leetcode：844. Backspace String Compare_given two聽strings聽s聽and t,聽return if they are equa-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_42522262/article/details/101031922

本文探讨了在存在退格符('#')的情况下，判断两个字符串是否相等的问题。通过从后向前遍历的方法，实现O(N)时间复杂度和O(1)空间复杂度的高效算法。

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题目：

Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.

Example 1:

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 "".
Example 3:

Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".
Example 4:

Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".
Note:

1 <= S.length <= 200
1 <= 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?

时间：2019.9.19

分析：本题想做的话还是很简单的。但如果考虑 follow up 的要求：O(N) time and O(1) space ，就不简单了。说实话，一开始我真的没有做出来，没有通过所有的案例，且代码写得比较复杂。
我是从下标 0 开始遍历的，分别赋予字符串下标 i， j，考虑字符串相等，不等或遇到 ‘#’ 等情况，自以为写的蛮好，测试的时候却出现各种之前没有想过的数据。于是发现自己的思路有问题。也许缝缝补补最终也能做出来吧。但这样终究不是好办法。后来看了别人的代码，实在佩服。短短几行代码就能搞定。关键在于从后向前遍历。
代码先不贴了，下次重新做一遍再贴出来。