首页 > 其他 > 详细

[LC] 24. Swap Nodes in Pairs

时间:2019-12-29 12:45:58      阅读:63      评论:0      收藏:0      [点我收藏+]

Given a linked list, swap every two adjacent nodes and return its head.

You may not modify the values in the list‘s nodes, only nodes itself may be changed.

 

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode nxt = head.next;
        ListNode newHead = swapPairs(nxt.next);
        nxt.next = head;
        head.next = newHead;
        return nxt;
    }
}

[LC] 24. Swap Nodes in Pairs

原文:https://www.cnblogs.com/xuanlu/p/12114589.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!