LeetCode 344. 反转字符串 Reverse String

Table of Contents

一、中文版

二、英文版

三、My answer

四、解题报告

一、中文版

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 char[] 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。

你可以假设数组中的所有字符都是 ASCII 码表中的可打印字符。

示例 1：

输入：["h","e","l","l","o"]

输出：["o","l","l","e","h"]

示例 2：

输入：["H","a","n","n","a","h"]

输出：["h","a","n","n","a","H"]

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/reverse-string

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

二、英文版

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.

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"]

三、My answer

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        tmp = ''
        for i in range(len(s)//2):
            tmp = s[i]
            s[i] = s[len(s)-1-i]
            s[len(s)-1-i] = tmp
        return s

四、解题报告

数据结构：无额外数据结构，空间复杂度 O(1)

算法：双指针变形

实现：两个指针一头一尾，相向而行，每一步交换时都用一个临时变量。

其实 Python 里有自带函数，可以一行搞定：

class Solution:
    def reverseString(self, s):
        s.reverse()