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LeetCode 24. 两两交换链表中的节点 #60

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Chocolate1999 opened this issue Oct 4, 2020 · 0 comments
Open

LeetCode 24. 两两交换链表中的节点 #60

Chocolate1999 opened this issue Oct 4, 2020 · 0 comments
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链表 数据结构-链表

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@Chocolate1999
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仰望星空的人,不应该被嘲笑

题目描述

给定一个链表,两两交换其中相邻的节点,并返回交换后的链表。

你不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。

示例:

给定 1->2->3->4, 你应该返回 2->1->4->3.

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/swap-nodes-in-pairs
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

解题思路

非递归解法

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var swapPairs = function(head) {
    if(head == null || head.next == null) return head;
    let hummyHead = new ListNode(); // 虚拟节点
    hummyHead.next = head;
    let p = hummyHead;
    let node1,node2; // 当前要交换的两个节点
    while((node1 = p.next) && (node2 = p.next.next)){
        // 进行交换操作
        node1.next = node2.next;
        node2.next = node1;
        // 将链表串起来
        p.next = node2;
        p = node1;
    }
    return hummyHead.next;
};

递归解法

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var swapPairs = function (head) {
    if (!head || !head.next) return head;
    let node1 = head, node2 = head.next;
    node1.next = swapPairs(node2.next);
    node2.next = node1;
    return node2;
};

最后

文章产出不易,还望各位小伙伴们支持一波!

往期精选:

小狮子前端の笔记仓库

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学如逆水行舟,不进则退
@Chocolate1999 Chocolate1999 added the 链表 数据结构-链表 label Oct 4, 2020
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