[LeetCode]344. Reverse String

Algorithm I 筆記撰寫計畫

敘述

這是 Algorithm I 的第四天第一個題目，總共有兩題。

• 難度: Easy
• 花費時間: 30min
• 題目

傳入一個包含字母的陣列，把陣列原地(in-place)反轉。

空間複雜度必須為 O(1)

點我開啟限制與範例

限制:

1 <= s.length <= 105
s[i] is a printable ascii character.

Example 1:
Input: s = ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]
Example 2:
Input: s = ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]

筆記

這道題使用 Two Pointer 非常的簡單，作法如下:

• Pointer 1: 遍歷陣列到中間
• Pointer 2: 反向遍歷陣列到中間

每次迴圈把指針 element 交換，再推進 Pointer 。

/**
 * @param {character[]} s
 * @return {void} Do not return anything, modify s in-place instead.
 */
var reverseString = function (s) {
    let temp;
    for (let i = 0; i < Math.floor(s.length / 2); i++) {
        temp = s[i];
        s[i] = s[s.length - i - 1];
        s[s.length - i - 1] = temp;
    }

    // console.log(s)
};

下面也有更簡單的做法，一行完成，哈哈哈。

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

成績

Two Pointers 成績

內建函式 reverse() 寫法成績

點我開啟舊寫法/失敗寫法