# 【LeetCode】 344. Reverse String

## Description
> Write a function that reverses a string. The input string is given as an array of characters `char[]`.

> Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

> You may assume all the characters consist of [printable ascii characters](https://en.wikipedia.org/wiki/ASCII#Printable_characters).

> 寫一個函式能夠反轉一個字串。輸入字串會使用字元陣列`char[]`。

> 不要配置任何額外的空間在其他陣列上，你應該在額外記憶體O(1)的原地演算法條件下，改變輸入字串完成它。

> 你可以假設所有字串都包含在[可顯示字元](https://zh.wikipedia.org/wiki/ASCII#%E5%8F%AF%E6%98%BE%E7%A4%BA%E5%AD%97%E7%AC%A6)中。

## Example:

Example 1:

Input: ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]


Example 2:

Input: ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]

## Solution

* 這題雖然題目說給予的是`char[]`，但實際上是`vector<char>`，應該是有改過題目的型別但忘了改敘述。
* 也因為這樣，執行時間會輸給很多人，推測是因為以前不用使用`vector`速度較快。
* 反轉字串，將第一個和最後一個交換、第二個和倒數第二交換......第`i`個和第`n - i`個交換即可。

### Code

class Solution {
public:
    void reverseString(vector<char>& s) {
        char temp;
        int S = s.size();
        for(int i = 0; i < S / 2; i++)
        {
            temp = s[i];
            s[i] = s[S - 1 - i];
            s[S - 1 - i] = temp;
        }
    }
};

###### tags: `LeetCode` `C++`