暴力替换 时间复杂度 O(N^2)

class Solution:
    def isValid(self, s: str) -> bool:
        if len(s) % 2 == 1:
            return False
        while True:
            startLen = len(s)
            s =  s.replace('[]','')
            s =  s.replace('{}','')
            s =  s.replace('()','')
            # 替换后和替换前长度相同 要么没有括号对 要么是空字符串
            if len(s) == startLen:
                return s == ''

利用栈 时间复杂度 O(N)

class Solution:
    def isValid(self, s: str) -> bool:
        if len(s) % 2 == 1:
            return False
        # 加一个 ? 来规避栈为空的情况
        dic = {'(':')', '[':']', '{':'}', '?':'?'}
        stack = ['?']
        for char in s:
            if char in dic:
                stack.append(char)
            elif dic[stack.pop()] != char:
                return False
        return len(stack) == 1