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chore: remove useless comment
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problems/24.swapNodesInPairs.md

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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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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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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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![24.swap-nodes-in-pairs](../assets/24.swap-nodes-in-pairs.gif)
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(图片来自: https://github.com/MisterBooo/LeetCodeAnimation)
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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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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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* @lc app=leetcode id=24 lang=javascript
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*
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* [24] Swap Nodes in Pairs
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*
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* https://leetcode.com/problems/swap-nodes-in-pairs/description/
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*
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* algorithms
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* Medium (43.33%)
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* Total Accepted: 287.2K
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* Total Submissions: 661.3K
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* Testcase Example: '[1,2,3,4]'
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*
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* Given a linked list, swap every two adjacent nodes and return its head.
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*
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* You may not modify the values in the list's nodes, only nodes itself may be
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* changed.
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*
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*
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*
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* Example:
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*
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*
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* Given 1->2->3->4, you should return the list as 2->1->4->3.
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*
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*/
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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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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// 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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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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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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# 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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