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

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