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344. 反转字符串

English Version

题目描述

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"]

示例 2：

输入：s = ["H","a","n","n","a","h"]
输出：["h","a","n","n","a","H"]

 

提示：

• 1 <= s.length <= 105
• s[i] 都是 ASCII 码表中的可打印字符

解法

Python3

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        s[:] = s[::-1]

Java

class Solution {
    public void reverseString(char[] s) {
        for (int i = 0, j = s.length - 1; i < j; ++i, --j) {
            char t = s[i];
            s[i] = s[j];
            s[j] = t;
        }
    }
}

C++

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

Go

func reverseString(s []byte) {
	for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
		s[i], s[j] = s[j], s[i]
	}
}

JavaScript

/**
 * @param {character[]} s
 * @return {void} Do not return anything, modify s in-place instead.
 */
var reverseString = function (s) {
    for (let i = 0, j = s.length - 1; i < j; ++i, --j) {
        [s[i], s[j]] = [s[j], s[i]];
    }
};

Rust

impl Solution {
    pub fn reverse_string(s: &mut Vec<char>) {
        let n = s.len();
        let mut l = 0;
        let mut r = n - 1;
        while l < r {
            s.swap(l, r);
            l += 1;
            r -= 1;
        }
    }
}

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