LeetCode 344. Reverse String

摘要： Description Write a function that reverses a string. The input string is given as an array of characters char[]. Do not allocate extra space fo...

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

編寫一個函式，其作用是將輸入的字串反轉過來。輸入字串以字元陣列 char[] 的形式給出。

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

你可以假設陣列中的所有字元都是 ASCII 碼錶中的可列印字元。

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]

原始碼檔案在這裡 。

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