leetcode刷题python之有效的括号

class Solution:
    def isValid(self, s: str) -> bool:
        dict={')':'(',']':'[','}':'{'}
        stack=[]
        for i in s:
            if i in dict:
                top=stack.pop() if stack else '%'
                if dict[i]!=top:
                    return False
            else:
                stack.append(i)
        return not bool(stack)

题解里的答案简直是太好用了，pop()在这里非常适合，可获取栈顶的值，并且同时将其从栈中退出，一举两得

class Solution:
    def isValid(self, s: str) -> bool:
        if s=="":return True
        list_open={"(":1,"[":2,"{":3}
        list_close={")":1,"]":2,"}":3}
        num_open=0
        num_close=0
        result=True
        for i in s:
            if i in list_open:num_open+=1
            if i in list_close:num_close+=1
        if num_open==num_close:
            half_window=0
            for idx,j in enumerate(s):
                if j in list_open:
                    half_window+=1
                else:
                    left=idx-half_window
                    right=idx+half_window-1
                    while left

反观自己一开始写的，废了半天的劲儿，写出来一个超时的/。。。。