LeetCode1021.Remove Outermost Parentheses（删除最外层的括号）

Description

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.

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有效括号字符串为空 ("")、"(" + A + ")" 或 A + B，其中 A 和 B 都是有效的括号字符串，+ 代表字符串的连接。例如，""，"()"，"(())()" 和 "(()(()))" 都是有效的括号字符串。

如果有效字符串 S 非空，且不存在将其拆分为 S = A+B 的方法，我们称其为原语（primitive），其中 A 和 B 都是非空有效括号字符串。

给出一个非空有效字符串 S，考虑将其进行原语化分解，使得：S = P_1 + P_2 + ... + P_k，其中 P_i 是有效括号字符串原语。

对 S 进行原语化分解，删除分解中每个原语字符串的最外层括号，返回 S 。

题目链接：https://leetcode.com/problems/remove-outermost-parentheses/

个人主页：http://redtongue.cn or https://redtongue.github.io/

Difficulty: easy

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:

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

分析

• 用index记录匹配的左右括号的数目，负数代表左括号“（”比右括号“）”多，绝对值代表多的数目；
• 遍历S，res存储删除最外层括号的结果，若index=0且当前字符为“（”，index-=1，继续；
• 若index=-1，且当前字符为“）”，index+=1，继续，（这是匹配到最外层括号的情况）；
• 反之当前字符加到res中，若当前字符为“（”，index-=1，反之，index+=1；
• 返回res。

参考代码

class Solution(object):
def removeOuterParentheses(self, S):
    res=''
    index=0
    for s in S:
        if(index==0 and s=='('):
            index-=1
            continue
        elif(index==-1 and s==')'):
            index+=1
            continue
        else:
            res+=s
        if(s=='('):
            index-=1
        else:
            index+=1
    return res