reverse the linked list (2-4-8-10) by- stack- iteration- recursion- U.docx

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reverse the linked list (2,4,8,10) by: stack; iteration; recursion. Using C++ Solution #include<iostream> #include<stdio.h> #include<stdlib.h> #includ<stack> using namespace std; struct Node { int value; struct Node* next; }node; struct Node* reverse(struct Node* head) {Â Â // Iteration Method struct Node* current,*prev; current = head; prev = NULL; while(current!=NULL) { next = current->next; current->next = prev; prev = current; current = next; } head = prev;- return head; } struct Node* insert(Node* head,int value) { Node* temp = (struct Node*)malloc(sizeof(struct Node)); temp->value = value; temp->next = NULL; if(head = = NULL) head = temp; else { Node*temp1=head; while(temp1->next!=NULL) temp1 = temp1->next; temp1->next = temp; } return head; } public void print(Node* head) { while(head!=NULL) { cout<<head->value; head = head->next; } } struct Node*recursiveReverse(struct Node* &head) {Â Â //Recursion method if(head==NULL || head->next ==NULL) return; Node*first = head; Node *rest = head->next; if(rest !=NULL) return; recursiveReverse(rest); first ->next->next = first; first->next =NULL; head = rest; return head; } void Reverse(struct Node* head)Â Â // stack Implementation { stack<node*> Stack; node *traverse = head; while(traverse!=NULL) { Stack.push(traverse); traverse = traverse->next;Â Â } node *temp = Stack.top(); head = temp; Stack.pop(); while(!Stack.empty()) { temp->next = Stack.top(); Stack.pop(); temp = temp->next;Â Â } temp->next = NULL; return Stack; } int main() { struct Node* head =NULL; head = insert(head,2); head = insert(head,4); head = insert(head,8); head = insert(head,10); print(head); head= reverse(head); print(head); head = recursiveReverse(head); print(head); head = void Reverse(head); void print(head); return 0; } .

Education

prev = NULL;
while(current!=NULL) {
next = current->next;
current->next = prev;
prev = current;
current = next; }
head = prev;-
return head; }
struct Node* insert(Node* head,int value) {
Node* temp = (struct Node*)malloc(sizeof(struct Node));
temp->value = value;
temp->next = NULL;
if(head = = NULL)
head = temp;
else {
Node*temp1=head;
while(temp1->next!=NULL)
temp1 = temp1->next;
temp1->next = temp;
} return head;
}
public void print(Node* head) {
while(head!=NULL) {

cout<<head->value;
head = head->next; }
}
struct Node*recursiveReverse(struct Node* &head) {Â Â //Recursion method
if(head==NULL || head->next ==NULL) return;
Node*first = head;
Node *rest = head->next;
if(rest !=NULL) return;
recursiveReverse(rest);
first ->next->next = first;
first->next =NULL;
head = rest;
return head; }
void Reverse(struct Node* head)Â Â // stack Implementation
{
stack<node*> Stack;
node *traverse = head;
while(traverse!=NULL) {
Stack.push(traverse);
traverse = traverse->next;Â Â }
node *temp = Stack.top();
head = temp;
Stack.pop();

while(!Stack.empty()) {
temp->next = Stack.top();
Stack.pop();
temp = temp->next;Â Â }
temp->next = NULL;
return Stack; }
int main() {
struct Node* head =NULL;
head = insert(head,2);
head = insert(head,4);
head = insert(head,8);
head = insert(head,10);
print(head);
head= reverse(head);
print(head);
head = recursiveReverse(head);
print(head);
head = void Reverse(head);
void print(head);
return 0;
}

reverse the linked list (2-4-8-10) by- stack- iteration- recursion- U.docx

coding in C; Create a function called reverseList that takes the head of a linked list, reverses the order of all the nodes. For example, if the list contained 1, 2, 3, 4 in its nodes, the list will now contain 4, 3, 2, 1. please add a main test. Solution Please find the required program below: #include<stdio.h> #include<stdlib.h> struct node { int data; struct node* next; }; void push(struct node** head, int new_element) { struct node* new_node = (struct node*) malloc(sizeof(struct node)); new_node->data = new_element; new_node->next = (*head); (*head) = new_node; } void printList(struct node *head) { struct node *temp = head; while(temp != NULL) { printf(\"%d \", temp->data); temp = temp->next; } } void reverseList(struct node** head) { struct node* prev = NULL; struct node* current = *head; struct node* next; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } *head = prev; } int main() { struct node* head = NULL; push(&head, 12); push(&head, 14); push(&head, 5); push(&head, 8); printf(\"\ Linked list Before Reversing :\ \"); printList(head); reverseList(&head); printf(\"\ Linked list After Reversing :\ \"); printList(head); } -------------------------------------------------------------------------------- OUTPUT: Linked list Before Reversing : 8Â Â 5Â Â 14Â Â 12 Linked list After Reversing : 12Â Â 14Â Â 5Â Â 8 .

Implement of c & its coding programming by sarmad baloch

Sarmad Baloch

#includeiostream struct node { char value; struct no.pdf

ankitmobileshop235

#include struct node { char value; struct node *next; }; class StringOfNode { struct node * head; public: StringOfNode() { head=NULL; } int size() { struct node *p=head; int count=1; while(p) { count++; p=p->next; } return count; } struct node * begin() { return head; } char &operator[](int index) { int i=0; struct node *p=head; while(index!=i) { p=p->next; i++; } return p->value; } void insertFront(char valuee) { struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(head==NULL) head=temp; else { temp->next=head; head=temp; } } void insert(char valuee, int pos) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; int i=0; if(p==NULL) { head=temp; } else { while(i!=pos-1) { p=p->next; i++; } temp->next=p->next; p->next=temp; } } void insertBack(char valuee) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(p==NULL) { head=temp; } else { while(p->next!=NULL) p=p->next; p->next=temp; } } void print() { struct node *p=head; while(p) { std::cout<value<<\"\ \"; p=p->next; } } StringOfNode operator+(const StringOfNode& b) { StringOfNode *S; struct node *p=head; while(p->next!=NULL) p=p->next; p->next=S->begin(); } }; int main() { StringOfNode *p1=new StringOfNode; p1->insertBack(\'b\'); p1->insertBack(\'c\'); p1->insertBack(\'e\'); p1->insert(\'d\',2); p1->insertFront(\'a\'); p1->print(); system(\"pause\"); return 0; } Solution #include struct node { char value; struct node *next; }; class StringOfNode { struct node * head; public: StringOfNode() { head=NULL; } int size() { struct node *p=head; int count=1; while(p) { count++; p=p->next; } return count; } struct node * begin() { return head; } char &operator[](int index) { int i=0; struct node *p=head; while(index!=i) { p=p->next; i++; } return p->value; } void insertFront(char valuee) { struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(head==NULL) head=temp; else { temp->next=head; head=temp; } } void insert(char valuee, int pos) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; int i=0; if(p==NULL) { head=temp; } else { while(i!=pos-1) { p=p->next; i++; } temp->next=p->next; p->next=temp; } } void insertBack(char valuee) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(p==NULL) { head=temp; } else { while(p->next!=NULL) p=p->next; p->next=temp; } } void print() { struct node *p=head; while(p) { std::cout<value<<\"\ \"; p=p->next; } } StringOfNode operator+(const StringOfNode& b) { StringOfNode *S; struct node *p=head; while(p->next!=NULL) p=p->next; p->next=S->begin(); } }; int main() { StringOfNode *p1=new StringOfNode; p1->insertBack(\'b\'); p1->insertBack(\'c\'); p1->insertBack(\'e\'); p1->insert(\'d\',2.

#include iostream #includestdlib.h using namespace std;str.pdf

lakshmijewellery

#include #include using namespace std; struct node { //Task 1 char val; node *next; }; void printList(node *head) { //Task 2 if (head == NULL) { cout << \"Empty list\" << endl; return; } node *temp = head; int count = 0; while(temp != NULL) { cout << \"Node \" << count << \": \" << temp ->val << endl; count++; temp = temp->next; } } struct node *globalhead = NULL; void deepCopy(node *head1, node *head2) { //Task 3 if (head1 == NULL) { cout << \"Empty list\" << endl; return; } node *temp = head1; node *curr = NULL; node *prev = NULL; while(temp != NULL) { curr = (struct node *)malloc(sizeof(struct node)); curr->val = temp->val; if(prev==NULL){ head2 = curr; globalhead = curr; }else{ prev->next = curr; } prev = curr; temp = temp->next; } } int main() { //Task 4 // Example #1 node p00, p01, p02, p03, p10, p11, p12, p13; // List 1: A B B B p00.val = \'A\'; p00.next = &p01; p01.val = \'B\'; p01.next = &p02; p02.val = \'B\'; p02.next = &p03; p03.val = \'B\'; p03.next = NULL; p10.next = NULL; cout<<\"First List:\"< Solution //Tested on Windos OS Dev c++ /*********************LinkedList.cpp********************/ #include #include using namespace std; struct node { //Task 1 char val; node *next; }; void printList(node *head) { //Task 2 if (head == NULL) { cout << \"Empty list\" << endl; return; } node *temp = head; int count = 0; while(temp != NULL) { cout << \"Node \" << count << \": \" << temp ->val << endl; count++; temp = temp->next; } } struct node *globalhead = NULL; void deepCopy(node *head1, node **head2) { //Task 3 if (head1 == NULL) { cout << \"Empty list\" << endl; return; } node *temp = head1; node *curr = NULL; node *prev = NULL; while(temp != NULL) { curr = (struct node *)malloc(sizeof(struct node)); curr->val = temp->val; curr->next = NULL; if(*head2==NULL){ *head2=curr; }else{ prev=*head2; while(prev->next!=NULL){ prev=prev->next; } prev->next=curr; } temp = temp->next; } } int main() { //Task 4 // Example #1 node p00, p01, p02, p03, p10, p11, p12, p13; // List 1: A B B B p00.val = \'A\'; p00.next = &p01; p01.val = \'B\'; p01.next = &p02; p02.val = \'B\'; p02.next = &p03; p03.val = \'B\'; p03.next = NULL; p10.next = NULL; cout<<\"First List:\"<.

Lab-2.2 717822E504.pdf

21E135MAHIESHWARJ

137 Lab-2.2.pdf

21E135MAHIESHWARJ

C++Write a method Node Nodereverse() which reverses a list..pdf

arjunenterprises1978

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(); }.

Help explain the code with line comments public class CompletedList implements ListADT, Iterable { //Three instance variables protected int count; protected int modChange; protected DoubleLinearNode head, tail; //head and tail set to empty count set to zero public CompletedList() { head = tail = null; count = 0; } //Override for head and tail, remove first number of list @Override public T removeFirst() throws NoSuchElementException { //if head is empty return error if(isEmpty()) throw new NoSuchElementException("The collection is empty"); //Temp for each number set to head item T temp = head.getItem(); //if first number equals last they are null if(head == tail) { head = tail = null; } //else the ne else if(head.getNext() == tail) { head.remove(); head = tail; } else { head = head.getNext(); head.getPrev().remove(); } count--; modChange++; return temp; } @Override public T removeLast() throws NoSuchElementException { if(isEmpty()) throw new NoSuchElementException("The collection is empty"); T temp = tail.getItem(); if(head == tail) { head = tail = null; } else if(head.getNext() == tail){ tail.remove(); tail = head; } else { tail = tail.getPrev(); tail.getNext().remove(); } count--; modChange++; return temp; } @Override public T remove( T element ) { if(isEmpty()) throw new NoSuchElementException("The target is not in the collection"); if(tail.getItem().equals(element)) { return removeLast(); } for (DoubleLinearNode current = head; current != null; current = current.getNext()) { if(current.getItem().equals(element)) { if(current == head) { return removeFirst(); } else { count--; modChange++; return current.remove(); } } } throw new NoSuchElementException("The target is not in the collection"); } @Override public T first() { if (isEmpty()) throw new NoSuchElementException("The collection is empty"); return head.getItem(); } @Override public T last() { if (isEmpty()) throw new NoSuchElementException("The collection is empty"); return tail.getItem(); } @Override public boolean contains( T target ) { if(isEmpty()) return false; if(tail.getItem().equals(target)) { return true; } for (DoubleLinearNode current = head; current != null; current = current.getNext()) { if(current.getItem().equals(target)) { return true; } } return false; } @Override public boolean isEmpty() { return count == 0; } @Override public int size() { return count; } @Override public Iterator iterator() { return new ListIterator(head); } @Override public String toString() { StringBuilder sb = new StringBuilder(); for (DoubleLinearNode current = head; current != null; current = current.getNext()) { sb.append(current.getItem()).append(' '); } return sb.toString(); } private class ListIterator implements Iterator { private DoubleLinearNode current; private int modNum; public ListIterator( DoubleLinearNode node) { this.modNum = modChange; this.current = node; } @Override public boolean hasNext() { if(this.modNum != modChange) throw new ConcurrentModificationException("The collection.

Solution#includestdio.h#includeconio.h#includealloc.h.pdf

poddaranand1

Solution : #include #include #include //------------------------------------------------- struct node { int data; struct node *next; }*start=NULL; //------------------------------------------------------------ void creat() { char ch; do { struct node *new_node,*current; new_node=(struct node *)malloc(sizeof(struct node)); printf(\"nEnter the data : \"); scanf(\"%d\",&new_node->data); new_node->next=NULL; if(start==NULL) { start=new_node; current=new_node; } else { current->next=new_node; current=new_node; } printf(\"nDo you want to creat another : \"); ch=getche(); }while(ch!=\'n\'); } //------------------------------------------------------------------ void display() { struct node *new_node; printf(\"The Linked List : n\"); new_node=start; while(new_node!=NULL) { printf(\"%d--->\",new_node->data); new_node=new_node->next; } printf(\"NULL\"); } //---------------------------------------------------- void main() { create(); display(); } //---------------------------------------------------- Output : #include #include #include #include struct node { int data; int key; struct node *next; struct node *prev; }; //this link always point to first Link struct node *head = NULL; //this link always point to last Link struct node *last = NULL; struct node *current = NULL; //is list empty bool isEmpty(){ return head == NULL; } int length(){ int length = 0; struct node *current; for(current = head; current != NULL; current = current->next){ length++; } return length; } //display the list in from first to last void displayForward(){ //start from the beginning struct node *ptr = head; //navigate till the end of the list printf(\"\ [ \"); while(ptr != NULL){ printf(\"(%d,%d) \",ptr->key,ptr->data); ptr = ptr->next; } printf(\" ]\"); } //display the list from last to first void displayBackward(){ //start from the last struct node *ptr = last; //navigate till the start of the list printf(\"\ [ \"); while(ptr != NULL){ //print data printf(\"(%d,%d) \",ptr->key,ptr->data); //move to next item ptr = ptr ->prev; printf(\" \"); } printf(\" ]\"); } //insert link at the first location void insertFirst(int key, int data){ //create a link struct node *link = (struct node*) malloc(sizeof(struct node)); link->key = key; link->data = data; if(isEmpty()){ //make it the last link last = link; }else { //update first prev link head->prev = link; } //point it to old first link link->next = head; //point first to new first link head = link; } //insert link at the last location void insertLast(int key, int data){ //create a link struct node *link = (struct node*) malloc(sizeof(struct node)); link->key = key; link->data = data; if(isEmpty()){ //make it the last link last = link; }else { //make link a new last link last->next = link; //mark old last node as prev of new link link->prev = last; } //point last to new last node last = link; } //delete first item struct node* deleteFirst(){ //save reference to first link struct node *tempLink = head; //if only one link if(head->next == .

DSA(1).pptx

DaniyalAli81

Ds 2 cycleChaitanya Kn

Lab-2.4 101.pdf

21E135MAHIESHWARJ

data structure3.pptx

SajalFayyaz

Linked lists

GowriKumar Chandramouli

I need to implment a function that can reverse a single linked list..pdf

rohit219406

I need to implment a function that can reverse a single linked list. Not just print it, but actually change the order of the list. THIS MUST BE DONE USING RECURSION EX: a, b, c, d -> d, c, b, a You have access to the start of the list. Code must be in C++ Solution #include using namespace std; struct node { char data; struct node* next; }; void reverse(struct node* head) { if (head == NULL) //base case return; reverse(head->next); //calling function recursively cout<data<<\",\"; } //add data to the front of the linked list void add(struct node** head_ref, char data) { //create new node struct node* newnode=new node(); //adding data newnode->data = data; //adding node to the front newnode->next = (*head_ref); //making new node head (*head_ref) = newnode; } int main() { // create linked list a, b, c, d struct node* head = NULL; add(&head, \'d\'); add(&head, \'c\'); add(&head, \'b\'); add(&head, \'a\'); reverse(head); return 0; } //OUTPUT*********** d,c,b,a, //OUTPUT*********** //This code has been tested on g++ compiler,Please ask in case of any doubt,Thanks..

Data structures cs301 power point slides lecture 03Nasir Mehmood

Hi,you covered mostly things.there are issue to point and link poi.pdf

aryan9007

Hi, you covered mostly things.there are issue to point and link pointer.I will try my best to solve your issue. I have added below functions which are wrong or missing. city* addCity(city *head, city *previous, string cityName ) { if (head == NULL) { head = new city; head->name = cityName; return head; } city *newNode = new city; newNode->name = cityName; if (previous == NULL) { newNode->next = head; head = newNode; return head; } newNode->next = previous->next; previous->next = newNode; cout << \"prev: \" << previous->name << \" new: \" << cityName << endl; return head; } newNode->name = cityName; city *current = new city; current = searchNetwork(head, previous->name); newNode->next = current->next; previous->next = newNode; cout << \"prev: \" << previous->name << \" new: \" << cityName << endl; city *searchNetwork(city *ptr, string cityName) { city *current = ptr; while (current->name != cityName) { if (current->name != cityName && current->next == NULL) return NULL; current = current->next; } return current; } city* deleteCity(city *head, string cityName) { city *current = head; cout << \"Checkpoint 1\" << endl; while (current->next->name != cityName) { if (current->next == NULL) { cout << \"City does not exist.\" << endl; return head; } current = current->next; } cout << \"Checkpoint 2\" << endl; current->next = current->next->next; return head; } void printPath(city *ptr) { cout << \"== CURRENT PATH ==\" << endl; // If the head is NULL if (ptr == NULL) cout << \"nothing in path\" << endl; else { city *temp2 = ptr; while( temp2!=NULL ) { cout << temp2->name << \" -> \";// show the data in the linked list temp2 = temp2->next; // tranfer the address of \'temp->next\' to \'temp\' } cout << \"NULL\" << endl; } cout << \"===\" << endl; } city* loadDefaultSetup(city *head) { head = deleteEntireNetwork(head); head = addCity(head,NULL,\"Los Angeles\"); city *two = new city; city *three = new city; city *four = new city; city *five = new city; city *six = new city; two->name = \"Phoenix\"; three->name = \"Denver\"; four->name = \"Dallas\"; five->name = \"Atlanta\"; six->name = \"New York\"; head->next = two; two->next = three; three->next = four; four->next = five; five->next = six; six->next = NULL; return head; } Solution Hi, you covered mostly things.there are issue to point and link pointer.I will try my best to solve your issue. I have added below functions which are wrong or missing. city* addCity(city *head, city *previous, string cityName ) { if (head == NULL) { head = new city; head->name = cityName; return head; } city *newNode = new city; newNode->name = cityName; if (previous == NULL) { newNode->next = head; head = newNode; return head; } newNode->next = previous->next; previous->next = newNode; cout << \"prev: \" << previous->name << \" new: \" << cityName << endl; return head; } newNode->name = cityName; city *current = new city; current = searchNetwork(head, previous->name); newNode->next = current->next; previous->next = newNode; c.

C++Write a function void headEnqueue(Queue q, int key) which enqu.pdf

arjuncp10

C++ Write a function void headEnqueue(Queue *q, int key) which enqueues the key such that, at function exit, it is positioned at the head of the queue rather than the tail; and the order and integrity of the queue is otherwise maintained. Solution struct node { int data; node* next; }; class Queue //linked_list class { private: node* front; node* rear; public: Queue(); void insert(int); headEnqueue(Queue *q, int key); }; Queue::Queue() { front = rear = NULL; } void Queue::insert(int val) { node* temp = new node; temp->data = val; temp->next = NULL; if (front == NULL) front= rear = temp; else { node* temp1 = front; while (temp1->next) temp1 = temp1->next; temp1->next = temp; } } void Queue::headEnqueue(Queue *q, int key) { node* temp = new node; temp->data = key; temp->next=q->front; q->front=temp; }.

Help I keep getting the same error when running a code. Below is the.pdf

mail931892

Help I keep getting the same error when running a code. Below is the code and the instructions about the code. import java.util.Arrays; public class Album implements Comparable { private final int id; private final String[] artists; private final int numSongs; public Album(int id, String title, String[] artists, int numSongs) { this.id = id; this.artists = new String[]{Arrays.toString(artists)}; this.numSongs = numSongs; } public int getId() { return id; } @Override public int compareTo(Album other) { return Integer.compare(this.numSongs, other.numSongs); } @Override public String toString() { StringBuilder artistNames = new StringBuilder(); for (String artist : artists) { artistNames.append(artist).append(", "); } artistNames.delete(artistNames.length() - 2, artistNames.length()); return "ID: " + id + " -- " + numSongs + " songs -- [" + artistNames + "]"; } } import java.util.Random; public class DoublyLinkedList { private static class Node { T data; Node prev; Node next; Node(T data) { this.data = data; } } private Node head; private Node tail; private int size; public DoublyLinkedList() { head = null; tail = null; size = 0; } public Node append(T data) { Node newNode = new Node<>(data); if (head == null) { head = newNode; } else { tail.next = newNode; newNode.prev = tail; } tail = newNode; size++; return newNode; } public Node insert(int location, T data) { if (location < 0 || location > size) { throw new IllegalArgumentException("Invalid location"); } Node newNode = new Node<>(data); if (location == 0) { newNode.next = head; if (head != null) { head.prev = newNode; } head = newNode; if (tail == null) { tail = newNode; } } else if (location == size) { tail.next = newNode; newNode.prev = tail; tail = newNode; } else { Node current = head; for (int i = 0; i < location; i++) { current = current.next; } newNode.prev = current.prev; newNode.next = current; current.prev.next = newNode; current.prev = newNode; } size++; return newNode; } public Node delete(int location) { if (location < 0 || location >= size) { throw new IllegalArgumentException("Invalid location"); } Node deletedNode; if (location == 0) { deletedNode = head; head = head.next; if (head != null) { head.prev = null; } else { tail = null; } } else if (location == size - 1) { deletedNode = tail; tail = tail.prev; tail.next = null; } else { Node current = head; for (int i = 0; i < location; i++) { current = current.next; } deletedNode = current; current.prev.next = current.next; current.next.prev = current.prev; } size--; return deletedNode; } public int getIndex(T data) { Node current = head; for (int i = 0; i < size; i++) { if (current.data.equals(data)) { return i; } current = current.next; } return -1; } @Override public String toString() { StringBuilder result = new StringBuilder(); Node current = head; while (current != null) { result.append(current.data).append(" -> "); current = current.next; } result.append("NULL"); return result.toString(); } public Node shuffle() { Random r.

Linked Stack program.docx

kudikalakalabharathi

operating system linux,ubuntu,Mac#include iostream #include .pdf

aquacareser

//operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << \" \"; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<\"Please Enter 3 argument\"< Solution //operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << \" \"; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<\"Please Enter 3 argument\"<.

Please find the answer to the above problem as follows- Program.pdf

angelfragranc

Please find the answer to the above problem as follows:- /* Program to insert in a sorted list */ #include #include /* Link list node */ struct node { int data; struct node* next; }; void sortedDelete(struct node** head_ref, int number) { struct node *temp = *head_ref; struct node *prev = NULL; if(*head_ref != NULL && (*head_ref)->data == number){ *head_ref = (*head_ref)->next; return; } while(temp != NULL) { if(temp->data == number){ prev->next = temp->next; free(temp); return; } prev = temp; temp = temp->next; } } /* function to insert a new_node in a list. Note that this * function expects a pointer to head_ref as this can modify the * head of the input linked list (similar to push())*/ void sortedInsert(struct node** head_ref, struct node* new_node) { struct node* current; /* Special case for the head end */ if (*head_ref == NULL || (*head_ref)->data >= new_node->data) { if(*head_ref != NULL&&(*head_ref)->data==new_node->data){ return; } new_node->next = *head_ref; *head_ref = new_node; } else { /* Locate the node before the point of insertion */ current = *head_ref; while (current->next!=NULL && current->next->data < new_node->data) { current = current->next; } if(current->next!=NULL && current->next->data == new_node->data){ return; } new_node->next = current->next; current->next = new_node; } } /* BELOW FUNCTIONS ARE JUST UTILITY TO TEST sortedInsert */ /* A utility function to create a new node */ struct node *newNode(int new_data) { /* allocate node */ struct node* new_node = (struct node*) malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; new_node->next = NULL; return new_node; } /* Function to print linked list */ void printList(struct node *head) { struct node *temp = head; if(temp == NULL) printf(\"\ \"); while(temp != NULL) { printf(\"%d \", temp->data); temp = temp->next; } } /* Drier program to test count function*/ int main(int argc, char *argv[]) { /* Start with the empty list */ struct node* head = NULL; struct node *new_node; FILE *fp; fp = fopen(argv[1], \"r+\"); if(!fp){ printf(\"error\"); return 1; } char op = \' \'; int number = 0; while(fscanf(fp, \"%c\\t%d\ \", &op, &number)!=EOF){ if(op==\'i\'){ new_node = newNode(number); sortedInsert(&head, new_node); }else if(op==\'d\'){ sortedDelete(&head, number); }else{ printf(\"error\"); return; } } printList(head); printf(\"\ \"); return 0; } Solution Please find the answer to the above problem as follows:- /* Program to insert in a sorted list */ #include #include /* Link list node */ struct node { int data; struct node* next; }; void sortedDelete(struct node** head_ref, int number) { struct node *temp = *head_ref; struct node *prev = NULL; if(*head_ref != NULL && (*head_ref)->data == number){ *head_ref = (*head_ref)->next; return; } while(temp != NULL) { if(temp->data == number){ prev->next = temp->next; free(temp); return; } prev = temp; temp = temp->next; } } /* function to insert a new_node in a list. Note that this * function expects a point.

#include iostream#include d_node.h #include d_nodel.h.docx

ajoy21

Write a C++ function that delete nodes in a doubly linkedlist- It shou.docx

noreendchesterton753

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 to fill-in TODOs in .cpp file and in .h file Could some.pdf

forladies

I need to fill-in //TODO\'s in .cpp file and in .h file Could someone help me at least with few of them to give me an idea how deal with it. ***SinglyLinkedList.cpp #include #include #include \"SinglyLinkedList.h\" void test_constructor() { SinglyLinkedList lst = {100, 200, 300, 400, 500}; assert(*lst.at(0) == 100); assert(*lst.at(1) == 200); assert(*lst.at(2) == 300); assert(*lst.at(3) == 400); assert(*lst.at(4) == 500); assert(lst.size() == 5); } void test_remove() { SinglyLinkedList lst = {100, 200, 300, 400, 500}; lst.remove(2); assert(*lst.at(0) == 100); assert(*lst.at(1) == 200); assert(*lst.at(2) == 400); assert(*lst.at(3) == 500); assert(lst.size() == 4); } void test_insert() { // TODO } void test_push_back() { // TODO } void test_push_front() { // TODO } void test_append() { // TODO } void test_sum() { // TODO } int main() { test_constructor(); test_remove(); test_insert(); test_push_back(); test_push_front(); test_append(); test_sum(); } ***SinglyLinkedList.h #include #include template class SinglyLinkedList { // Nested class representing the nodes in the list. class SinglyLinkedListNode { public: // The value stored in this node. T value; // The next node in the sequence. SinglyLinkedListNode *next; SinglyLinkedListNode(T value) : value(value), next(nullptr) {} SinglyLinkedListNode(T value, SinglyLinkedListNode *next) : value(value), next(next) {} // Return the size (length) of the linked list. std::size_t size(); }; SinglyLinkedListNode *head; SinglyLinkedListNode *tail; public: // Constructs a new SinglyLinkedList from an initializer_list of type T[]. // This is mostly for convenience, especially when testing. SinglyLinkedList(std::initializer_list items) : head(nullptr), tail(nullptr) { if (items.size() == 0) { return; } // initializer_lists were designed to be used iteratively, // so thats what we do. // Can you think of how to write this recursively? auto it = items.begin(); while (it != items.end()) { this->push_back(*it++); } } // Return a pointer to the value at the given index. // If the index is larger than the size of the list, // return a nullptr. // // ASIDE: We will cover exceptions later. T* at(std::size_t i); // Pushes a new node onto the back of the list. void push_back(T value); // Pushes a new node onto the front of the list. void push_front(T value); // Return the size (length) of the linked list. std::size_t size(); // Remove the specified node from the list. void remove(std::size_t i); // Insert the value at the index. void insert(std::size_t i, T value); // Append the given list to this one. void append(SinglyLinkedList list); }; template T* SinglyLinkedList::at(std::size_t i) { // TODO } template void SinglyLinkedList::push_back(T value) { // TODO // Make sure that this is a O(1) operation! } template void SinglyLinkedList::push_front(T value) { // TODO // Make sure that this is a O(1) operation! } template void SinglyLinkedList::remove(std::size_t i) { // TODO // Don\'t forget to not only unlink the node, b.

Research the issues that you might face as a network administrator whe.docx

acarolyn

Research the issues that you might face as a network administrator when moving an enterprise network from IPv4 to IPv6. Discuss the changes you will need to make in your day-to-day activities managing the network or the tools you will need. Assume the addressing scheme has been worked out. How will this change impact your troubleshooting process, your tools, and the process you use to identify network errors? Solution The task of switching from IPv4 to IPv6 is a daunting task since virtually every device needs to support it. Itâ€™s best to approach a cost-effective implementation in steps or phases, breaking down your network security and systemâ€™s environment into smaller components. There are several important factors to consider before implementing an IPv6 strategy. These include: IP Address Management Address Management is one of the most important factors to understand before you implement an IPv6 strategy. Many of the ideas that governed the design of IPv4 networks are not necessary in IPv6, and in some cases hinder it. When designing IPv4 networks, conservation is a driving factor. With IPv6, this is no longer a concern.For this reason, transitioning from IPv4 to IPv6 may feel uncomfortable at first. The designers of IPv6 have recommended ISP internet service allocate /48 networks to all customers with the exception of very large or mobile customers. It is highly recommended to run a /64 on all LANs, regardless of size. Many protocols in IPv6 such as Stateless Address Auto-configuration (SLAAC) require the use of /64 on all LAN interfaces. It is highly recommended not to use more specific networks such as /96 for your LANs. The exception to this would be for assigning addresses to WAN point to point links and loopback interfaces. WAN links can be assigned /126 networks, and loopback interfaces should have /128 addresses. Dual Stack Support Dual Stack Support is a fundamental requirement for implementing IPv6. Your network security design may include the typical core, distribution, and access layers. Implementing IPv6 in the core first, then through the distribution and access layers is optimal. Prepare a plan of action to research a network problem. Conduct the research needed to isolate the problem. Use a structured approach to identify a network problem. Identify the severity of a network problem based on its initial symptoms. Detecting errors like dropped packets or retransmissions on the network level is relatively easy. Figuring out if those errors affect the performance and connectivity of your services is however another matter. Some network errors are mitigated and compensated for by network protocols and active networking components, like network interfaces. Meanwhile, other network errors lead to performance problems that negatively affect your services. .

Reset Help By the Brnsted-Lowry definition- acids are proton donors an.docx

acarolyn

Reset Help By the BrÃ¸nsted-Lowry definition, acids are proton donors and bases are proton acceptors. By the Lewis definition, acids are electron-pair acceptors and bases are electron-pair donors. For bases, the two definitions are equivalent such that all Lewis bases are BrÃ¸nsted-Lowry bases and vice versa However, it is possible to have a Lewis acid that is not a BrÃ¸nsted-Lowry acid. This is because Lewis acids include molecules and cations that have a vacant valence orbital, regardless of whether they have a proton to donate Common examples of Lewis acids (that are not BrÃ¸nsted-Lowry acids) are metal ions, such as Al+and Cu2+. The following is an example ofa Lewis acid-base reaction Lewis acids Lewis bases NH3 donates the electron pair to Cu2+. Therefore NH3 is a Lewis base, and Cu2 is a Lewis acid SubmitPrev Previous Answer Solution 1. Lewis acids: Lewis acids are the species which accepts electrons. SO 3 since S has only three electron domains. Hence it accepts electrons. Fe 2+ is electron deficient. Hence it accepts electrons too. CO 2 where C accepts electrons since it is bound to two highly electronegative oxygen atoms. Lewis bases are species that donates electrons. I - , CO, H - , CH 3 NH 2 2. The order of decreasing pH is: Ba(OH) 2 > NaOH > N 2 H 4 > HOCl > HCl Ba(OH) 2 decomposes to give 2 OH - ions. This is why it has higher pH than NaOH which decomposes to give only 1 OH - ion. Hydrazine N 2 H 4 is basic however lesser than NaOH and Ba(OH) 2 HOCl is a weak acid that partially dissociates to give H + and ClO - HCl is the strongest acid that completely dissociates into H + and Cl - and hence has the lowest pH. More the H + ions in solution, lesser the pH. More the OH - ions in solution, higher the pH. 3. Given, [ OH - ] = 2.3 x 10 -2 M We know that at 25 0 C, [H + ] [OH - ] = Kw = Ionic product of water = 10 -14 Putting the value of [OH - ] in the above equation, we have, [H + ] x 2.3 x 10 -2 = 10 -14 [H + ] = 10 -14 / (2.3 x 10 -2 ) [H + ] = 0.44 x 10 -12 M .

Reed Pentak a finance major has been following globalization and made.docx

acarolyn

Reed Pentak a finance major has been following globalization and made the following observation concerning accounting convergence. \" I do not see many obstacles concerning development of a single accounting standard for inventories. Prepare a response to Reed to explain the main obstacle to achieving convergence in the area of inventory accounting. Solution The greater emphasis on convergence is a result of the wave of globalization that has swept our world. Today more and more companies are becoming multinational. Companies have operations in different parts of the world with increasing number of subsidiaries, joint ventures and franchises. The capital markets of the United States, European Union, Eastern Europe, Asia, and Australia have become intertwined. Investors now have more access than ever to financial markets and foreign companies. It is imperative to continue to improve the efficiency of such markets so that they remain healthy However, a major obstacle that lingers in these markets is that the U.S., Europe and Japan have different regulatory standards for financial reporting. The translation of financial statements from one type of standard to another is burdensome because of time and cost required. Some companies prefer not to list on U.S. stock exchanges, to avoid the complexities associated with the U.S. reporting. Also, investors have to be aware of the differences when evaluating foreign companies, and this distorts comparability and consistency. Financial statements are less relevant to its users when they are not fully comparable. The consequence is that some investors choose not to invest in foreign companies to avoid potential misunderstanding of financial statements, and hence making a wrong investment decision. In the short term the disadvantages are not so obvious, but in the long run this can impede integration of capital markets, and efficient allocation of capital which is vital for a strong world economy. In 2000, FASB stated that it .

Redesign- Make a prediction relating heat to the movement of molecules.docx

acarolyn

Redesign: Make a prediction relating heat to the movement of molecules. How does this relate to fluorescent lighting? Gas Liquid Solution generally the , Solids - vibrational (and rotational); moving very slowly. Liquids - vibrational, rotational, and translational; moving quickly. Gases - vibrational, rotational, and translational; moving very quickly. Explanation: Particles are always moving. No matter what. There is a certain point where they move so slowly it is considered minuscule, but they are in fact moving. Thus, in solids, liquids, and gases, you can expect particles to move in a certain trend. That trend is average kinetic energy of the particles and the forces holding them together (attraction). Solids When you have little to no heat, the particles are moving very slowly, so slowly that they atomic attractions result in bonds that are difficult to break without adding energy. At this stage, this is known as a solid. In a solid, the particles are vibrating. Only some solids have the ability to rotate. They cannot translate regardless. Liquids When you add energy (heat) to a solid, you \"excite\" the particles so much that they break their bond; their attraction is too weak to hold them together - yet strong enough to keep a form As a result, you get a liquid. In a liquid, the particles are not only vibrating (their bonds are not present but they still vibrate), but they are also rotating and translating. The particles are moving much faster than in a solid. Gases When you add even more energy to the substance, you increase the kinetic energy of those particles so much, that they lose their state form, becoming a gas. In a gas, particles have vibrational, rotational, and translational motion, allowing them to \"bounce\" off of one another. Particles in a gas state move much faster than a liquid. Light is a form of energy that can be released by an atom. It is made up of many small particle-like packets that have energy and momentum but no mass. These particles, called light photons, are the most basic units of light. When the atom gains or loses energy, the change is expressed by the movement of electrons. When something passes energy on to an atom -- heat , for example -- an electron may be temporarily boosted to a higher orbital (farther away from the nucleus). The electron only holds this position for a tiny fraction of a second; almost immediately, it is drawn back toward the nucleus, to its original orbital. As it returns to its original orbital, the electron releases the extra energy in the form of a photon, in some cases a light photon .

Recycling and composting are important parts of an overall management.docx

acarolyn

Recycling and composting are important parts of an overall management scheme, but their connections to other management practices must be understood. Discuss how a major increase in domestic-waste recycling and composting affects management schemes that rely on: Bulk incineration Strictly landfilling Solution Bulk Incineration One of the most effective methods of reducing the volume and weight of solid waste is by burning it in a well designed furnace.\' When the available land is scarce, disposal requirements are string out and destruction of toxic water is necessary, incineration is the best method for treatment of solid waste. There are mainly stages involved in the process of incineration first drying and them combustion. Drying and combustion may be accomplished either in separate mits as successively in the same unit depending on temperature constraints and control parameter. Estimation of oxygen requirement and heat balance are very vital for efficient functioning of the incineration process. This method is widely used in countries like USA where the calorific value of refuse is high and hence are quite suitable for burning. Large scale air pollution particularly due to the emission of dioxins remains a serious problem with the incinerators. The process is a bit expensive and requires a lot of technical know how. Strictly Land Filling Important aspect in the implementation of sanitary landfills include : (1) site selection, (2) land filling methods (3) occurrence of gases and leachate in landfills and (4) movement and control of landfill gases and leachate.\' The following event occur when major increase in domestic-waste recycling and composting affects management schemes : Biological decay of organic materials (aerobically/ anaerobically) with evolution of gases and liquids. Chemical oxidation of waste materials. Escape of gases from the fill. Movement of liquids caused by differential heads. dissolving and leaching of organic and inorganic materials by water and leachate moving through the fills. movement of dissolved material by concentration gradients and osmosis. uneven settlement caused by consolidation of material into voids. Gases found in landfills include air, ammonia, carbon dioxide, carbon monoxide, hydrogen, hydrogen, sulfide, methane, nitrogen and oxygen. Carbon dioxide and methane are the principal gases produced from the anaerobic decomposition of the organic solid-waste components. The movement of gases in landfills can be controlled by constructing vents and barriers and by gas recovery. The movement of landfill gases can be controlled by landfill sealants. Compacted clay is most commonly used as landfill sealants. .

R5232- RS422- 20mA Current Loop and 1EE488 are typical standard interf.docx

acarolyn

R5232, RS422, 20mA Current Loop and 1EE488 are typical standard interfaces used in communication. Compare and describe these protocols. Solution RS-232 is the most common serial interface and ships as a standard component on most Windows-compatible desktop computers. RS-232 only allows for one transmitter and one receiver on each line. RS-232 also uses a Full-Duplex transmission method. Some RS-232 boards sold by National Instruments support baud rates up to 1 Mbit/s, but most devices are limited to 115.2 kbits/s. Note that RS-422/RS-485 interface is not available on most IBM PCs. RS-422 (EIA RS-422-A Standard) is the serial connection used on Apple computers. It provides a mechanism for transmitting data up to 10 Mbits/s. RS-422 sends each signal using two wires to increase the maximum baud rate and cable length. RS-422 is also specified for multi-drop applications where only one transmitter is connected to, and transmits on, a bus of up to 10 receivers. 20 mA Current loop: The current loop interface for serial data communications is used in a number of applications. As the name implies, the current loop serial data communications scheme uses the level of current instead of voltage to provide the signalling method. Often the scheme is called a 20mA current loop interface in view of the standard level of current that is often used. Line losses are not usually significant, Can be used for long distances, Can be isolated from ground, Provided a simple form of networking IEEE488 or GPIB (General Purpose Interface Bus) : GPIB is a standardized communications bus consisting of 16 signal lines (wires) used to operate up to 15 instruments simultaneously via a parallel connection. It Can be used to control up to 15 instruments simultaneously from one control device (PC). But, It is only good over short distances 60 ft and requires the purchase of extensive hardware to setup and extend operating range. Highly susceptible to noise over long distances. .

RADIUS provides three services- authentication- authorization- and acc.docx

acarolyn

RADIUS provides three services: authentication, authorization, and accounting. RADIUS facilitates centralized user administration and keeps all user profiles in one location that all remote services share. Describe what happens at each stage of RADIUS Solution RADIUS:- Remote Authentication Dial-In User Service (RADIUS) is a networking protocol that provides centralized Authentication, Authorization, and Accounting management for users who connect and use a network service. Authentication : The user or machine sends a request to a Network Access Server (NAS) to gain access to a particular network resource using access credentials. The credentials are passed to the NAS device via the link-layer protocol. This request includes access credentials, typically in the form of username and password or security certificate provided by the user. The RADIUS server checks that the information is correct using authentication schemes such as PAP, CHAP or EAP. The user\'s proof of identification is verified, along with, optionally. RADIUS servers checked the user\'s information against a locally stored flat file database. Modern RADIUS servers can do this, or can refer to external sources to verify the user\'s credentials. The RADIUS server then returns one of three responses to the NAS : 1) Access Reject, 2) Access Challenge,3) Access Accept. Authorization: Authorization attributes are conveyed to the NAS stipulating terms of access to be granted. For example, the following authorization attributes may be included in an Access-Accept -The specific IP address to be assigned to the user -The address pool from which the user\'s IP should be chosen -The maximum length of time that the user may remain connected -Quality of Service (QoS) parameters Accounting:- When network access is granted to the user by the NAS, an Accounting Start is sent by the NAS to the RADIUS server to signal the start of the user\'s network access. \"Start\" records typically contain the user\'s identification, network address, point of attachment and a unique session identifier.Periodically, Interim Update records may be sent by the NAS to the RADIUS server, to update it on the status of an active session. \"Interim\" records typically convey the current session duration and information on current data usage. Finally, when the user\'s network access is closed, the NAS issues a final Accounting Stop record to the RADIUS server, providing information on the final usage in terms of time, packets transferred, data transferred, reason for disconnect and other information related to the user\'s network access. .

Question- Federal funding agencies must form committees to decide whic.docx

acarolyn

Question : Federal funding agencies must form committees to decide which telescope projects will receive funds for construction. When deciding which projects wil be funded, the committees must consider: -that certain wavelengths of light are blocked from reaching Earth\'s surface by the atmosphere, -how efficiently telescopes work at different wavlengths, and -that telescopes in space are much more expensive to construct than Earth-based telescopes. Using these three criteria, consider each pairing of telescope proposals listed below. For each pair, state which of the two you would choose to fund and explain your reasoning. Pair A) Project Beta: An X-ray wavelength telescope, located near the North Pole, which will be used to examine the Sun. Project Alpha : An infrared wavelength telescope, placed on a satellite in orbit around Earth, which will be used to view supernovae. Pair B) Project Rho : A UV wavelength telescope, placed high atop Mauna Kea in Hawaii at 14,000 ft above sea level, which will be used to look at distant galaxies. Project Sigma : A visible wavelength telescope, placed on a satellite orbit around Earth, which will be used to observe a pair of binary stars located in the constellation Ursa Major. Pair C) Project Zeta: A radio wavelength telescope, placed on the floor of the Mojave Desert, which will be used to detect potential communications from distant civilizations outside our solar system. Project Epsilon: An infrared wavelength telescope, located in the high-elevation mountains of Chile, which will be used to view newly forming stars (protostars) in the Orion nebula. Solution From pair A, I will choose project alpha as some infrared waves can reach earth\'s surface. On the other hand, x-rays will be absorbed in the outer atompshere. So, I will choose project alpha. From pair B, I will select project rho as project sigma is space based and willl be expensive. so, project rho is more feasible. From pair C, Project epsilon because, .

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reverse the linked list (2-4-8-10) by- stack- iteration- recursion- U.docx

1. reverse the linked list (2,4,8,10) by: stack; iteration; recursion. Using C++ Solution #include<iostream> #include<stdio.h> #include<stdlib.h> #includ<stack> using namespace std; struct Node { int value; struct Node* next; }node; struct Node* reverse(struct Node* head) {Â Â // Iteration Method struct Node* current,*prev; current = head;

2. prev = NULL; while(current!=NULL) { next = current->next; current->next = prev; prev = current; current = next; } head = prev;- return head; } struct Node* insert(Node* head,int value) { Node* temp = (struct Node*)malloc(sizeof(struct Node)); temp->value = value; temp->next = NULL; if(head = = NULL) head = temp; else { Node*temp1=head; while(temp1->next!=NULL) temp1 = temp1->next; temp1->next = temp; } return head; } public void print(Node* head) { while(head!=NULL) {

3. cout<<head->value; head = head->next; } } struct Node*recursiveReverse(struct Node* &head) {Â Â //Recursion method if(head==NULL || head->next ==NULL) return; Node*first = head; Node *rest = head->next; if(rest !=NULL) return; recursiveReverse(rest); first ->next->next = first; first->next =NULL; head = rest; return head; } void Reverse(struct Node* head)Â Â // stack Implementation { stack<node*> Stack; node *traverse = head; while(traverse!=NULL) { Stack.push(traverse); traverse = traverse->next;Â Â } node *temp = Stack.top(); head = temp; Stack.pop();

4. while(!Stack.empty()) { temp->next = Stack.top(); Stack.pop(); temp = temp->next;Â Â } temp->next = NULL; return Stack; } int main() { struct Node* head =NULL; head = insert(head,2); head = insert(head,4); head = insert(head,8); head = insert(head,10); print(head); head= reverse(head); print(head); head = recursiveReverse(head); print(head); head = void Reverse(head); void print(head); return 0; }

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