844. Backspace String Compare

leetcode

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?

class Solution {
public:
    bool backspaceCompare(string s, string t) {
        int i = s.size() - 1;
        int j = t.size() - 1;

        int s_backspace_count = 0;
        int t_backspace_count = 0;

        // 只要其中一個字串沒有掃完，就繼續掃
        while(i >= 0 || j >= 0) {
            while(i >= 0) {
                if(s[i] == '#') {
                    --i;
                    // 要扣掉的數量 + 1
                    ++s_backspace_count;
                } else if(s_backspace_count > 0) {
                    --i;
                    --s_backspace_count;
                } else break;
            }
            while(j >= 0) {
                if(t[j] == '#') {
                    --j;
                    ++t_backspace_count;
                } else if(t_backspace_count > 0) {
                    --j;
                    --t_backspace_count;
                } else break;
            }
            // 如果發現 char 不一樣，直接 return false
            if (i >= 0 && j >= 0 && s[i] != t[j])
                return false;

            // 如果都被扣光光了，檢查 i 跟 j 是否相等
            // 這行可以讓程式 early return
            if(i < 0 || j < 0) return i == j;
            --i; --j;
        }
        return i == j;
    }
};
  • T: O(M+N)O(M + N)
  • S: O(1)O(1)