题目名称

844. Backspace String Compare

题目描述

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”.

Example 2:
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”.

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?

初试代码

// 我的代码
class Solution {
public:bool backspaceCompare(string s, string t) {process(s);process(t);return s==t;}void process(string &s){//必须传引用int len = size(s);int j=0;for(int i=0; i0){j--;}}}//s[j] = '\0'; //c++中的string并不是以'\0'作为结束标志，只有c语言中的char *是这样的s = s.substr(0, j); //substr字符串切片切片}
};

学到了啥

见注释