LeetCode 344. 反转字符串 Reverse String
Table of Contents
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()
