Leetcode 921 Minimum Add to Make Parentheses Valid

思想：建立空棧，遍歷字串，遇到左括號壓入棧中，遇到有括號進行判斷，如果棧空直接壓入棧中，如果與棧頂元素匹配則棧頂元素出棧，否則壓棧，返回棧中元素的個數即為所求

class Solution(object):
    def minAddToMakeValid(self, S):
        """
        :type S: str
        :rtype: int
        """
        stack=[]
        for x in S:
            if x=='(':
                stack.append(x)
            elif x==')':
                if not stack:
                    stack.append(x)
                elif stack[-1]=='(':
                    stack.pop()                
                else:
                    stack.append(x)
        return len(stack)