[LeetCode]25. Reverse Nodes in k-Group k个一组翻转链表

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

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

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list‘s nodes, only nodes itself may be changed.

题目要求我们把给出的链表按K个一组翻转，不足K个的不翻转

这个可以看做是上一个题目两两翻转节点的扩展，我们还是用递归来操作，先来确定下大致的处理过程

1.先从链表头开始数k个节点，记录个数，不满k个的按k个处理

2.判断上一步数出的节点个数，小于k则说明不用翻转，直接返回head就行

3.节点个数等于k，说明需要反转，利用3指针翻转链表的做法，把这个k节点链表翻转

4.前面几步把k个节点翻转了，剩下的节点递归调用该方法，然后返回翻转后链表的头结点

浓缩一下就是，先翻转k个节点，然后通过递归把k个节点之后的链表也翻转

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode prev = null;
        ListNode cur = head;
        ListNode next = null;
        ListNode check = head;
        int canProceed = 0;
        int count = 0;
        // 检查链表长度是否满足翻转
        while (canProceed < k && check != null) {
            check = check.next;
            canProceed++;
        }
        // 满足条件，进行翻转
        if (canProceed == k) {
            while (count < k && cur != null) {
                next = cur.next;
                cur.next = prev;
                prev = cur;
                cur = next;
                count++;
            }
            if (next != null) {
                // head 为链表翻转后的尾节点
                head.next = reverseKGroup(next, k);
            }
            // prev 为链表翻转后的头结点
            return prev;
        } else {
            // 不满住翻转条件，直接返回 head 即可
            return head;
        }
    }
}

原文地址：https://www.cnblogs.com/jchen104/p/10268380.html