921. Minimum Add to Make Parentheses Valid

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最新推荐文章于 2024-06-22 09:00:00 发布

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分类专栏： leetcode

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本文介绍了一种算法，用于计算使给定的包含‘(’和‘)’的字符串变为有效括号字符串所需的最小添加括号数量。通过使用栈来跟踪左括号，当遇到右括号时，若栈中有左括号则弹出，否则增加右括号计数。最终返回未匹配左括号及额外右括号的总数。

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Given a string S of ‘(’ and ‘)’ parentheses, we add the minimum number of parentheses ( ‘(’ or ‘)’, and in any positions ) so that the resulting parentheses string is valid.

Formally, a parentheses string is valid if and only if:

It is the empty string, or
It can be written as AB (A concatenated with B), where A and B are valid strings, or
It can be written as (A), where A is a valid string.
Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.

Example 1:

Input: “())”
Output: 1
Example 2:

Input: “(((”
Output: 3
Example 3:

Input: “()”
Output: 0
Example 4:

Input: “()))((”
Output: 4

分析：
直接把左括号进栈，遇到右括号时，如果栈里有左括号就弹出，否则需要加上右括号的个数。最后还要加上栈里面左括号的个数。
python:

class Solution(object):
    def minAddToMakeValid(self, S):
        """
        :type S: str
        :rtype: int
        """
        res = 0
        stack = []
        for c in S:
            if c == '(':
                stack.append('(')
            else:
                if stack:
                    stack.pop()
                else:
                    res += 1
        return res + len(stack)