LeetCode 25. K 个一组翻转链表

1、题目

2、分析

方法一：就按照第92题，反转链表的题目来写就好了。每次传的left和right的参数 + k就好了

方法二：递归法：
https://labuladong.github.io/zgnb/2/4/

3、代码

方法一的代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode listHead = head;
        ListNode cur = head;
        int j = 1;
        while(cur != null){
            //查看是否已经到链表尾部
            int i = 0;
            for(i = 0; i < k; i++){
                if(cur == null) break;
                cur = cur.next;
            }
            if(cur == null && i != k) break; //到达链表尾部，i = k的情况，是刚好整数的情况
            //按照区间，反转链表
            if(listHead == head){
                listHead = reverseList(listHead, j, j + k - 1);
            }
            else{
                reverseList(listHead, j, j + k - 1);
            }
            j = j + k;
        }
        return listHead;
    }

    private ListNode reverseList(ListNode head, int left, int right){
        ListNode dummyNode = new ListNode(-1);
        dummyNode.next = head;
        ListNode pre = dummyNode;
        ListNode cur = dummyNode.next;

        for(int i = 0; i < left - 1; i++){
            pre = pre.next;
            cur = cur.next;
        }

        for(int i = 0; i < right - left; i++){
            ListNode removeNode = cur.next;
            cur.next = removeNode.next;
            removeNode.next = pre.next;
            pre.next = removeNode;
        }
        return dummyNode.next;
    }
}