Reverse a Linked List in groups of given size.

PROBLEM :

Given a linked list, write a function to reverse every k nodes (where k is an input to the function).If a linked list is given as 1->2->3->4->5->6->7->8->NULL and k = 3 then output will be 3->2->1->6->5->4->8->7->NULL.
Example:
Inputs:  1->2->3->4->5->6->7->8->NULL and k = 3
Output:  3->2->1->6->5->4->8->7->NULL.

Inputs:   1->2->3->4->5->6->7->8->NULL and k = 5
Output:  5->4->3->2->1->8->7->6->NULL


Input:
In this problem, method takes two argument: the head of the linked list and int k. You should not read any input from stdin/console.
The struct Node has a data part which stores the data and a next pointer which points to the next element of the linked list.
There are multiple test cases. For each test case, this method will be called individually.

Output:
Reverse the linked list in the group of given size and return the reference of starting node(head) of the reversed Linked list .

Note: If you use "Test" or "Expected Output Button" use below example format
Example:
Input:
1
8
1 2 2 4 5 6 7 8
4

Output:
4 2 2 1 8 7 6 5

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SIMPLE c++ IMPLEMENTATION :
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/*
  Reverse a linked list
  The input list will have at least one element
  Return the node which points to the head of the new LinkedList
  Node is defined as
  struct node
  {
     int data;
     struct Node *next;
  }
*/
struct node *reverse (struct node *head, int k)
{
    if(head==NULL)
    return head ;
   
    if(head->next==NULL)
    return head;
   
    if(k==1)
        return head ;
       
    struct node *one,*two,*third ;
    int i=2 ;
    one=head ;
   
    one=head ;
    two=one->next ;
    third=two->next ;
   
   
    one->next=NULL ;
    two->next=one ;
   
    while(i!=k&&third!=NULL)
    {
        one=two ;
        two=third ;
        third=third->next ;
        i++ ;
        two->next=one ;
    }
 
    if(third!=NULL)
        head->next=reverse(third,k) ;
   
    return two ;
}

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