ollowing function reverse() is supposed to reverse a singly linked list. There is one line missing at the end of the function. /* Link list node */ struct node { int data; struct node* next; }; /* head_ref is a double pointer which points to head (or start) pointer of 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; } /*ADD A STATEMENT HERE*/ } What should be added in place of “/*ADD A STATEMENT HERE*/”, so that the function correctly reverses a linked list. a) *head_ref = prev; b) *head_ref = current; c) *head_ref =

ollowing function reverse() is supposed to reverse a singly linked list. There is one line missing at the end of the function. /* Link list node / struct node { int data; struct node next; }; /* head_ref is a double pointer which points to head (or start) pointer of 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; } /ADD A STATEMENT HERE/ } What should be added in place of “/ADD A STATEMENT HERE/”, so that the function correctly reverses a linked list. a) head_ref = prev; b) head_ref = current; c) *head_ref =

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter17: Linked Lists

Section: Chapter Questions

Problem 18SA

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Suppose you have two objects of Doubly Linked List D1 and D2. Each object is representing a different Doubly Linked List. Write a function to concatenate both doubly linked lists. The concatenation function must return the address of head node of concatenated Doubly Linked List, and it must take two Node* parameters. You need to keep in mind all exceptional cases for example what if either one of the doubly linked lists is empty? Perform this task on paper, take a clear picture of solution and paste it in answer section.
Consider the following function that takes reference to head of a Doubly Linked List as parameter. Assume that anode of doubly linked list has previous pointer as prev and next pointer as next.void fun(struct node **head_ref){struct node *temp = NULL;struct node *current = *head_ref;while (current != NULL){temp = current->prev;current->prev = current->next;current->next = temp;current = current->prev;}if(temp != NULL )*head_ref = temp->prev;}Assume that reference of head of following doubly linked list is passed to above function 1 <--> 2 <--> 3 <--> 4 <--> 5 <-->6. What should be the modified linked list after the function call? Note: solve as soon as possible
ANSWER THAT MCQ QUESTION IN 2 MINUTES PLEASE.. The following function reverse() is supposed to reverse a singly linked list. There is one line missing at the end of the function. /* Link list node */ struct node { int data; struct node* next; }; /* head_ref is a double pointer which points to head (or start) pointer of 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; } /*ADD A STATEMENT HERE*/ }
Suppose a node of a linked list is defined as follows in your program: typedef struct{ char data; struct Node* link; } Node; Write the definitions of two functions (i) Node* pop(Node* top) – which removes the top element of the stack, and (ii) bool compare(Node* topA, Node* topB) - which returns true if the both stacks have identical data. The function returns false if the data in both stacks are not identical. In the function definitions, top, topA, and topB point the top elements of corresponding stacks. You must use the following function definitions in your answer. The peek function has already been defined. You need to write the definition of pop and compare functions. You are not allowed to modify the parameters and/or return types of the functions. Node* pop(Node* top){ //write the function definition } char peek(Node* top){ return top->data; } bool compare(Node* topA, Node* topB){ //write the function definition } Write the time complexity in Big-O notation…
Using C languge, implement programmer defined-data types with linked lists. A set of integers may be implemented using a linked list.Implement the following functions given the definition:typedef struct node* nodeptr;typedef struct node{int data;nodeptr next;}Node;typedef Node* Set;Set initialize();- simply initialize to NULLvoid display(Set s);- display on the screen all valid elements of the listSet add(Set s, elem);- simply store elem in the listint contains(Set s, int elem);- search the array elements for the value elemSet getUnion(Set result, Set s1, Set s2);- store in the set result the set resulting from the union of s1 and s2- x is an element of s1 union s2 if x is an element of s1 or x is an element of s2Set intersection(Set result, Set s1, Set s2);- store in the set result the set resulting from the intersection of s1 and s2- x is an element of s1 intersection s2 if x is an element of s1 and x is an element of s2Set difference(Set result, Set s1, Set s2);- store in the set…
Suppose a node of a linked list is defined as follows in your program: typedef struct{ int data; struct Node* link; } Node; Write the definition of a function, printOdd(), which displays all the odd numbers of a linked list. What will be the time complexity of printOdd()? Express the time complexity in terms of Big-O notation.
Consider the below code of circular linked list. Your work is to update the same code, by providing the functionalities with tail pointer only. #include <iostream> using namespace std; class Node{ public: int data; Node *next; }; class List{ public: Node *head; List(){ head = NULL; } void insert(int data){ Node *nn = new Node; nn->data = data; if(head==NULL){ head = nn; nn->next = head; cout<<"Node inserted successfully at head"<<endl; } else{ Node *temp = head; while(temp->next!=head) temp = temp->next; nn->next = head; temp->next=nn; cout<<"Node inserted successfully at end"<<endl; } } void insertInBetween(Node *node, int data){ Node *nn = new Node; nn->data = data; Node *temp = node->next;…
This is in c++ Given the MileageTrackerNode class, complete main() to insert nodes into a linked list (using the InsertAfter() function). The first user-input value is the number of nodes in the linked list. Use the PrintNodeData() function to print the entire linked list. Don't print the dummy head node. Example input : 3 2.2 7/2/18 3.2 7/7/18 4.5 7/16/18 output: 2.2, 7/2/18 3.2, 7/7/18 4.5, 7/16/18 main.cpp: #include "MileageTrackerNode.h"#include <string>#include <iostream>using namespace std; int main (int argc, char* argv[]) {// References for MileageTrackerNode objectsMileageTrackerNode* headNode;MileageTrackerNode* currNode;MileageTrackerNode* lastNode; double miles;string date;int i; // Front of nodes listheadNode = new MileageTrackerNode();lastNode = headNode; // TODO: Read in the number of nodes // TODO: For the read in number of nodes, read// in data and insert into the linked list // TODO: Call the PrintNodeData() method// to print the entire linked list //…
This is in c++ Given the MileageTrackerNode class, complete main() to insert nodes into a linked list (using the InsertAfter() function). The first user-input value is the number of nodes in the linked list. Use the PrintNodeData() function to print the entire linked list. Don't print the dummy head node. Example input : 3 2.2 7/2/18 3.2 7/7/18 4.5 7/16/18 output: 2.2, 7/2/18 3.2, 7/7/18 4.5, 7/16/18 main.cpp: #include "MileageTrackerNode.h"#include <string>#include <iostream>using namespace std; int main (int argc, char* argv[]) {// References for MileageTrackerNode objectsMileageTrackerNode* headNode;MileageTrackerNode* currNode;MileageTrackerNode* lastNode; double miles;string date;int i; // Front of nodes listheadNode = new MileageTrackerNode();lastNode = headNode; // TODO: Read in the number of nodes // TODO: For the read in number of nodes, read// in data and insert into the linked list // TODO: Call the PrintNodeData() method// to print the entire linked list //…
1. Suppose you have already developed an SLL ADT. You have to search a specificelement of the linked list. If you find the element in the list then you have to swapits previous node with its next node. Otherwise, simply print the message “Datanot found”. Now write a function in your SLL ADT with this functionality.
Write a void function that takes a linked list of integers and removes all the duplicate elements from the list. The function will have one call-by-reference parameter that is a pointer to the head of the list. After the function is called, this pointer will point to the head of a linked list that has the nodes that have all the distinct integers. Place your function in a suitable test program.
Write a function that gets a linked list of ints, and reverses it. For example - on input 1 -> 2 -> 3 -> 4, after the function finishes the execution, the list becomes 4 -> 3 -> 2 -> 1 - If the list has one element, then it doesn’t change - If the list is empty, then it doesn’t change You may use the data fields in the struct and the functions provided in LL.h and LL.c. // reverses a linked list void LL_reverse(LL_t* list); Test for the Function: void test_q3() { LL_t* lst = LLcreate(); LL_add_to_tail(lst, 1); LL_add_to_tail(lst, 3); LL_add_to_tail(lst, 8); LL_add_to_tail(lst, 4); LL_add_to_tail(lst, 3); LL_reverse(lst); intcorrect[] = {3,4,8,3,1}; inti; node_t* n = lst->head; for(i=0;i<5;i++) { if (n==NULL) { printf("Q3 ERROR: node %d==NULL unexpected\n", i); return; } if (n->data != correct[i]) { printf("Q3 ERROR: node%d->data==%d, expected %d\n", i, n->data, correct[i]); return; } n = n->next; } if (n==NULL) printf("Q3 ok\n"); } Support…