每日 LeetCode] 1021. Remove Outermost Parentheses

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.

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

思路：本题要求去掉字符串中外层的括号。可以使用栈结构，依次进栈，遇到第一个 ( 就进栈，第二个 ( 就加入到返回的字符串中，遇到 ) 时判断此时栈中的元素个数，如果个数为 1 则将已有的栈元素弹出，如果个数为 2 则加入返回字符串后弹出。

C++ 代码

class Solution {
public:
    string removeOuterParentheses(string S) {
        string ret = "";
        stack<char> st;
        int size = S.size();
        for (int i=0; i<size; i++){
            if (S[i] == '('){
                if (st.size() > 0)
                    ret += '(';
                st.push('(');
            }
            else{
                if (st.size() > 1)
                    ret += ')';
                st.pop();
            }
        }
        return ret;
    }
};
​

运行时间：8ms

运行内存：9.2M