---
# System prepended metadata
title: Leetcode 344. Reverse String 練習
tags: [java, Leetcode, Algorithm]
---
# Leetcode 344. Reverse String 練習
> Difficulty : `Easy`
> `Java` `演算法Algorithm`
> 撰寫人[name=KVJK_2125] [time=SAT, May 18, 2024 17:40]
## 題目
### 說明 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 with `O(1)` extra memory.
翻譯：
寫一個函數，其作用是將輸入的字串反轉過來。輸入字串以字元陣列s的形式給出。
不要給另外的陣列分配額外的空間，你必須原地修改輸入陣列、使用O(1)的額外空間解決這個問題。
### 範例 Example
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"]
### 限制條件 Constraints
- `1 <= s.length <= 10^5`
- `s[i]` is a printable ascii character.
## 解題
### 思路 Intuition/Approach
- Step1.
- Step2.
for 迴圈前，設定一個存放`s.length`的變數：
`int lengths = s.length`
- Step3.
for迴圈中，設定一個變數`char boxtemp = s[i]`，存放左邊要被交換的值;
將左邊要被交換的＂位置＂放入右邊要交換的值`s[i] = s[--lengths]`;
將右邊要被交換的＂位置＂放入存放在`boxtemp`中的值
### 程式碼 Code（加註解）
```clink=
class Solution {
public void reverseString(char[] s) {
int lengths = s.length; //建立長度變數
//交換
for(int i = 0; i <= (s.length-1) / 2; i++)
{
char boxtemp = s[i]; //暫時存放左邊要被交換的值
s[i] = s[--lengths]; //左邊要被交換的'位置'存放右邊要交換的值
s[lengths] = boxtemp; //右邊要被交換的'位置'存放左邊原本的值
}
}
}
```