95 lines
2.2 KiB
Markdown
95 lines
2.2 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/swap-nodes-in-pairs/description/
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## 题目描述
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Given a linked list, swap every two adjacent nodes and return its head.
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You may not modify the values in the list's nodes, only nodes itself may be changed.
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Example:
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Given 1->2->3->4, you should return the list as 2->1->4->3.
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## 思路
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设置一个 dummy 节点简化操作，dummy next 指向 head。
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1. 初始化 first 为第一个节点
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2. 初始化 second 为第二个节点
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3. 初始化 current 为 dummy
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4. first.next = second.next
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5. second.next = first
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6. current.next = second
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7. current 移动两格
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8. 重复
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（图片来自： https://github.com/MisterBooo/LeetCodeAnimation)
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## 关键点解析
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1. 链表这种数据结构的特点和使用
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2. dummyHead 简化操作
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## 代码
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* 语言支持：JS，Python3
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```js
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* @param {ListNode} head
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* @return {ListNode}
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*/
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var swapPairs = function(head) {
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const dummy = new ListNode(0);
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dummy.next = head;
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let current = dummy;
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while (current.next != null && current.next.next != null) {
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// 初始化双指针
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const first = current.next;
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const second = current.next.next;
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// 更新双指针和 current 指针
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first.next = second.next;
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second.next = first;
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current.next = second;
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// 更新指针
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current = current.next.next;
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}
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return dummy.next;
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};
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```
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Python3 Code:
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```python
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class Solution:
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def swapPairs(self, head: ListNode) -> ListNode:
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"""
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用递归实现链表相邻互换：
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第一个节点的 next 是第三、第四个节点交换的结果，第二个节点的 next 是第一个节点；
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第三个节点的 next 是第五、第六个节点交换的结果，第四个节点的 next 是第三个节点；
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以此类推
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:param ListNode head
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:return ListNode
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"""
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# 如果为 None 或 next 为 None，则直接返回
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if not head or not head.next:
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return head
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_next = head.next
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head.next = self.swapPairs(_next.next)
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_next.next = head
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return _next
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```
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