8 Nov 2013

Reverse Linked List in Chunks

Problem Statement:
                   Reverse a given linked list in chunks of given size.

Given below is the linked list which is reversed in chunks of size 3.

Solution : 

The above problem can be solved both by 
  1. Recursion 
  2. Iteration

Full Source Code: LINK

1. Below is recursive implementation: 
       The below sub routine use a another sub-routine which helps in reversing the list of the chunk size.

public Node reverseInchunks(Node root, int k){

   return null;

  Node first=root;
  Node current=first;

  int count = k-1;
  while(current.next!=null && count > 0)

  Node  last=reverseInchunks(current.next,k);
  Node subHead=reverse(first);


  return subHead;

 public Node  reverse(Node head) {

  Node result;
  if(head.next ==null) 
   return head;

  result = reverse(head.next);

  head.next.next = head;

  head.next = null;

  return result;

2. You can also be asked to reverse the list iteratively, below is the solution.

  public Node reverseChunkIterative(Node head, int k){

   return null;
  Node first, previous,last;
  first= previous=last=null;
  Node current=head;

   Node temp=current; 
   previous =null;
   Node next=null;
   int count =k;
   while(current!=null  && count>0){
    next = current.next;
    current.next = previous ;
    previous = current ;
    current = next;
   if(first == null)
    first = previous;

  return first;

For the simulation of the above procedures on ideone.com follow the link given above.
Thanks to Rohit Jain for suggesting the recursive solution.

Your suggestions, queries , critiques are most welcome , feel free to comment.

Happy Coding !! :)


  1. This comment has been removed by the author.

  2. Please explain to me one thing. You are not explicitly changing the head pointer of the linked list. How does it manage to change after reversing first k nodes?


    1. Applogies for the delay , the comment went unnoticed,
      1. Recursive Case:
      The variable "subHead" is returned in reverse manner of the linked list.
      So if u consider the chunk 4-> 5-> 6 , for the example given above
      here 'first' = 4 , last = '7' and 'subhead' = 6 , all this is done in the reverse manner ,
      this 'subhead' becomes 'last' for the chunk 1->2->3 , and here
      'first' =1 ; 'last' = 6 and 'subhead'=3
      so when the subroutine call is over u get last element of the first chunk, that is "3" is returned as the 'subhead'
      which is nothing but the new head of the linked list in the above example.

      2. Iterative Case:
      This is pretty and u must have already figured it out,
      i have taken a pointer 'first' which is initially null, and after reversal of the
      first chunk 1->2->3 "first" pointer is updated.

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