【LeetCode】844. Backspace String Compare 解答・解説【Python】

LeetCode

2023.04.222023.04.19

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問題
1. 原文
2. 内容
解答
1. 解答1：Python
2. 解答2:
メモ・参考・感想

問題

原文

Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character.

Note that after backspacing an empty text, the text will continue empty.

Example 1:

Input: s = "ab#c", t = "ad#c" Output: true Explanation: Both s and t become "ac".

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Input: s = "ab#c", t = "ad#c"
Output: true
Explanation: Both s and t become "ac".

Example 2:

Input: s = "ab##", t = "c#d#" Output: true Explanation: Both s and t become "".

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Input: s = "ab##", t = "c#d#"
Output: true
Explanation: Both s and t become "".

Example 3:

Input: s = "a#c", t = "b" Output: false Explanation: s becomes "c" while t becomes "b".

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Input: s = "a#c", t = "b"
Output: false
Explanation: s becomes "c" while t becomes "b".

Constraints:

1 <= s.length, t.length <= 200

s and t only contain lowercase letters and '#' characters.

Follow up: Can you solve it in O(n) time and O(1) space?

内容

2つの文字列sとtが与えられたとき、空のテキストエディタに両者を入力し、

等しければ真を返してください。

‘#’はバックスペース文字を意味します。

空のテキストをバックスペースした後、テキストは空のままであることに注意してください。

※正しくない可能性があります。

解答

解答1：Python

class Solution:
    def backspaceCompare(self, S, T):
        def back(res, c):
            if c != '#': res.append(c)
            elif res: res.pop()
            return res
        return reduce(back, S, []) == reduce(back, T, [])