leetcode--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

[java]  view plain  copy

1. /** 
2.  * Definition for singly-linked list. 
3.  * public class ListNode { 
4.  *     int val; 
5.  *     ListNode next; 
6.  *     ListNode(int x) { val = x; } 
7.  * } 
8.  */  
9. public class Solution {  
10.     public ListNode reverseKGroup(ListNode head, int k) {  
11.         if(head==null) return head;  
12.         ListNode tail = head;  
13.         for(int i=1;i<k;i++){  
14.             tail = tail.next;  
15.             if (tail == null){  
16.                 return head;  
17.             }  
18.         }  
19.         ListNode nextBegin = tail.next;  
20.         ListNode first = head;  
21.         ListNode second = head.next;   
22.         for(int i=1;i<k;i++){//翻转指针  
23.             ListNode secondNext = second.next;  
24.             second.next = first;  
25.               
26.             first = second;  
27.             second = secondNext;  
28.         }  
29.         head.next = reverseKGroup(nextBegin, k);//本质是尾续接(因为链表已经翻转)  
30.         return tail;//本质是返回头  
31.     }  
32. }  

原文链接http://blog.youkuaiyun.com/crazy__chen/article/details/45583171