GATE-CS-2015 (Set 3) - Section 65

This is a sample question from the GATE-CS-2015 (Set 3) paper, section 65.

Problem Statement

Consider the following C function. The function takes a single-linked list of integers as input parameter. It modifies the list by moving the last element to the front of the list, and returns the modified list. Assume that the list has at least one element.

struct node {
  int value;
  struct node *next;
};

struct node * modify_list(struct node *head) {
  struct node *last = head, *second_last = NULL;
  while (last->next != NULL) {
    second_last = last;
    last = last->next;
  }
  second_last->next = NULL;
  last->next = head;
  head = last;
  return head;
}

Explanation

The given code takes a single-linked list of integers as input parameter and modifies the list by moving the last element to the front of the list.

The modify_list function starts by pointing two pointers last and second_last to the first node of the linked list. It then traverses the linked list until last pointer points to the last node of the linked list and second_last points to the second last node of the linked list.

Once last pointer points to the last node, second_last->next is set to NULL, breaking the link between the last and second last node. last->next is then set to point to the head of the linked list, effectively moving the last node to the front of the list. Finally, the head pointer is updated to point to the new head of the list and returned.

Conclusion

This code can be used to move the last element of a single-linked list to the front of the list. It can be a useful utility function for modifying linked lists in programming problems.

Code Snippet (Markdown)

struct node {
  int value;
  struct node *next;
};

struct node * modify_list(struct node *head) {
  struct node *last = head, *second_last = NULL;
  while (last->next != NULL) {
    second_last = last;
    last = last->next;
  }
  second_last->next = NULL;
  last->next = head;
  head = last;
  return head;
}