Question

The following C function takes a singly-linked list as input argument. It modified by the list by moving the last element to the front of the list and returns the modified list. Some part of the code is left blank.

typed of struct node
{
int value;
struct node * next;
} Node;
Node *mode_to_ front(Node *head)
{
Node *p, *q;
if((head == NULL)$\parallel$(head $\to$ next == NULL))
return head;
q = NULL;
p = head;
while (p$\to$next != NULL)
{
q = p;
p = p$\to$ next;
}
______________________
return head;
}

Choose the correct alternative to replace the blank line.

• A. head = p; p->next = q; q$\to$ next = NULL;
• B. q = NULL; p$\to$ next = head; head=p;
• C. q$\to$ next = NULL; p$\to$ next = head; head = p;
• D. q$\to$ next = NULL; head = p; p$\to$ next = head;

Answer 1

The correct option is C q$\to$ next = NULL; p$\to$ next = head; head = p;

This is the program by moving last element to front of list and returns modified list.

When while (P$\to$ next != NULL) execute then p pass to q and now p point to next. When "while" condition is false then we get q$\to$ next = NULL;
p$\to$ next = head; head = p; means next to q made null, and points p $\to$ next is head and then make p as the head.