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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 "()()" + "()" = "()()()".

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

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

 #include <iostream>  
 #include <string>  
 using namespace std;  
 string removeOuterParentheses(string s)  
 {  
      string res;  
      int opened = 0;  
      for (char c : s)  
      {  
           if (c == '(' && opened++ > 0) res += c;  
           if (c == ')' && opened-- > 1) res += c;  
      }  
      return res;  
 }  
 int main()  
 {  
      string strInput, strOutput;  
      strInput = "(()())(())";  
      strOutput = removeOuterParentheses(strInput);  
      cout << strOutput << endl;  
      getchar();  
 }