL) \{ newNode = new NodeType < Type > ; // create a node newNode > data = current > data; // copy the data newNode->next = NULL; // set the link of // newNode to NULL tail->next = newNode; //attach newNode after tail tail = newNode; // make tail point to //the actual tail node current = current > next; 1/ make current point //to the next node \}/ /end while \}/lend else \}//end copylist
.
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
Which asymmetric cryptosystem is used for digital signaturesA. DE.pdf
Which asymmetric cryptosystem is used for digital signatures?
A. DES
B. SHA1
C. Diffie-Hellman
D. ECC
Solution
D. Elliptic curve cryptosystems (ECC) is an asymmetric cryptosystem created in the 1980s to
create and store digital signatures in a small amount of memory. Answer A is incorrect because
DES is a symmetric algorithm. Answer B is incorrect because SHA1 is a hashing algorithm.
Answer C is incorrect because Diffie-Hellman is used for key exchange..
Which of these theories fit into the contextual world-viewQuestio.pdf
Which of these theories fit into the contextual world-view?
Question 13 options:
Stage theories, like Piaget\'s.
Learning theories, like Skinner\'s.
Universal theories, like Bowlby\'s.
Systems theories, like Bronfenbrenner\'s.
Save
Stage theories, like Piaget\'s.
Learning theories, like Skinner\'s.
Universal theories, like Bowlby\'s.
Systems theories, like Bronfenbrenner\'s.
Solution
ANS: D. Systems theories, like Bronfenbrenner\'s.
Explanation:
Developmental psychology is mainly influenced by three major world views. They are 1.
Organismic, 2. Mechanistic and 3. contextual world views.
Contextual world views:
Development is the result of reciprocal social interactions between an active person and their
social partners from in a series of nested contexts.
Systems theories, like Bronfenbrenner\'s theory fit into the contextual world-view.
Wrong theories:
A. Stage theories, like Piaget\'s. (Organismic)
B. Learning theories, like Skinner\'s. (Mechanistic)
C. Universal theories, like Bowlby\'s. (Maturational).
More Related Content
Similar to Use C++class Node{public Node ( int = 0 ); constru.pdf
How do you stop infinite loop Because I believe that it is making a.pdf
How do you stop infinite loop? Because I believe that it is making an infinite circular list.
c++ code:
Here is the list class:
#ifndef LIN_J_LIST
#define LIN_J_LIST
typedef unsigned int uint;
#include
#include
using namespace std;
/**
* a simplified generic singly linked list class to illustrate C++ concepts
* @author Jerry Lin
* @version 2/17/17
*/
template< typename Object >
class List
{
private:
/**
* A class to store data and provide a link to the next node in the list
*/
class Node
{
public:
/**
* The constructor
* @param data the data to be stored in this node
*/
explicit Node( const Object & data )
: data{ data }, next{ nullptr } {}
Object data;
Node * next;
};
public:
/**
* The constructor for an empty list
*/
List()
: size{ 0 }, first{ nullptr }, last{ nullptr } {}
/**
* the copy constructor-creates and copy the list
*/
List( List && rhs ) = delete;
List( const List & rhs )
{
count = 0;
size = 0;
if(rhs.size != 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first;
size++;
while(current->next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op.
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode->next;
size++;
}
}
// you document and implement this method
}
/**
* the operator= method-copies the list
*/
List & operator=( List && rhs) = delete;
List & operator=( const List & rhs )
{
count = 0;
size = 0;
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
if(rhs.size!= 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first; //create a temporary to store
size++;
while(current -> next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode -> next;
size++;
}
}
return *this;
// you document and implement this method
}
/**
* accessor
* @return count
*/
int get_count() const
{
return count;
}
/**
* The destructor that gets rid of everything that\'s in the list and
* resets it to empty. If the list is already empty, do nothing.
*/
~List()
{
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
}
/**
* Put a new element onto the beginning of the list
* @param item the data the new element will contain
*/
void push_front( const Object& item )
{
Node * new_node = new Node( item );//basic op.
if(is_empty())
{
last = new_node;
}
else
{
new_node->next = first;
}
first = new_node;
size++;
/* you complete the rest */
}
/**
* Remove the element that\'s at the front of the list. Causes an
* assertion error if the list is empty.
*/
void pop_front()
{
assert( !is_empty() );
Node * temp = first;
if( first == last )
{
first = last = nullptr;
}
else
{
first = first->next;
}
delete temp;
size--;
}
/**
* Accessor to return the da.
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\");
}
.
I need help completing this C++ code with these requirements.instr.pdf
I need help completing this C++ code with these requirements.
instructions: IN C++ LANGUAGE PLEASE
Update the comments for each prototype by filling in the pre and post conditions.
Remember that pre conditions indicate what conditions must be met BEFORE the user calls the
function. The function will not work properly unless the pre conditions are met.
Post conditions indicate what will be the result of the function call. In other words, what
conditions will exist after the function is called.
See class notes from Thursday on pre and post conditions for some examples and further
explanation.
Code begins here:
#include //for NULL
class List
{
private:
struct Node
{
int data;
Node* next;
Node(int data): data(data), next(NULL){}
};
typedef struct Node* Nodeptr;
Nodeptr first;
Nodeptr last;
int size;
public:
/**Constructors and Destructors*/
List();
//Default constructor; initializes and empty list
//Postcondition:
~List();
//Destructor. Frees memory allocated to the list
//Postcondition:
/**Accessors*/
int getFirst();
//Returns the first element in the list
//Precondition:
int getLast();
//Returns the last element in the list
//Precondition:
bool isEmpty();
//Determines whether a list is empty.
int getSize();
//Returns the size of the list
/**Manipulation Procedures*/
void removeLast();
//Removes the value of the last element in the list
//Precondition:
//Postcondition:
void removeFirst();
//Removes the value of the first element in the list
//Precondition:
//Postcondition:
void insertLast(int data);
//Inserts a new element at the end of the list
//If the list is empty, the new element becomes both first and last
//Postcondition:
void insertFirst(int data);
//Inserts a new element at the start of the list
//If the list is empty, the new element becomes both first and last
//Postcondition:
/**Additional List Operations*/
void printList();
//Prints to the console the value of each element in the list sequentially
//and separated by a blank space
//Prints nothing if the list is empty
};
Solution
PROGRAM CODE:
#include
#include //for NULL
using namespace std;
class List
{
private:
struct Node
{
int data;
Node* next;
Node(int data): data(data), next(NULL){}
};
typedef struct Node* Nodeptr;
Nodeptr first;
Nodeptr last;
int size;
public:
/**Constructors and Destructors*/
List()
{
first = (Node*) malloc(sizeof(Node));
last = (Node*) malloc(sizeof(Node));
size = 0;
}
//Default constructor; initializes and empty list
//Postcondition:
~List()
{
delete first;
delete last;
}
//Destructor. Frees memory allocated to the list
//Postcondition:
/**Accessors*/
int getFirst()
{
return first->data;
}
//Returns the first element in the list
//Precondition:
int getLast()
{
return last->data;
}
//Returns the last element in the list
//Precondition:
bool isEmpty()
{
if(first == NULL)
return true;
else return false;
}
//Determines whether a list is empty.
int getSize()
{
return size;
}
//Returns the size of the list
/**Manipulation Procedures*/
void removeLast()
{
last = .
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.
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
.
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
Answer:
Note: LinkedList.cpp is written and driver program main.cpp is also given.
//ListInterface.h
#ifndef _LIST_INTERFACE
#define _LIST_INTERFACE
template
class ListInterface
{
public:
virtual bool isEmpty() const = 0;
virtual int getLength() const = 0;
virtual bool insert(int newPosition, const ItemType& newEntry) = 0;
virtual bool remove(int position) = 0;
virtual void clear() = 0;
virtual ItemType getEntry(int position) const = 0;
virtual void replace(int position, const ItemType& newEntry) = 0;
}; // end ListInterface
#endif
//Node.h
#ifndef NODE_
#define NODE_
template
class Node
{
private:
ItemType item; // A data item
Node* next; // Pointer to next node
public:
Node();
Node(const ItemType& anItem);
Node(const ItemType& anItem, Node* nextNodePtr);
void setItem(const ItemType& anItem);
void setNext(Node* nextNodePtr);
ItemType getItem() const ;
Node* getNext() const ;
}; // end Node
#endif
//Node.cpp
#include \"Node.h\"
template
Node::Node() : next(nullptr)
{
} // end default constructor
template
Node::Node(const ItemType& anItem) : item(anItem), next(nullptr)
{
} // end constructor
template
Node::Node(const ItemType& anItem, Node* nextNodePtr) :
item(anItem), next(nextNodePtr)
{
} // end constructor
template
void Node::setItem(const ItemType& anItem)
{
item = anItem;
} // end setItem
template
void Node::setNext(Node* nextNodePtr)
{
next = nextNodePtr;
} // end setNext
template
ItemType Node::getItem() const
{
return item;
} // end getItem
template
Node* Node::getNext() const
{
return next;
} // end getNext
//PrecondViolatedExcept.h
#ifndef PRECOND_VIOLATED_EXCEPT_
#define PRECOND_VIOLATED_EXCEPT_
#include
#include
class PrecondViolatedExcept : public std::logic_error
{
public:
PrecondViolatedExcept(const std::string& message = \"\");
}; // end PrecondViolatedExcept
#endif
//PrecondViolatedExcept.cpp
#include \"PrecondViolatedExcept.h\"
PrecondViolatedExcept::PrecondViolatedExcept(const std::string& message)
: std::logic_error(\"Precondition Violated Exception: \" + message)
{
} // end constructor
//LinkedList.h
#ifndef LINKED_LIST_
#define LINKED_LIST_
#include \"ListInterface.h\"
#include \"Node.h\"
#include
#include \"PrecondViolatedExcept.h\"
template
class LinkedList : public ListInterface
{
private:
Node* headPtr; // Pointer to first node in the chain;
// (contains the first entry in the list)
int itemCount; // Current count of list items
// Locates a specified node in this linked list.
Node* getNodeAt(int position) const;
public:
LinkedList();
LinkedList(const LinkedList& aList);
virtual ~LinkedList();
bool isEmpty() const;
int getLength() const;
bool insert(int newPosition, const ItemType& newEntry);
bool remove(int position);
void clear();
ItemType getEntry(int position) const;
void replace(int position, const ItemType& newEntry);
}; // end LinkedList
#endif
//LinkedList.cpp
#include \"LinkedList.h\"
#include
#include
#include
#include
//Default constructor
template
LinkedList::LinkedList() : headPtr(nullptr), itemCount(0)
{
.
C Homework Help
I am Andrew O. I am a C Homework Expert at programminghomeworkhelp.com. I hold a master’s in Programming from, the University of Southampton, UK. I have been helping students with their homework for the past 10 years. I solve homework related to C.
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--INSTRUCTION- --It helps to first create if-then-else structure to fi.pdf
//INSTRUCTION:
//It helps to first create if-then-else structure to fill in later.
//- Make sure PURPOSE and PARAMETER comments are given in detail
// Add sufficient comments to your code body to describe what it does.
// - Make sure all if-then-else are commented describing which case it is
// - Make sure all local variables are described fully with their purposes
//Complete the TODO parts.
// ====================================================
//Complier: g++
//File type: linkedList.cpp implementation file
//=====================================================
#include<iostream>
#include"linkedlist.h"
using namespace std;
//cunstructor that initializes an empty list
LinkedList::LinkedList() {
front = nullptr ;
rear = nullptr ;
count = 0;
}
/**
* @brief Destructor to destroy all nodes and release memory
*/
LinkedList::~LinkedList() {
Node *p, *q;
p = front;
while (p != nullptr ) {
q = p->next;
delete p;
p = q;
}
}
/**
* @brief Purpose: Checks if the list is empty
* @return true if the list is empty, false otherwise
*/
bool LinkedList::isEmpty() const {
// TODO: Add code here
return count == 0;
}
/**
* @brief Get the number of nodes in the list
* @return int The number of nodes in the list
*/
int LinkedList::length() const {
return count;
}
/**
* @brief Convert the list to a string
*
*/
string LinkedList::toString() {
string str = "[";
Node *ptr = front;
if (ptr != nullptr ) {
// Head node is not preceded by separator
str += to_string(ptr->val);
ptr = ptr->next;
}
while (ptr != nullptr ) {
str += ", " + to_string(ptr->val);
ptr = ptr->next;
}
str += "]";
return str;
}
/**
* @brief Displays the contents of the list
*/
void LinkedList::displayAll() {
cout << toString() << endl;
}
//Add a node to the end of the list
void LinkedList::addRear(T val) {
Node *p = new Node;
p->val = val;
p->next = nullptr ;
if (isEmpty()) {
front = p;
}
else {
rear->next = p;
}
rear = p;
count++;
// consider the two cases of whether the list was empty
}
//Add a node to the front of the list
void LinkedList::addFront(T val) {
Node *p = new Node;
p->val = val;
p->next = front;
if (isEmpty()) {
rear = p;
}
front = p;
count++;
// consider the two cases of whether the list was empty
}
//Delete the first node of the list
bool LinkedList::deleteFront(T &OldNum) {
if (isEmpty()) {
return false ;
}
Node *p = front;
OldNum = p->val;
front = front->next;
if (front == nullptr ) {
rear = nullptr ;
}
delete p;
count--;
return true ;
// consider if the list was empty and return false if the list is empty
// consider the special case of deleting the only node in the list
}
//Delete the last node of the list
bool LinkedList::deleteRear(T &OldNum) {
if (isEmpty()) {
return false ;
}
Node *p = rear;
OldNum = p->val;
if (front == rear) {
front = rear = nullptr ;
}
else {
Node *q = front;
while (q->next != rear) {
q = q->next;
}
rear = q;
q->next = nullptr ;
}
delete p;
count--;
return true ;
// consider if the list was empty and return false if the list is empty
// consider the specia.
COMP2710: Software Construction - Linked list exercises
Goals:
• To learn how to use structures
• To learn how to use linked data structures
• To design and implement functions
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.
Program to insert in a sorted list #includestdio.h#include.pdf
* Program to insert in a sorted list */
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* function to insert a new_node in a list. Note that this
function expects a pointer to head_ref as this can modify the
head of the input linked list (similar to push())*/
void sortedInsert(struct node** head_ref, struct node* new_node)
{
struct node* current;
/* Special case for the head end */
if (*head_ref == NULL || (*head_ref)->data >= new_node->data)
{
new_node->next = *head_ref;
*head_ref = new_node;
}
else
{
/* Locate the node before the point of insertion */
current = *head_ref;
while (current->next!=NULL &&
current->next->data < new_node->data)
{
current = current->next;
}
new_node->next = current->next;
current->next = new_node;
}
}
/* BELOW FUNCTIONS ARE JUST UTILITY TO TEST sortedInsert */
/* A utility function to create a new node */
struct node *newNode(int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
new_node->next = NULL;
return new_node;
}
/* Function to print linked list */
void printList(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
printf(\"%d \", temp->data);
temp = temp->next;
}
}
/* Drier program to test count function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
struct node *new_node = newNode(5);
sortedInsert(&head, new_node);
new_node = newNode(10);
sortedInsert(&head, new_node);
new_node = newNode(7);
sortedInsert(&head, new_node);
new_node = newNode(3);
sortedInsert(&head, new_node);
new_node = newNode(1);
sortedInsert(&head, new_node);
new_node = newNode(9);
sortedInsert(&head, new_node);
printf(\"\ Created Linked List\ \");
printList(head);
return 0;
}
public class Listnode {
//*** fields ***
private Object data;
private Listnode next;
//*** methods ***
// 2 constructors
public Listnode(Object d) {
this(d, null);
}
public Listnode(Object d, Listnode n) {
data = d;
next = n;
}
// access to fields
public Object getData() {
return data;
}
public Listnode getNext() {
return next;
}
// modify fields
public void setData(Object ob) {
data = ob;
}
public void setNext(Listnode n) {
next = n;
}
}
Thsi below program print out all the varibles in SLL
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* 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) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int .
C++ problemPart 1 Recursive Print (40 pts)Please write the recu.pdf
C++ problem
Part 1: Recursive Print (40 pts)
Please write the recursive List reverse print function, whose iterative version we wrote in class.
Below are the function signatures for the functions you are going to need:
public:
/**Additional Operations*/
void print_reverse();
//Wrapper function that calls the reverse helper function to print a list in reverse
//prints nothing if the List is empty
private:
void reverse(Nodeptr node);
//Helper function for the public printReverse() function.
//Recursively prints the data in a List in reverse.
Why do we need the private helper function here?
Since we are going to be reversing our list node by node, in a recursive fashion, we want to pass
a one node at a time to our reverse function.
However, since our nodes are private, we cannot access them if we call the function inside of
main.
Add these function signatures to your List.h file along with your other function prototypes inside
the class definition.
Make sure that you place the reverse function inside the private portion of your List class
definition and the print_reverse function prototype to the public portion of your List class
definition.
Now, implement these two functions inside of List.h, under your section for additional
operations.
Important: Test each function carefully inside of your ListTest.cpp to make sure that it is
working properly.
Part 2: Adding an Index to Your List Nodes (20 pts)
Next, you will add the following functions to your List.h
/**Accessor Functions*/
int get_index();
//Indicates the index of the Node where the iterator is currently pointing
//Nodes are numbered from 1 to length of the list
//Pre: length != 0
//Pre: !off_end()
...
int List::get_index()
{
//Implement the function here
}
/**Manipulation Procedures*/
void scroll_to_index(int index);
//Moves the iterator to the node whose index is specified by the user
//Pre: length != 0
...
void scroll_to_index(int index)
{
//Implement function here
}
Part 3: Implementing Search as Part of Your List (40 pts)
Now, we are going to add two search functions to our List so that we can search for elements in
our List.
The first of these functions is going to be a simple linear search function.
You will need to add the following function prototype and function definition to your List.h:
/**Additional Operations*/
int linear_search(listitem item);
//Searchs the list, element by element, from the start of the List to the end of the List
//Returns the index of the element, if it is found in the List
//Returns -1 if the element is not in the List
//Pre: length != 0
...
int List::linear_search(listitem item)
{
//Implement the function here
}
You are also going to add a function to perform recursive binary search on your List.
You will need to add the following function prototype and function definition to your List.h:
int binary_search(int low, int high, listitem item);
//Recursively searchs the list by dividing the search space in half
//Returns the index of the element, if it is fo.
Write java program using linked list to get integer from user and.docx
Write java program using linked list to get integer from user and
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.
Inspect the class declaration for a doubly-linked list node in Node-h-.pdf
Inspect the class declaration for a doubly-linked list node in Node.h. Access Node.h by clicking
on the orange arrow next to main.cpp at the top of the coding window. The Node class has three
member variables: a double data value, a pointer to the next node, and a pointer to the previous
node. Each member variable is protected. So code outside of the class must use the provided
getter and setter member functions to get or set a member variable. Node.h is read only, since no
changes are required. Step 2: Implement the Insert() member function A class for a sorted,
doubly-linked list is declared in SortedNumberList.h. Implement the SortedNumberList class's
Insert() member function. The function must create a new node with the parameter value, then
insert the node into the proper sorted position in the linked list. Ex: Suppose a
SortedNumberList's current list is 23 47.25 86, then Insert(33.5) is called. A new node with data
value 33.5 is created and inserted between 23 and 47.25, thus preserving the list's sorted order
and yielding: 23 35.5 47.25 86 Step 3: Test in develop mode Code in main() takes a space-
separated list of numbers and inserts each into a SortedNumberList. The list is displayed after
each insertion. Ex: If input is 77 15 -42 63.5 then output is: List after inserting 77: 77 List after
inserting 15: 15 77 List after inserting -42: -42 15 77 List after inserting 63.5: -42 15 63.5 77 Try
various program inputs, ensuring that each outputs a sorted list. Step 4: Implement the Remove()
member function Implement the SortedNumberList class's Remove() member function. The
function takes a parameter for the number to be removed from the list. If the number does not
exist in the list, the list is not changed and false is returned. Otherwise, the first instance of the
number is removed from the list and true is returned. Uncomment the commented-out part in
main() that reads a second input line and removes numbers from the list. Test in develop mode to
ensure that insertion and removal both work properly, then submit code for grading. Ex: If input
is 84 72 19 61 19 84 then output is: List after inserting 84: 84 List after inserting 72: 72 84 List
after inserting 19: 19 72 84 List after inserting 61: 19 61 72 84 List after removing 19: 61 72 84
List after removing 84: 61 72
main.cpp
#include
#include
#include
#include "Node.h"
#include "SortedNumberList.h"
using namespace std; void PrintList(SortedNumberList& list); vector SpaceSplit(string source);
int main(int argc, char *argv[]) { // Read the line of input numbers string inputLine; getline(cin,
inputLine); // Split on space character vector terms = SpaceSplit(inputLine); // Insert each value
and show the sorted list's contents after each insertion SortedNumberList list; for (auto term :
terms) { double number = stod(term); cout << "List after inserting " << number << ": " << endl;
list.Insert(number); PrintList(list); } /* // Read the input line with numbers to remove getline(cin,
inputLi.
C++ please put everthing after you answer it- thanks Complete the stub.docx
C++ please put everthing after you answer it, thanks
Complete the stub file for the DoublyLinkedList implementation.
Please note: Any changes to the NodeType struct, not using the NodeType struct, or changing from template Type to using a specific data type will all be cause for deductions.
You are required to implement the methods:
insertFirst
insertLast
insertNode - (insert an item at a given index)
deleteNode - (remove a specific item)
The stub file contains the basic class format to implement the virtual methods from ListType. You should only need to complete the code for these four functions. Your code must maintain (or add) the node connections both forward and backward. In the double link list class these are called next and prev.
Your code must also maintain the count variable.
\} template Type ListType : : back ( ) const \{ assert(tail != NULL); //confirm tail exists return tail->data; //return the data of the tail node \} template void ListType >: :copyList (const ListType > \& otherList) \{ NodeType *newNode; //pointer to create a node NodeType current; //pointer to traverse the list if (head ! = NULL) //if the list is nonempty, make it empty destroyList(); if (otherList. head == NULL) //otherList is empty \{ head = NULL; tail = NULL; count = 0 ; \} else \{ current = otherList. head; // current points to the count = otherList . count; // list to be copied //copy the head node head = new NodeType < Type > ; / / create the node head->data = current > data; // copy the data head->next = NULL; // set the link field of tail = head; // the node to NULL // the next node //copy the remaining list while (current != NULL) \{ newNode = new NodeType < Type > ; // create a node newNode > data = current > data; // copy the data newNode->next = NULL; // set the link of // newNode to NULL tail->next = newNode; //attach newNode after tail tail = newNode; // make tail point to //the actual tail node current = current > next; 1/ make current point //to the next node \}/ /end while \}/lend else \}//end copylist
.
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
Similar to Use C++class Node{public Node ( int = 0 ); constru.pdf (20)
How do you stop infinite loop Because I believe that it is making a.pdf
How do you stop infinite loop Because I believe that it is making a.pdf
Write a C++ function that delete nodes in a doubly linkedlist- It shou.docx
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I need help completing this C++ code with these requirements.instr.pdf
I need help completing this C++ code with these requirements.instr.pdf
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf
C++ Please write the whole code that is needed for this assignment- wr.docx
C++ Please write the whole code that is needed for this assignment- wr.docx
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
--INSTRUCTION- --It helps to first create if-then-else structure to fi.pdf
--INSTRUCTION- --It helps to first create if-then-else structure to fi.pdf
COMP2710: Software Construction - Linked list exercises
COMP2710: Software Construction - Linked list exercises
Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
Program to insert in a sorted list #includestdio.h#include.pdf
Program to insert in a sorted list #includestdio.h#include.pdf
C++ problemPart 1 Recursive Print (40 pts)Please write the recu.pdf
C++ problemPart 1 Recursive Print (40 pts)Please write the recu.pdf
Write java program using linked list to get integer from user and.docx
Write java program using linked list to get integer from user and.docx
Write a program to implement below operations with both singly and d.pdf
Write a program to implement below operations with both singly and d.pdf
Inspect the class declaration for a doubly-linked list node in Node-h-.pdf
Inspect the class declaration for a doubly-linked list node in Node-h-.pdf
C++ please put everthing after you answer it- thanks Complete the stub.docx
C++ please put everthing after you answer it- thanks Complete the stub.docx
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned in.pdf
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Which asymmetric cryptosystem is used for digital signaturesA. DE.pdf
Which asymmetric cryptosystem is used for digital signatures?
A. DES
B. SHA1
C. Diffie-Hellman
D. ECC
Solution
D. Elliptic curve cryptosystems (ECC) is an asymmetric cryptosystem created in the 1980s to
create and store digital signatures in a small amount of memory. Answer A is incorrect because
DES is a symmetric algorithm. Answer B is incorrect because SHA1 is a hashing algorithm.
Answer C is incorrect because Diffie-Hellman is used for key exchange..
Which of these theories fit into the contextual world-viewQuestio.pdf
Which of these theories fit into the contextual world-view?
Question 13 options:
Stage theories, like Piaget\'s.
Learning theories, like Skinner\'s.
Universal theories, like Bowlby\'s.
Systems theories, like Bronfenbrenner\'s.
Save
Stage theories, like Piaget\'s.
Learning theories, like Skinner\'s.
Universal theories, like Bowlby\'s.
Systems theories, like Bronfenbrenner\'s.
Solution
ANS: D. Systems theories, like Bronfenbrenner\'s.
Explanation:
Developmental psychology is mainly influenced by three major world views. They are 1.
Organismic, 2. Mechanistic and 3. contextual world views.
Contextual world views:
Development is the result of reciprocal social interactions between an active person and their
social partners from in a series of nested contexts.
Systems theories, like Bronfenbrenner\'s theory fit into the contextual world-view.
Wrong theories:
A. Stage theories, like Piaget\'s. (Organismic)
B. Learning theories, like Skinner\'s. (Mechanistic)
C. Universal theories, like Bowlby\'s. (Maturational).
Which of the following distinguishes prophase of meiosis I from proph.pdf
Which of the following distinguishes prophase of meiosis I from prophase of mitosis?
homologous chromosomes pair up spindle forms nuclear membrane breaks down
chromosomes become visible
Solution
In meiosis there are two successive divion occure, meiosis I(which is reduction division) and
meiosis II (similar division). While meiosis II is similar to mitosis,Meiosis I is a eventfull cell
divion, divided into 4 phases....Prophase1, metaphase 1,Anaphase 1 and telophase 1.
Prophase 1 of MEIOSIS 1 divided into following subphases- LEPTONEMA, ZYGONEMA,
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In zygotene or Zygonema homologous chromosome(paternally derrived and maternally derrived)
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Solution
1. Primary growth refers to the growth that occurs due to cell divisions at the tips of roots and
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Secondary growth refers to the one that occurs due to cell division in cambium or lateral
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What to the characteristic X-ray radiation What is the originSo.pdf
What to the characteristic X-ray radiation? What is the origin?
Solution
- Characteristics x- ray radiations are emitted from heavy elements when their electrons make
transitions between the lower atomic energy levels.
- the high energy electron can also cause an electron close to the neclues in a metal atom to be
knocked out from its place
ORIGIN
Discovery. German physicist Wilhelm Röntgen is usually credited as the discoverer of x rays in
1895, because he was the first to systematically study them, though he is not the first to have
observed their effects.
Rountgen called them x rays after the mathematical symbol X for unknown.
What other blood cell related problems are often seen in patients wi.pdf
What other blood cell related problems are often seen in patients with chronic leukemia?
Citereferencesplease
Solution
The term leukemia is referred to neoplasm of hematopoietic tissue. Unlike the solid tumors the
leukemic cells diffuse into bone marrow and lymphoid tissues and spill over into blood stream to
spread to various organs of the body. The leukemic cells are either mature or extremely
primitive. Rich amounts of white blood cell counts in leukemia reveals the spread of tumors to
peripheral blood. However, the proliferation of white blood cells is mostly confined to bone
marrow and no significant increase in number of white blood cells in the blood stream can be
observed.
Leukemia is classified depending on the type of the cell and maturity of proliferating cell. Any
type of hematopoietic cell may produce leukemia but most common types are granulocytic,
monocytic and lymphocytic. The leukemia developing from stem cells produce the leukocytes
containing specific granules such as neutrophils, eosinophils and basophils. This type of
leukemia is called granulocytic leukemia. Monocytic leukemia is developed from the precursor
cells of monocytes. Lymphocytic leukemia is produced by lymphoid precursor cells.
When the leukemic cells are most primitive forms the neoplasm is called acute leukemia. The
leukemia with mature cells is called chronic leukemia. Chronic granulocyte leukemia is caused
by circulating matured granulocytes and neutrophils along with few primitive cells. Chronic
lymphocytic leukemia is caused by circulating mature lymphocytes.
The clinical features of leukemia are either caused by impairment of function of bone marrow or
caused by infiltration of leukemic cells into viscera. The overgrowth of leukemic cells in bone
marrow crowds out the normal cells leading to anemia because of inadequate production of red
blood cells. Bleeding is caused by thrombocytopenia and infection resulting from insufficient
numbers of white blood cells to provide protection by immune defense mechanism against
pathogens.
Leukemic cells spreading to other tissues may invade into spleen, liver, lymph nodes and other
tissues. The chronic leukemia proceeds at a slow rate and can be controlled by treatment over
long periods of time. Therefore, the patient may survive for several years. In an acute leukemia
the disease is rapidly progressive and leukemic cells infiltrate bone marrow and viscera with
conspicuous appearance.
1.American Cancer Society. Cancer Facts and Figures 2016. Atlanta, Ga: American Cancer
Society; 2016.
2. Byrd JC, Brown JR, O\'Brien S, et al. Ibrutinib versus ofatumumab in previously treated
chronic lymphoid leukemia. N Engl J Med. 2014 Jul 17;371(3):213-23. Epub 2014 May 31.
3. Byrd JC, Furman RR, Coutre SE, et al. Targeting BTK with ibrutinib in relapsed chronic
lymphocytic leukemia. N Engl J Med. 2013 Jul 4;369(1):32-42. Epub 2013 Jun 19.
4. Byrd JC, Flynn JM. Chronic lymphocytic leukemia. In: Niederhuber JE,Armitage JO,
Doroshow .
what is inductive and deductive reasoningSolutionDeductive and.pdf
what is inductive and deductive reasoning
Solution
Deductive and Inductive Reasoning
Deductive and inductive are basically types of arguments and where argument generally supports
conclusion with premises.
Deductive reasoning is basically a type of argument which supports conclusion logically for
example arguments which are based on rules, facts, truth, laws and principles and are accepted
widely.
It is the process of reasoning which involves to draw conclusion based on the relationship of two
premises in which one assertion is broad(major premises and valid) and other is specific(minor
premises and invalid). As major premises are broad it is to be supported by values and laws.
Like for instance:
Reena is a woman
All women are mortal
Therefore it proves that reena is mortal.
It is a top to down approach.
Inductive Reasoning is basically a type of argument which supports conclusion on the basis of
judgment, pattern, trend, experience and observation. These arguments do not prove truth as
these conclusions are based on probable support. The argument which support successfully is a
strong argument whereas that fails to support is considered as weak. It is a bottom to up
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What are the major evolutionary feats shared by all 3 phyla( Platyhe.pdf
What are the major evolutionary feats shared by all 3 phyla( Platyhelminthes, Nematoda, and
Annelida)? What are the implications of these evolutionary feats? What two additional
evolutionary feats are exhibited by Nemotoda? And 2 feats by Annelida?
Solution
Answer:
The major evolutionary feats shared by Platyhelminthes, Nematoda and Annelida are:
1. Level of Organization- Organ system level of organization
The organisms possesses distinct organ systems which marks the evolution of organ systems in
animals.
2. Symmetry - Bilateral
This is often a result of moving in one direction, thus having a “head” end.
3. Triploblastic
They share common body wall organization, consisting of 3 layers. Three germ layers are
evidence of slowly adapting layers of developmental control during embryogenesis as cells
specialized for greater efficiency.
Two additional evolutionary feats exhibited by Nematoda are:
1. Coelom- Pseudocoelomate
2. Digestive system -Complete
Two feats of Annelida are:
1. Segmentation - Presented
2. Circulatory system -Present.
What are some of the advantages in the use of AOA notation as oppose.pdf
What are some of the advantages in the use of AOA notation as opposed to AON? Under what
circumstances does it seem better to apply AON methodology in network development?
Solution
AOA means activity on arc. In network diagrams AOA notation uses arrow diagram where each
activity is represented by an arrow joining 2 circles (nodes). The circles or the nodes represents
the transition between activities, which are called events.
AON means activity on node. Here activities are represented by circles and the arc represents the
events.
The advantage of using AOA as opposed to AON is that AOA is more suitable for analytical
techniques. Path tracking in case of AOA is simplified by event numbering. Lastly key events
are easy to identify.
AON is better under the following circumstances:.
Verify the identity.StartFraction sine left parenthesis alpha plus.pdf
Verify the identity.
StartFraction sine left parenthesis alpha plus beta right parenthesis Over sine alpha cosine beta
EndFraction equals 1 plus cotangent alpha tangent beta
sin(+)
sincos
=1+cottan
Solution
sin(+)/sincos =1+cottan
sin( + ) = sin()cos() + cos()sin()
LHS ======> sin()cos()/sincos + cos()sin()/sincos = 1 + cot()*tan() = RHS.
This is a lab for a java program that I am very unsure of, it has to.pdf
This is a lab for a java program that I am very unsure of, it has to do with Binary trees, and I am
very confused on how to work with them, if anyone can help I will be very grateful, thank you!!
You are to create a balanced binary tree which means that at each node n, the size of the left
subtree of n is within one of the size of the right subtree of n
Your program shall read in from user input a list of positive integers into an array (at most 100
elements). Your input stops when it reads a 0 (sentinel ).
Your program will then sort the elements in the array. You MUST write your own sort routine
You are then to create a balanced binary tree from this data.
You are then to print out your balanced tree using pre-order, in-order, and post-order routines
Requirements
1) Program must read integers from user input into an array
2) Program must sort integers with sort routine written by user
3) Program must implement a binary tree object using links (not arrays)
4) Program must implement at least methods PREORDER, INORDER, POSTORDER, ADD
(create a balanced binary search tree from the given array).
5) Output demonstrating input and output
Solution
import java.io.*;
import java.util.*;
class Nodetype
{
int info;
int ht;
Nodetype left;
Nodetype right;
Nodetype(int i)
{
info=i;
ht=0;
left=null;
right=null;
}
}
class Functions
{
Nodetype AVLroot;
void create()
{
int x,n,i;
AVLroot=null;
int[] a=new int[100];
String str;
int counter=0;
System.out.println(\"\ Enter numbers:\");
do
{
x=getNumber();
a[counter]=x;
counter++;
}while(x!=0);
int[] finalarr=new int[100];
finalarr=sort(a);
int tcntr=0;
while(finalarr[tcntr]!=0)
{
AVLroot=insert(AVLroot,finalarr[tcntr]);
tcntr++;
}
System.out.println(\"\ Preorder sequene:\");
preorder(AVLroot);
System.out.println(\"\ Inorder sequence:\");
inorder(AVLroot);
System.out.println(\"\ Postorder sequence:\");
postorder(AVLroot);
}
int[] sort(int[] arr)
{
int n = arr.length;
int temp = 0;
for(int i=0; i < n; i++){
for(int j=1; j < (n-i); j++){
if(arr[j-1] > arr[j]&&arr[j]!=0){
//swap elements
temp = arr[j-1];
arr[j-1] = arr[j];
arr[j] = temp;
}
}
}
return (arr);
}
int getNumber()
{
String str;
int ne=0;
InputStreamReader input=new InputStreamReader(System.in);
BufferedReader in=new BufferedReader(input);
try
{
str=in.readLine();
ne=Integer.parseInt(str);
}
catch(Exception e)
{
System.out.println(\"\ I/O error.\");
}
return ne;
}
Nodetype insert(Nodetype T,int x)
{
if(T==null)
{
Nodetype node=new Nodetype(x);
T=node;
}
else
if(x>T.info)
{
T.right=insert(T.right,x);
if(BF(T)==-2)
if(x>T.right.info)
T=RR(T);
else
T=RL(T);
}
else
if(xrh)
return(lh);
return(rh);
}
Nodetype rotateright(Nodetype x)
{
Nodetype y;
y=x.left;
x.left=y.right;
y.right=x;
x.ht=height(x);
y.ht=height(y);
return(y);
}
Nodetype rotateleft(Nodetype x)
{
Nodetype y;
y=x.right;
x.right=y.left;
y.left=x;
x.ht=height(x);
y.ht=height(y);
return(y);
}
Nodetype RR(Nodetype T)
{
T=rotateleft(T);
return(T);
}
Nodetype LL(Nodetype T)
{
T=rotateright(T);
return(T);
.
The concentration of carbon dioxide in the atmosphere today is about .pdf
The concentration of carbon dioxide in the atmosphere today is about Select one: a. 170 parts
per million. b. 300 parts per million. c. 400 parts per million d. 800 parts per million. How
would a shift in land use from forest to agriculture affect atmospheric carbon dioxide? Select
one: a. It would have no effect. b. It would result in a large release of CO_2 c. It would result
in a decrease in release of CO_2 d. Neither forests nor crops contribute significantly to
atmospheric carbon dioxide Charles Keeling chose to monitor atmospheric carbon dioxide levels
at Mauna Loa Observatory because Select one: a. uptake of carbon dioxide would be greatest in
warm tropical forests b. it was located far from industrial sources of pollution that might make
interpretation of the data more difficult c. the climate in Hawaii was very different from the
climate of Big Sur (California) where he did his previous work on carbon dioxide d. the relative
contribution of fossil fue
Solution
The concentration of carbondioxide in the atmosphere today is around 400 Parts per million
(ppm). The safe level of carbondioxide in atmosphere is around 350ppm. The shift in land use
from forest to agrivulture results in large release of carbondioxide in atmosphere. Also release of
other greenhouse gasses results in global warming. Its because of the relative contribution of
fossil fuel burning to the release of atmospheric carbondioxide was debated. Its one of the major
reason for the increase in global warminh..
Research and discuss two operating systems and how incident response.pdf
Research and discuss two operating systems and how incident response processes may differ in
properly shutting down the systems for transport to the laboratory.
Solution
Computer Incident Response and Computer Forensics Overview carried out in different
operating system. As follows below:
Introduction
When a compromise of security or an unauthorized/illegal action associated with a
computer is suspected, it is important that steps are taken to ensure the protection of the
data within the computer and/or storage media. The stored data is needed to determine
the level of security compromise and location of potential evidence concerning the
unauthorized or illegal act.
The initial response to a computer security incident may be more important than later
technical analysis of the computer system because of the actions taken by incident
response team members. Actions taken by the incident response team impact subsequent
laboratory examinations of the computer and/or media. Of most importance is that the
first responder act appropriately.
In the event of a suspected computer incident, care must be taken to preserve evidence in
its original state. While it may seem that simply viewing files on a system would not
result in alteration of the original media, opening a file changes it. From a legal sense, it
is no longer the original evidence and may be inadmissible in any subsequent legal or
administrative proceedings.
This paper will focus on the incident response and computer forensics on the personal or
desktop computers. The incident response and forensic procedures and techniques for
servers may additional knowledge and tools.
Incident Response
Every organization should have an incident response team. This team may consist of one
person in an organization or several persons. In the event of suspected computer crime or
violations of user policies, the team should be activated. The team should have written
procedures for incident response, including what conditions warrant calling in local
and/or federal law enforcement authorities. Violations of user policies may result in
administrative actions whereas suspected computer crimes may require that law
enforcement authorities be called in. The incident team needs to protect evidence for
either situation. For administrative actions, the procedures described in this paper may be
sufficient. However, for suspected computer crimes, the law enforcement officials may
instruct the incident team to wait for their arrival.
The activities/procedures for securing a suspected computer incident scene include
· Securing the scene
· Shutting down the computer
· Labeling the evidence
· Documenting the evidence
· Transporting the evidence
· Providing chain-of-custody documentation
The computer incident response team should keep in mind that what begins as a
collection of evidence for violation of administrative policy violations may escalate into
collection of evidence for more serious violations. The computer may hav.
Proteins can be attached to a membrane in a variety of ways. How doe.pdf
Proteins can be attached to a membrane in a variety of ways. How does the mode of attachment
relate to their function?
Solution
Membrane proteins are very important for different type of cellular functions like a transport
signalling pathways and cell adhesion etc…. proteins are interacting with membrane in different
types
Some proteins are associated with cytoplasmic face of the cell membrane; some proteins are
associated with extracellular face of the plasma membrane (peripheral proteins) and integral or
intrinsic proteins. Extrinsic proteins do not interact with the hydrophobic core of the
phospholipids bilayer.
Example porins are transmembrane proteins that play a major role in water transport
Prostaglandin H2 extracellular membrane bounded protein that is associated with surface of the
plasma membrane. Play major role in inflammation..
Problem 1-What is environmental engineering (20 pts) Below you wil.pdf
Problem 1-What is environmental engineering? (20 pts) Below you will see a list of four
different scenarios which to field of environmental engineering. For each one of ihena may
involve the area you need (1) Write out if th have defined it in our course e scenario involves
environmental engineering as wo for (1) by explaining what area of environmental engineering
the scenario fits into and why decide YES, the scenario does involve the field of specific an
environmental engineering, then describe something environmental engineer might do as part o
scenario. If you decide scendor of environmental engineering. you do not have to do anything t f
their job in thot e NQ, the scenario does not involve the field this step. northern Scenario 1:
Unsustainable deforestation in an old growth Canadian forest the Congo. Scenario 3: Assessment
of the level of exposure of indoor air pollutants in a school building for children that go to the
school. Scenario 4: Assisting a Native American tribe in legal arguments to justity to a state
court that their water rights are being violated. Scenario 2: Right handling of acid mine water
drainage for a gold mine in
Solution
1.
1.At 0.02% of its forested area, deforestation in Canada is among the world’s lowest, yet many
myths exist about the state of our forests. The reality is that Canada is a world leader in
sustainable forest management. Canadian forests are healthy, productive and thriving.
Deforestation is an important issue, since shrinking forest cover reduces biodiversity, affects soil
and water quality, impacts wildlife habitat and influences climate change. The Canadian
government carefully monitors and regularly publishes reports on deforestation. Our scientists
combine satellite and aerial images with information about regional development, forest
ecosystems, natural processes and local conditions to help monitor and manage the health of
Canadian forests.
Here are some key facts about Canada’s low levels of deforestation.
The annual deforestation rate in Canada in 2010 was less than 0.02% of our forests and the rate
has been declining for over 25 years. In 1990, 63,100 hectares were lost to deforestation and in
2014 this figure dropped to 34,200 hectares.
Today, Canada’s 348 million hectares of forest lands represent about 9% of the world’s forest
cover, but account for only 0.3% of global deforestation.
Deforestation only occurs when forests are permanently removed so the land can be used for
something else. Harvesting, forest fires and insect infestations do not constitute deforestation,
since the affected areas will grow back. According to laws, regulations and policies in place
across Canada, all areas harvested on public land must be reforested, either by replanting or
through natural regeneration. About 94% of Canada’s forests are on public land.
The conversion of forest to agricultural land is decreasing but it remains the largest contributor to
deforestation in Canada. The small contribution the.
NetWork Design Question2.) How does TCP prevent Congestion Dicuss.pdf
NetWork Design Question
2.) How does TCP prevent Congestion? Dicuss the information identifying congestion in the
network as well as the mechanism for reducing congestion?
Solution
Congestion is a problem that occurs on shared networks when multiple users contend for access
to the same resources (bandwidth, buffers, and queues).
Transmission Control Protocol (TCP) uses a network congestion-avoidance algorithm that
includes various aspects of an additive increase/multiplicative decrease (AIMD) scheme, with
other schemes such as slow-start to achieve congestion avoidance.
The TCP congestion-avoidance algorithm is the primary basis for congestion control in the
Internet.
Congestion typically occurs where multiple links feed into a single link, such as where internal
LANs are connected to WAN links. Congestion also occurs at routers in core networks where
nodes are subjected to more traffic than they are designed to handle.
TCP/IP networks such as the Internet are especially susceptible to congestion because of their
basic connection- less nature. There are no virtual circuits with guaranteed bandwidth. Packets
are injected by any host at any time, and those packets are variable in size, which make
predicting traffic patterns and providing guaranteed service impossible. While connectionless
networks have advantages, quality of service is not one of them.
Shared LANs such as Ethernet have their own congestion control mechanisms in the form of
access controls that prevent multiple nodes from transmitting at the same time.
Identifying:
Congestion is primarily reflected by a conventional user feeling-- slowness. This statement
reflects the change in the network effective flow, that is the time required to transmit an entire
data from one point to another. The effective flow doenot exist as such, it consists in reality of
three seperate indicators:
*Latency:the effective flow is inversely proportional to the latency.
*Jitter:it is latency variation over time, impacts by influencing the flow latency
*Loss Rate:the theoritical bandwidth is inversely proportional to the square root of the loss rate
These Congestion symtoms allow us to rely on objective indicators to characterize it.
Mechanism to reduce congestion:
The standard fare in TCP implementations today has four standard congestion control algorithms
that are now in common use. Their usefulness has passed the test of time.
The four algorithms, Slow Start, Congestion Avoidance, Fast Retransmit and Fast Recovery are
described below. (a) Slow Start
Slow Start, a requirement for TCP software implementations is a mechanism used by the sender
to control the transmission rate, otherwise known as sender-based flow control. This is
accomplished through the return rate of acknowledgements from the receiver. In other words, the
rate of acknowledgements returned by the receiver determine the rate at which the sender can
transmit data. When a TCP connection first begins, the Slow Start algorithm initializes a
.
Mosses and liverworts are early colonizers during ecological success.pdf
Mosses and liverworts are early colonizers during ecological succession and make the
environment more amenable for other species. How do these plants contribute to nitrogen levels
in the soil?
Solution
Answer:
Mosses and liverworts have been found to be the first organisms to colonize either a primary
succession or secondary succession. Their spores are carried by birds, insects or wind.
Studies indicate that mosses and liverworts make the soil more amenable to colonization by other
plants. This can be attributed to the fact that mosses and liverworts establish symbiotic
association with nitrogen fixing cyanobacteria, thus replenishing the soil with nitrogen..
Many programming languages, especially older ones, provide no langua.pdf
Many programming languages, especially older ones, provide no language support for
concurrency. C and C++ are examples of such languages. Is it essential that a language include
syntax for concurrency to be able to write concurrent programs in that language? If not, how is it
accomplished? What are the advantages and disadvantages of including support for concurrency
in a language?
Solution
Advantages:
1).Although the tasks are running and the user can relate with the applications.
2).It has the availability of services. means long running the tasks will not delay tiny running
ones.
3). complex programs make better use of many resources.
4).tasks are requiring certain pre conditions can hang and wait untill the pre conditions are seize
and then restart implementation properly.
Disadvantages:
1). a chance of corrupting consistent state of program by concurrent tasks.
2).Deadlock may occur.
3).It would be masters exponential number of interleavings due to different schedules..
Inferior nasal concha Lacrimal bone Mandible Maxilla Nasal bone .pdf
Inferior nasal concha Lacrimal bone Mandible Maxilla Nasal bone Palatine bone Vomer
bone Zygomatic bone Forms bridge of nose Only movable bone in the facial skeleton Contains
coronoid process Creates prominence of cheek inferior and lateral to the eye Contains sockets
of upper teeth Forms inferior portion of nasal septum Forms anterior portion of zygomatic arch
Scroll-shaped bone in nasal passage Forms anterior roof of mouth Contains mental foramen
Forms posterior roof of mouth Small bone in medial wall of orbit Forms a very small portion of
inferior orbit Contains large sinus inferior to orbit
Solution.
import java.util.;public class Program{public static void.pdf
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
Solution
Hi,
I have modified the code. it is working as expected now. Highlighted the code changes below.
Issue here is with you have written this statement twice System.out.println(x); which causes the
issue.
Program.java
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println();
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Output:
Enter amount
10
Amount is: 10
X
XX
XXX
XXXX
XXXXX
XXXXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
XXXXXXXXXX.
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Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
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