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*/ }
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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*/ }
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- 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…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 File…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
- 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;…1-Write a C program to show operations on a singly linked where each node consists of integers. (a) Insert (At beginning, At end) (b) Delet (At beginning, At end) (c) Search (d)reverse (e) Print middle element 2 Write a C program using dynamic variable & pointers to construct a singly linked list consisting of the following information in each node. Student id (integer), student name (character string) & semester (integer). 3 Write a C program using dynamic variables & pointers to construct an ordered (ascending) singly linked list based on the rank of the student, where each node consists of the following information student id( integer) student name(character), rank(integer).Write a function that changes the places of the first and second nodes of a linked list. I.e. the First node will become the second node, and the second node will become the first node. THE FUNCTION WILL NOT EXCHANGE DATA, IT WILL CHANGE LINKS.
- 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.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 //…