Reverse a Linked List

Objective: Reverse the given linked list.

Input: A Linked List

Output: Reversed Linked List

Example:

Input : ->30->25->20->15->10->5

Reversed : ->5->10->15->20->25->30

NOTE : Click Reverse a Linked List – Part 2 to see the another implementation of this problem.

Approach:

Iterative:

  • Create 3 nodes, currNode, PrevNode and nextNode.
  • Initialize them as currNode = head; nextNode = null;prevNode = null;
  • Now keep reversing the pointers one by one till currNode!=null.
   while(currNode!=null){
     nextNode = currNode.next;
     currNode.next = prevNode;
     prevNode = currNode;
     currNode = nextNode;
}
  • At the end set head = prevNode;
  • See Example:

Linked List Reversal

Note: If Image above is not clear, either zoom your browser or right click on image and open in a new tab.

  • Recursive Approach:
    • Take 3 nodes as Node ptrOne,Node ptrTwo, Node prevNode
    • Initialize them as ptrOne = head; ptrTwo=head.next, prevNode = null.
    • Call reverseRecursion(head,head.next,null)
    • Reverse the ptrOne and ptrTwo
    • Make a recursive call for reverseRecursion(ptrOne.next,ptrTwo.next,null)

Complete Code:


Output:

->30->25->20->15->10->5
Reverse Through Iteration
->5->10->15->20->25->30
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Original Link List 2 : ->36->35->34->33->32->31
Reverse Through Recursion
->31->32->33->34->35->36

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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