算法19（力扣244）反转字符串
1、问题
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。
2、示例
（1）
示例 1：
输入：s = ["h","e","l","l","o"]
输出：["o","l","l","e","h"]
（2）
示例 2：
输入：s = ["H","a","n","n","a","h"]
输出：["h","a","n","n","a","H"]
3、思路
遍历数组，将数组的第一个和最后一个交换，以此往复，直至，左边的>=右边的，结束循环
4、具体步骤
5、完整代码
<!DOCTYPE html>
<html lang="en"><head><meta charset="UTF-8" /><meta name="viewport" content="width=device-width, initial-scale=1.0" /><title>反转字符串</title>
</head><body><p><p>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。</p><p>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。</p><p><p>示例 1：输入：s = ["h","e","l","l","o"]输出：["o","l","l","e","h"]</p><p>示例 2：输入：s = ["H","a","n","n","a","h"]输出：["h","a","n","n","a","H"]</p></p></p>
<script>const s = ["H","a","n","n","a","h"]reverseString(s)function reverseString(s) {for(let left = 0,right = s.length-1;left<right;left++,right--){[s[left],s[right]] = [s[right],s[left]]}// console.log(s);return s};
</script>
</body></html>6、力扣通过代码
/*** @param {character[]} s* @return {void} Do not return anything, modify s in-place instead.*/
var reverseString = function(s) {for(let left = 0,right = s.length-1;left<right;left++,right--){[s[left],s[right]] = [s[right],s[left]]}console.log(s);return s
};