[leetcode] 844. Backspace String Compare

题目

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

从右到左遍历两个字符串，当遇到“#”符号时，删去相应的字符，随后对比字符串的值。

智慧算法：遍历字符串的同时，构造结果字符串，正常时复制字符，遇到#时弹出最后一位。

代码

python版本：

class Solution:
    def backspaceCompare(self, s: str, t: str) -> bool:
        def back(s, p):
            count = 0
            while p>=0 and count != 0 or s[p] == '#':
                count = count+1 if s[p] == '#' else count-1
                p -= 1
            return p

        p1, p2 = len(s)-1, len(t)-1
        while p1 >= 0 and p2 >= 0:
            # print(p1, p2, s[p1], t[p2])
            p1 = back(s, p1)
            p2 = back(t, p2)
            # print(p1, p2, s[p1], t[p2])
            # print()
            if (p1 < 0 and p2 >= 0) or (p1 >= 0 and p2 < 0) or p1 >= 0 and p2 >= 0 and s[p1] != t[p2]:
                return False
            p1 -= 1
            p2 -= 1
        p1 = back(s, p1) if p1 >= 0 else p1
        p2 = back(t, p2) if p2 >= 0 else p2
        return True if p1 < 0 and p2 < 0 else False

    # 智慧算法。。。
    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, [])

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