题解 | 牛群的秘密通信

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param s string字符串 
# @return bool布尔型
#
class Solution:
    def is_valid_cow_communication(self , s: str) -> bool:
        # write code here
        a = []
        for i in s:
            if not a:
                a.append(i)
            else:
                if i == ')' and a[-1] == '(':
                    a.pop(-1)
                elif i == ']' and a[-1] == '[':
                    a.pop(-1)
                elif i == '}' and a[-1] == '{':
                    a.pop(-1)
                else:
                    a.append(i)
        return len(a) == 0