# 24. 两两交换链表中的节点 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解
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---
访问原文链接：[24. 两两交换链表中的节点 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解](https://leetcode.blog/zh/leetcode/24-swap-nodes-in-pairs)，体验更佳！
力扣链接：[24. 两两交换链表中的节点](https://leetcode.cn/problems/swap-nodes-in-pairs), 难度等级：**中等**。
## LeetCode “24. 两两交换链表中的节点”问题描述
给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题（即，只能进行节点交换）。
### [示例 1]
**输入**: `head = [1,2,3,4]`
**输出**: `[2,1,4,3]`
### [示例 2]
**输入**: `head = []`
**输出**: `[]`
### [示例 3]
**输入**: `head = [1]`
**输出**: `[1]`
### [示例 4]
**输入**: `head = [1,2,3]`
**输出**: `[2,1,3]`
### [约束]
- 链表中节点的数目在范围 `[0, 100]` 内
- `0 <= Node.val <= 100`
## 思路
在做本题前，建议先完成简单题目[206. Reverse Linked List](206-reverse-linked-list.md)。
1. 解决这个问题，依然至少需要定义两个变量：`current`和`previous`。
2. 循环条件应是`while (current.next != null)`，而不应该是`while (current != null)`，因为需要操作`current.next.next`。
## 步骤
1. 遍历所有节点。
```java
var previous = null;
var current = head;
while (current != null) {
current = current.next;
}
```
2. 因为每两个节点进行一次位置互换，所以需要改为一次走两步。
```java
var previous = null;
var current = head;
while (current != null && current.next != null) { // 1
var nextNext = current.next.next; // 2
current = nextNext; // 3
}
```
3. 交换 `current` 和 `current.next` 的位置。
```java
var previous = null;
var current = head;
while (current != null && current.next != null) {
var nextNext = current.next.next;
current.next.next = current; // 1
current.next = nextNext; // 2
current = nextNext;
}
```
4. 处理 `previous`。
```java
var previous = null;
var current = head;
while (current != null && current.next != null) {
var nextNext = current.next.next;
previous.next = current.next; // 1
current.next.next = current;
current.next = nextNext;
previous = current; // 2
current = nextNext;
}
```
5. 确定返回值。因为`head`节点在节点数量超过1时，会被交换到第二个节点的位置，为了统一方便处理，最好加入`dummy_head`节点。
```java
var dummyHead = new ListNode(); // 1
dummyHead.next = head; // 2
var previous = dummyHead; // 3
var current = head;
while (current != null && current.next != null) {
var nextNext = current.next.next;
previous.next = current.next;
current.next.next = current;
current.next = nextNext;
previous = current;
current = nextNext;
}
return dummyHead.next; // 4
```
## 复杂度
- 时间复杂度: `O(N)`.
- 空间复杂度: `O(1)`.
## Java
```java
/**
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode swapPairs(ListNode head) {
var dummyHead = new ListNode(0, head);
var previous = dummyHead;
var current = head;
while (current != null && current.next != null) {
var nextNext = current.next.next;
previous.next = current.next;
current.next.next = current;
current.next = nextNext;
previous = current;
current = nextNext;
}
return dummyHead.next;
}
}
```
## Python
```python
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
dummy_head = ListNode(next=head)
previous = dummy_head
current = head
while current and current.next:
next_next = current.next.next
previous.next = current.next
current.next.next = current
current.next = next_next
previous = current
current = next_next
return dummy_head.next
```
## C++
```cpp
/**
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
auto dummy_head = new ListNode(0, head);
auto previous = dummy_head;
auto current = head;
while (current != nullptr && current->next != nullptr) {
auto next_next = current->next->next;
previous->next = current->next;
current->next->next = current;
current->next = next_next;
previous = current;
current = next_next;
}
auto result = dummy_head->next;
delete dummy_head;
return result;
}
};
```
## JavaScript
```javascript
/**
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
var swapPairs = function (head) {
const dummyHead = new ListNode(0, head)
let previous = dummyHead
let current = head
while (current != null && current.next != null) {
const nextNext = current.next.next
previous.next = current.next
current.next.next = current
current.next = nextNext
previous = current
current = nextNext
}
return dummyHead.next
};
```
## C#
```csharp
/**
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
public class Solution
{
public ListNode SwapPairs(ListNode head)
{
var dummyHead = new ListNode(0, head);
var previous = dummyHead;
var current = head;
while (current != null && current.next != null)
{
var nextNext = current.next.next;
previous.next = current.next;
current.next.next = current;
current.next = nextNext;
previous = current;
current = nextNext;
}
return dummyHead.next;
}
}
```
## Go
```go
/**
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func swapPairs(head *ListNode) *ListNode {
dummyHead := &ListNode{0, head}
previous := dummyHead
current := head
for current != nil && current.Next != nil {
nextNext := current.Next.Next
previous.Next = current.Next
current.Next.Next = current
current.Next = nextNext
previous = current
current = nextNext
}
return dummyHead.Next
}
```
## Ruby
```ruby
# class ListNode
# attr_accessor :val, :next
#
# def initialize(val = 0, _next = nil)
# @val = val
# @next = _next
# end
# end
def swap_pairs(head)
dummy_head = ListNode.new
dummy_head.next = head
previous = dummy_head
current = head
while current && current.next
next_next = current.next.next
previous.next = current.next
current.next.next = current
current.next = next_next
previous = current
current = next_next
end
dummy_head.next
end
```
## Other languages
```java
// Welcome to create a PR to complete the code of this language, thanks!
```
> 🚀 **打造你的开发者个人IP**
>
> 掌握算法是成功的基石，而全方位展示你的才华则是获得垂青的关键。
>
> 我的另一个项目 [**leader.me**](https://www.leader.me) —— 专为程序员打造的“全能型”个人品牌展示平台。
>
> **三位一体（All-In-One）的职场利器：**
> - 📄 **简历 + 作品集 + 博客：** 将你的 GitHub 项目、技术心得与职场经历完美融合。
> - 🌐 **永久免费自定义域名：** 支持绑定你自己的独立域名，且该功能永久免费。
> - ✨ **顶级行业子域名：** 提供 `name.leader.me`，极具职业含金量的专属域名。
>
> [**立即前往 leader.me 打造你的个人品牌 →**](https://www.leader.me)
---
访问原文链接：[24. 两两交换链表中的节点 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解](https://leetcode.blog/zh/leetcode/24-swap-nodes-in-pairs)，体验更佳！
GitHub 仓库: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).