【leetcode】1021. Remove Outermost Parentheses

题目如下：

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

解题思路：括号配对的题目在leetcode出现了很多次了，从左往右遍历数组，分别记录左括号和右括号出现的次数，当两者相等的时候，即为一组括号。

代码如下：

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