Reverse Nodes in k-Group

题目：

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

分析：

简单；翻转链表变形

参考代码：
http://www.cnblogs.com/lichen782/p/leetcode_Reverse_Nodes_in_kGroup.html

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
/**
* Reverse a link list between pre and next exclusively
* an example:
* a linked list:
* 0->1->2->3->4->5->6
* |           |
* pre        next
* after call pre = reverse(pre, next)
*
* 0->3->2->1->4->5->6
*          |  |
*          pre next
* @param pre
* @param next
* @return the reversed list's last node, which is the precedence of parameter next
*/
private static ListNode reverse(ListNode pre, ListNode next){
ListNode last = pre.next;//where first will be doomed "last"
ListNode cur = last.next;
while(cur != next){
last.next = cur.next;
cur.next = pre.next;
pre.next = cur;
cur = last.next;
}
return last;
}

public static ListNode reverseKGroup(ListNode head, int k) {
if(head == null || k == 1) return head;
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode pre = dummy;
int i = 0;
while(head != null){
i++;
if(i % k ==0){
pre = reverse(pre, head.next);
head = pre.next;
}else {
head = head.next;
}
}
return dummy.next;
}

}