LeetCode 344. Reverse String

Description

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"]

描述

編寫一個函數，其做用是將輸入的字符串反轉過來。輸入字符串以字符數組 char[] 的形式給出。

不要給另外的數組分配額外的空間，你必須原地修改輸入數組、使用 O(1) 的額外空間解決這一問題。

你能夠假設數組中的全部字符都是 ASCII 碼錶中的可打印字符。

示例 1：

輸入：["h","e","l","l","o"]
輸出：["o","l","l","e","h"]
示例 2：

輸入：["H","a","n","n","a","h"]
輸出：["h","a","n","n","a","H"]

思路

• 第一個位置的元素和嘴後一個位置的元素交換，第二個和倒數第二個，第三個和倒數第三個 ...

# -*- coding: utf-8 -*-
# @Author:             何睿
# @Create Date:        2019-04-08 21:47:07
# @Last Modified by:   何睿
# @Last Modified time: 2019-04-08 21:54:18


class Solution:
    def reverseString(self, s: [str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        # 中間位置的索引，最後一個位置的索引
        half, count = len(s) // 2, len(s) - 1
        for i in range(half):
            s[i], s[count - i] = s[count - i], s[i]

源代碼文件在 這裏 。
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