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

思路1：模拟题，仔细处理即可。

代码（太乱，有空重写）：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {	public ListNode reverseKGroup(ListNode head, int k) {		if (head == null)			return null;		ListNode dummyHead = new ListNode(-1);		dummyHead.next = head;		ListNode p = dummyHead;		ListNode[] s = new ListNode[k];		while (hasNext(p, k)) {			ListNode tmp = p.next;			for (int i = 0; i < k; i++) {				s[i] = tmp;				tmp = tmp.next;			}			s[0].next = s[k - 1].next;			for (int i = k - 1; i > 0; i--)				s[i].next = s[i - 1];			p.next = s[k - 1];			for (int i = 0; i < k; i++) {				p = p.next;				if (p == null)					break;			}		}		return dummyHead.next;	}	private boolean hasNext(ListNode p, int k) {		for (int i = 0; i <= k; i++) {			if (p == null)				return false;			p = p.next;		}		return true;	}	public static void main(String[] args) {		ListNode list = new ListNode(1);		list.next = new ListNode(2);		list.next.next = new ListNode(3);		list.next.next.next = new ListNode(4);		System.out.println(list);		System.out.println(new Solution().reverseKGroup(list, 3));	}}

参考：

http://rleetcode.blogspot.com/2014/01/reverse-nodes-in-k-group-java.html

http://blog.csdn.net/linhuanmars/article/details/19957455