【LeetCode】1021. Remove Outermost Parentheses 解题报告（Python）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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目录

• • 题目描述
  • 题目大意
  • 解题方法
  • • 遍历
  • 日期

题目地址：https://leetcode.com/problems/remove-outermost-parentheses/

题目描述

A valid parentheses string is either empty (""), "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.

A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.

Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings.

Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.

Example 1:

Input: "(()())(())"
Output: "()()()"
Explanation:
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

Example 2:

Input: "(()())(())(()(()))"
Output: "()()()()(())"
Explanation:
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".

Example 3:

Input: "()()"
Output: ""
Explanation:
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".

Note:

1. S.length <= 10000
2. S[i] is “(” or “)”
3. S is a valid parentheses string

题目大意

找出字符串中每对括号匹配的结果去除最外层的()之后的结果之和。

解题方法

遍历

看到括号匹配，大家一般都想到用一个栈，其实不用栈也可以。

我们只需要一个变量count保存左括号数-右括号数即可。即遇到左括号则自增1，遇到右括号则自减1.当count为0的时候，说明在这一段中左括号和右括号相等，是个完美匹配的括号串了。

我们使用一个变量previ保存的是上一次括号完全匹配之后，下一个括号匹配开始位置。由于这个题需要让我们把最外层的括号去掉，所以，当count==0的时候，我们把结果增加的是[previ + 1, i)，左闭右开区间。

Python代码如下：

class Solution(object):
    def removeOuterParentheses(self, S):
        """
        :type S: str
        :rtype: str
        """
        previ = 0
        res = ""
        count = 0
        for i, s in enumerate(S):
            if s == '(':
                count += 1
            else:
                count -= 1
            if count == 0:
                res += S[previ + 1: i]
                previ = i + 1
        return res

日期

2019 年 4 月 7 日 —— 周赛bug了3次。。