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) {
            int groups = GetLongth(head) / k;
            if (groups == 0 || k == 1) return head;
            ListNode pre = null;
            ListNode cur = head;
            ListNode nex;
            ListNode pf = null;//上一组的第一个
            ListNode pl;//这一组的第一个
            ListNode newHead = null;
            for (int i = 0; i < groups; i++) {
                pl = cur;
                pre = cur;
                cur = cur.next;
                for (int j = 1; j < k; j++) {
                    nex = cur.next;
                    cur.next = pre;
                    pre = cur;
                    cur = nex;
                }
                pl.next = cur;
                if (pf != null) pf.next = pre;
                pf = pl;
                newHead = newHead == null ? pre : newHead;
            }
            return newHead;
        }
        public int GetLongth(ListNode head) {
            int length = 0;
            while (head != null) {
                head = head.next;
                length++;
            }
            return length;
        }
    }