题解 | #括号序列#

两种题型：

只有一种括号匹配可以用count

    def matchParentheses(self, string):

        count = 0

        for i in range(len(string)):
            if string[i] == '(':
                count += 1
            else:
                count -= 1

            if count < 0:
                return False

        return True if count == 0 else False

有多种括号匹配时 用count做没法完成“({)}”这样的edge case，所以要用stack来match

class Solution:
    def isValid(self , s ):
        # write code here
        stack = []

        for i in range(len(s)):
            if s[i] == '(':
                stack.append(')')
            elif s[i] =='[':
                stack.append(']')
            elif s[i] =='{':
                stack.append('}')
            elif stack == []:
                return False
            elif s[i] != stack[-1]:
                return False
            else:
                stack.pop()

        return True if stack == [] else False