Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

The key of this problem, is to calculate k-1 times each time you do "in place" operation!

The way of "in place" is very smart!!!

 public class Solution {  
   public ListNode reverseKGroup(ListNode head, int k) {  
     if (head == null || head.next == null || k < 2)  
       return head;  
     ListNode newHead = new ListNode(0);  
     newHead.next = head;  
     ListNode pre = newHead;  
     ListNode curr = head;  
     while (curr != null) {  
       // make sure the rest of the List is enough   
       int counter = k;         // calculate k-1 times
         curr = curr.next;  
         counter--;  
       }  
       if (curr != null) {  
         curr = pre.next;  
         counter = k;  
         while (counter > 1) {  // calculate k-1 times
ListNode tmp = curr.next;  
           curr.next = tmp.next;  
           tmp.next = pre.next;  
           pre.next = tmp;  
           counter--;  
         }  
         pre = curr;  
         curr = pre.next;  
       }  
     }  
     return newHead.next;  
   }  
 }