题解 | #链表中的节点每k个一组翻转#

using System;
using System.Collections.Generic;

/*
public class ListNode
{
    public int val;
    public ListNode next;

    public ListNode (int x)
    {
        val = x;
    }
}
*/

class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类
     * @param k int整型
     * @return ListNode类
     */
    public ListNode reverseKGroup (ListNode head, int k)
    {
        // write code here
        if (k == 1) return head;
        ListNode from = head;//第一个要反转的Node
        ListNode to = head;//反转List的下一个
        ListNode res = head;//结果
        ListNode p1 = null;//from的前一个，用来连接。
        int index = 1;//计数器
        while (to != null) 
        {
            from = head;
            while (index < k) // k + 1 会 越 界 
            {
                head = head.next;
                index++;
                if (head == null) break;
            }
            if(head == null)break;//return也行
            to = head.next;
            index = 1;
            if (p1 == null) //设置初始值
            {
                res = process(from, to);
            } 
            else 
            {
                Console.WriteLine(p1.val);//测试用的不用管
                p1.next = process(from, to);//链接反转后的List
            }
            p1 = from;
            head = to;
        }
        return res;
    }
    //逆序
    public ListNode process(ListNode from, ListNode to) {
        ListNode p1 = null;
        ListNode p2 = to;
        while (from != to) {
            p1 = from.next;
            from.next = p2;
            p2 = from;
            from = p1;
        }
        return p2;
    }

}