344-reverse-string.md Separated Chinese solutions from English solutions.

gazeldx · gazeldx · commit ebbf9362e9bf · 2025-02-24T09:38:11.000+08:00
diff --git a/en/1-1000/344-reverse-string.md b/en/1-1000/344-reverse-string.md
@@ -1,53 +1,55 @@
 # 344. Reverse String - Best Practices of LeetCode Solutions
-LeetCode link: [344. Reverse String](https://leetcode.com/problems/reverse-string),
-[344. 反转字符串](https://leetcode.cn/problems/reverse-string)
 
-[中文题解](#中文题解)
+LeetCode link: [344. Reverse String](https://leetcode.com/problems/reverse-string), difficulty: **Easy**.
+
+## LeetCode description of "344. Reverse String"
 
-## LeetCode problem description
 Write a function that reverses a string. The input string is given as an array of characters `s`.
 
 You must do this by modifying the input array [in-place](https://en.wikipedia.org/wiki/In-place_algorithm) with `O(1)` extra memory.
 
-Difficulty: **Easy**
-
 ### [Example 1]
 
 **Input**: `s = ["h","e","l","l","o"]`
 
 **Output**: `["o","l","l","e","h"]`
 
 ### [Example 2]
+
 **Input**: `s = ["H","a","n","n","a","h"]`
 
 **Output**: `["h","a","n","n","a","H"]`
 
 ### [Constraints]
-- `1 <= s.length <= 105`
+
+- `1 <= s.length <= 10^5`
 - `s[i]` is a [printable ascii character](https://en.wikipedia.org/wiki/ASCII#Printable_characters).
 
+### [Hints]
+
 <details>
   <summary>Hint 1</summary>
   The entire logic for reversing a string is based on using the opposite directional two-pointer approach!
 </details>
 
 ## Intuition
-[中文题解](#中文题解)
 
 1. This problem can be solved in one line of code using the built-in `sort()` method of the programming language. If this question is asked in an interview, the questioner should be testing how to do it without the built-in method.
 2. Use **two pointers** with **opposite directions**, initially one pointer points to the index `0` and the other pointer points to the index `s.length - 1`.
 3. Traverse the elements of the array, and the loop condition is `while (left < right)`. In the loop body, `left += 1`, `right -= 1`.
 4. In the loop body, swap the two values.
 5. The above is the template for `two pointers` in `opposite directions`.
 
-## Approach
+## Steps
 1. Use two pointers with **opposite directions**, initially one pointer points to the index `0` and the other pointer points to the index `s.length - 1`.
+
 ```ruby
 left = 0
 right = s.length - 1
 ```
 
 2. Traverse the elements of the array, and the loop condition is `while (left < right)`. In the loop body, `left += 1`, `right -= 1`.
+
 ```ruby
 left = 0
 right = s.length - 1
@@ -59,6 +61,7 @@ end
 ```
 
 3. In the loop body, swap the two values.
+
 ```ruby
 left = 0
 right = s.length - 1
@@ -72,10 +75,12 @@ end
 ```
 
 ## Complexity
+
 * Time: `O(n)`.
 * Space: `O(1)`.
 
 ## Java
+
 ```java
 class Solution {
     public void reverseString(char[] s) {
@@ -95,6 +100,7 @@ class Solution {
 ```
 
 ## Python
+
 ```python
 class Solution:
     def reverseString(self, s: List[str]) -> None:
@@ -108,6 +114,7 @@ class Solution:
 ```
 
 ## C++
+
 ```cpp
 class Solution {
 public:
@@ -126,6 +133,7 @@ public:
 ```
 
 ## JavaScript
+
 ```javascript
 var reverseString = function (s) {
   let left = 0
@@ -141,6 +149,7 @@ var reverseString = function (s) {
 ```
 
 ## C#
+
 ```c#
 public class Solution
 {
@@ -161,6 +170,7 @@ public class Solution
 ```
 
 ## Go
+
 ```go
 func reverseString(s []byte)  {
     left := 0
@@ -176,6 +186,7 @@ func reverseString(s []byte)  {
 ```
 
 ## Ruby
+
 ```ruby
 def reverse_string(s)
   left = 0
@@ -191,77 +202,31 @@ end
 ```
 
 ## C
+
 ```c
 // Welcome to create a PR to complete the code of this language, thanks!
 ```
 
 ## Kotlin
+
 ```kotlin
 // Welcome to create a PR to complete the code of this language, thanks!
 ```
 
 ## Swift
+
 ```swift
 // Welcome to create a PR to complete the code of this language, thanks!
 ```
 
 ## Rust
+
 ```rust
 // Welcome to create a PR to complete the code of this language, thanks!
 ```
 
 ## Other languages
-```
-// Welcome to create a PR to complete the code of this language, thanks!
-```
-
-## 力扣“344. 反转字符串”问题描述
-力扣链接：[344. 反转字符串](https://leetcode.cn/problems/reverse-string) ，难度：**简单**。
-
-编写一个函数，其作用是将输入的字符串反转过来。输入字符串以字符数组 `s` 的形式给出。
-
-不要给另外的数组分配额外的空间，你必须 **原地** **修改输入数组**、使用 `O(1)` 的额外空间解决这一问题。
-
-### [示例 1]
-**输入**: `s = ["h","e","l","l","o"]`
 
-**输出**: `["o","l","l","e","h"]`
-
-# 中文题解
-## 思路
-1. 这种问题用编程语言内置的`sort()`方法可以一行代码解决。如果面试中出这种题目，出题人考察的应该是不用内置方法如何做。
-2. 用**方向相对**的两个指针，最初时一个指针指向下标`0`，另一个指针指向下标`s.length - 1`。
-3. 遍历数组元素，循环条件为`while (left < right)`。在循环体内，`left += 1`, `right -= 1`。
-4. 在循环体内，交换数组中两个下标对应的值。
-5. 以上为`方向相对`的**双指针**的模板。
-
-## 步骤
-1. 用**方向相对的两个指针**，最初时一个指针指向下标`0`，另一个指针指向下标`s.length - 1`。
-```ruby
-left = 0
-right = s.length - 1
 ```
-
-2. 遍历数组元素，循环条件为`while (left < right)`。在循环体内，`left += 1`, `right -= 1`。
-```ruby
-left = 0
-right = s.length - 1
-
-while left < right # 1
-  left += 1 # 2
-  right -= 1 # 3
-end
-```
-
-3. 在循环体内，交换数组中两个下标对应的值。
-```ruby
-left = 0
-right = s.length - 1
-
-while left < right
-  s[left], s[right] = s[right], s[left] # 1
-
-  left += 1
-  right -= 1
-end
+// Welcome to create a PR to complete the code of this language, thanks!
 ```
diff --git a/en/3001-4000/unorganized.md b/en/3001-4000/unorganized.md
@@ -200,6 +200,7 @@ Add a table to show the differences between A-Start and breadth-first search
 - 1514-path-with-maximum-probability.md
 - 1334 https://leetcode.com/problems/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance
 - 752 a star
+- 3464 https://leetcode.cn/problems/maximize-the-distance-between-points-on-a-square
 
 ## other finished problems
 - https://leetcode.com/problems/k-closest-points-to-origin/
diff --git a/zh/1-1000/344-reverse-string.md b/zh/1-1000/344-reverse-string.md
