[leetcode 25] k个一组翻转链表

LeetCode K反转链表

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本文介绍了一种解决LeetCode上K反转链表问题的方法，该问题要求将链表节点每K个一组进行反转，同时确保剩余节点保持原样。通过递归方式实现，仅使用常数额外内存。

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 change

package com.leetcode;

/**
 递归和反转链表
 * @author King
 * @date 2019/8/24 10:08
 */
public class K反转链表 {
    public static void main(String[] args) {
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);
        head.next.next.next = new ListNode(4);
        head.next.next.next.next = new ListNode(5);

        System.out.println(reverseKGroup(head,2));
    }

    public static ListNode reverseKGroup(ListNode head, int k) {
        int count = 0;
        ListNode cur = head,prev = null;
        while (cur != null && count < k) {
            ListNode next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
            count++;
        }
        if (count == k) {
            head.next = reverseKGroup(cur,k);
        }else if (count < k && count > 0 ) {
            return reverseKGroup(prev,count);
        }
        return prev;
    }
}