# 206. 反转链表 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解 > 🚀 **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：[206. 反转链表 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解](https://leetcode.blog/zh/leetcode/206-reverse-linked-list)，体验更佳！ 力扣链接：[206. 反转链表](https://leetcode.cn/problems/reverse-linked-list), 难度等级：**简单**。 ## LeetCode “206. 反转链表”问题描述 给你单链表的头节点 `head` ，请你反转链表，并返回反转后的链表。 ### [示例 1] ![](../../images/examples/206_1.jpg) **输入**: `head = [1,2,3,4,5]` **输出**: `[5,4,3,2,1]` ### [示例 2] ![](../../images/examples/206_2.jpg) **输入**: `[1,2]` **输出**: `[2,1]` ### [示例 3] **输入**: `[]` **输出**: `[]` ### [约束] - 链表中节点的数目范围是 `[0, 5000]` - `-5000 <= Node.val <= 5000` ## 思路 1. 解决这个问题，只需要定义**两**个变量：`current`和`previous`。
点击查看答案

`current.next = previous` 就是反转了。

2. 循环条件是`while (current != null)`，还是`while (current.next != null)`？
点击查看答案

是 `while (current != null)`，因为需要操作的是 `current.next = previous`。

## 步骤 1. 遍历所有节点。 ```javascript previous = null current = head while (current != null) { current = current.next } ``` 2. 加入`current.next = previous`。 ```javascript previous = null current = head while (current != null) { tempNext = current.next current.next = previous current = tempNext } ``` 3. `previous`目前始终是`null`，需要让它变化起来：`previous = current`。 ```javascript previous = null current = head while (current != null) { tempNext = current.next current.next = previous previous = current current = tempNext } ``` ## 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 reverseList(ListNode head) { ListNode previous = null; var current = head; while (current != null) { var tempNext = current.next; current.next = previous; previous = current; current = tempNext; } return previous; } } ``` ## Python ```python # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: previous = None current = head while current: temp_next = current.next current.next = previous previous = current current = temp_next return previous ``` ## 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* reverseList(ListNode* head) { ListNode* previous = nullptr; ListNode* current = head; while (current) { auto temp_next = current->next; current->next = previous; previous = current; current = temp_next; } return previous; } }; ``` ## JavaScript ```javascript /** * function ListNode(val, next) { * this.val = (val===undefined ? 0 : val) * this.next = (next===undefined ? null : next) * } */ var reverseList = function (head) { let previous = null let current = head while (current != null) { const tempNext = current.next current.next = previous previous = current current = tempNext } return previous }; ``` ## 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 ReverseList(ListNode head) { ListNode previous = null; ListNode current = head; while (current != null) { var tempNext = current.next; current.next = previous; previous = current; current = tempNext; } return previous; } } ``` ## Go ```go /** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */ func reverseList(head *ListNode) *ListNode { var previous *ListNode current := head for current != nil { tempNext := current.Next current.Next = previous previous = current current = tempNext } return previous } ``` ## Ruby ```ruby # class ListNode # attr_accessor :val, :next # # def initialize(val = 0, _next = nil) # @val = val # @next = _next # end # end def reverse_list(head) previous = nil current = head while current temp_next = current.next current.next = previous previous = current current = temp_next end previous 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：[206. 反转链表 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解](https://leetcode.blog/zh/leetcode/206-reverse-linked-list)，体验更佳！ GitHub 仓库: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).