1. 題目鏈接：

https://leetcode.com/problems/remove-outermost-parentheses/submissions/

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:
S.length <= 10000
S[i] is "(" or ")"
S is a valid parentheses string

2. 題目關(guān)鍵詞

• 難度等級(jí)：easy
• 關(guān)鍵詞：
• 語言： C++

3. 解題思路

遍歷字符串试溯，如果"("數(shù)量于 ")" 數(shù)量一致冀偶，時(shí)婉徘，拼接它們之間的字符串嵌灰。

class Solution {
public:
    string removeOuterParentheses(string S) {
        string sTmp;
        int num = 0;
        
        int index = 0;  // 元字符串
        
        for (int i = 0; i < S.length(); i++) {
            if (S[i] == '(') {
                num++;
            }
            
            if (S[i] == ')') {
                num--;
            }
            
            if (num == 0) {
                sTmp += S.substr(index + 1, i - (index + 1)); // 拼接 ：最外邊 "(" 到 ")" 之間的字符串
                index = i + 1;
            }
        }
        
        return sTmp;
    }
};