题解 | #链表中的节点每k个一组翻转#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    public ListNode reverseKGroup (ListNode head, int k) {
        if(k == 1 || head == null || head.next == null){
            return head;
        }
        // write code here
        ListNode newList = new ListNode(0);
        //head开始时指向一组内容的第一个
        newList.next = head;
        ListNode pre = newList;
        ListNode groupEnd = toNextEnd(head, k);
        while(head != null && groupEnd != head){
		  //方案①先移动pre，再移动head
            // ListNode tmp = pre.next;
            // pre.next = head.next;
            // head.next = head.next.next;
            // pre.next.next = tmp;
		  //方案②先移动head，再移动pre
            ListNode tmp = head.next;
            head.next = tmp.next;
            tmp.next = pre.next;
            pre.next = tmp;
            if(head.next == groupEnd){
                pre = head;
                head = groupEnd;
                groupEnd = toNextEnd(groupEnd, k);
            }
        }
        return newList.next;
    }
    //返回1、原链，不够一组；2、null或者下一组的第一个节点，说明够一组
    public ListNode toNextEnd(ListNode head, int k){
        ListNode tmp = head;
        for(int i = 0; i< k; i++){
            if(head == null){
                return tmp;
            }
            head = head.next;

        }
        return head;
    }
}