25. Reverse Nodes in k-Group(23.53%)

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
题目解释：给你一个链表和一个常数k，在这个链表上，每k个节点就做一次倒序，然后接回去，如果节点数少于k，则直接返回。要求不能改变节点的值，要使用常量级内存。

依然是使用递归，这个是Swap Nodes in Pairs的升级版本，每一次做k个节点的倒序，我们都用一个长度为k+1的数组来记录节点并完成倒序操作。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode reverseKGroup(ListNode head, int k) {        if (k < 2) {            return head;        }        if (head == null) {            return head;        }        ListNode judge = head;        ListNode[] tempNodes = new ListNode[k+1];        for (int i = 0; i <= k; i++) {            tempNodes[i] = judge;            if (i == k) {                break;            }            if (judge == null) {                return head;            }            judge = judge.next;        }        tempNodes[0].next = reverseKGroup(tempNodes[k], k);        for (int i = k-1; i >= 1;) {            tempNodes[i].next = tempNodes[--i];        }        return tempNodes[k-1];    }}