Reversing A Linked List


Reversing A Linked List

In this article let's learn how to reverse a  linked list(iterative method) in an efficient and easy manner. This problem is a part of data structures(Linked Lists). This reversing a linked list problem is frequently asked in many of the interviews.

Now let's see what a linked list is and how it works along with its advantages.

What is a linked list?

  • A linked list is a data structure that stores a linear collection of data elements, called nodes which are stored in memory at random locations.
  • In the linked list, each node is divided into two parts.
  • The first part contains the data of the element.
  • The second part contains the address of the next node in the list.

Advantages of a linked list

  • A linked list is a dynamic data structure, so it is easy to allocate and deallocate memory.
  • No memory wastage.
  • Insertion and deletion of elements are easier.
  • Using a linked list one can implement sack and queue.
This problem was frequently asked in big tech company interviews, so now let's get into the problem and its optimized solution.
Reverse A Linked List

Given the head of a singly linked list, reverse the list, and return the reversed list.

Input: 1->2->3->4->NULL

Output: 4->3->2->1->NULL


Iterative Method:

Let's focus on the function ReverseLinkedList():

->Initialize three-pointers prev, curr, nxt.

->Now assign head to curr and nxt, assign NULL to prev.

->Now iterate through the loop :

  • change the next of nxt pointer(nxt->next=next).
  • store prev node in next of curr node(curr->next=prev).
  • now assign curr node to prev node(prev=curr).
  • At last assign nxt node to curr(curr=nxt).

 ->Now assign node prev to head(head=prev)

->return head.

Program to reverse a linked list:

SinglyLinkedListNode* ReverseLinkedList(SinglyLinkedListNode* head)
     SinglyLinkedListNode *curr=head;
     SinglyLinkedListNode *nxt=head;
      SinglyLinkedListNode* prev=0;

     return head;



Recursive Method:
  • Divide the list into two parts first node and the rest of the linked list.
  • Call reverse for the rest of the linked list.
  • Now link the rest of the linked list to the first node.
  • At last fix head.

SinglyLinkedListNode* reverse(SinglyLinkedListNode* head)
        if (head == NULL || head->next == NULL)
            return head;
        SinglyLinkedListNode* rest = reverse(head->next);
        head->next->next = head;
        head->next = NULL;
        return rest;

Both the methods run in O(n) time complexity and O(1) space complexity.

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