算法自由之路

344.反转字符串

双指针
递归
字符串

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

1、双指针

使用两个指针，一个从左开始， 一个从右开始， 两两交换

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

  const swap = (s, left, right) => {
    const temp = s[left];
    s[left] = s[right];
    s[right] = temp;
  };

  while (left < right) {
    swap(s, left++, right--);
  }
};

2、递归

也可以是用递归求解

/**
 * @param {character[]} s
 * @return {void} Do not return anything, modify s in-place instead.
 */
var reverseString = function (s) {
  if (s == null || s.length === 0) {
    return;
  }

  const swap = (s, i, j) => {
    const temp = s[i];
    s[i] = s[j];
    s[j] = temp;
  };

  const reverseStringHelper = (s, left, right) => {
    if (left >= right) {
      return;
    }
    swap(s, left, right);
    reverseStringHelper(s, ++left, --right);
  };

  reverseStringHelper(s, 0, s.length - 1);
};

3、我的直觉解法（2021.11.22）

头尾交换

/**
 * @param {character[]} s
 * @return {void} Do not return anything, modify s in-place instead.
 */
var reverseString = function (s) {
  const swap = (s, i, j) => {
    const temp = s[i];
    s[i] = s[j];
    s[j] = temp;
  };

  for (let i = 0; i < s.length / 2; i++) {
    swap(s, i, s.length - 1 - i);
  }
};