844. Backspace String Compare

 844. Backspace String Compare  ( https://leetcode.com/problems/backspace-string-compare/ )

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?

思路
輸入兩個字串，要先處理成真實要比較的字串，所以用一個 function realString 來處理
最後比較兩個經 realString 處理後的結果
realString 內就用 stack 的觀念去做 push 與 pop
幸運的是 C++ 的 string 就有這樣的功能，很方便，不用自己 maintain stack
要注意的是 pop 的時候，string 不能是空的

class Solution {

public:

    bool backspaceCompare(string s, string t) {

        return realString(s) == realString(t);

    }

    string realString(string strInput) {

        string strOutput = "";

        for(char c : strInput) {

            if (c == '#') {

                if (strOutput != "")

                    strOutput.pop_back();

            } else {

                strOutput.push_back(c);

            }

            cout << strOutput << endl;

        }

        return strOutput;

    }

};

參考
https://www.youtube.com/watch?v=aILH2R7zjXg&list=PLY_qIufNHc29OLToHki4t0Z5KIhUzdYxD&index=9