LeetCode 1021. Remove Outermost Parentheses_leetcode 1021 remove outermost parentheses python-优快云博客

博客围绕有效括号字符串展开，介绍了有效括号字符串及原始有效括号字符串的定义，给定一个有效括号字符串，需对其进行原始分解，最终要去除每个原始字符串的最外层括号并返回结果，还给出了示例。

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

//Java
class Solution {
    public String removeOuterParentheses(String S) {
        StringBuilder s = new StringBuilder();
        int opened = 0;
        for (char c : S.toCharArray()) {
            if (c == '(' && opened++ > 0) s.append(c);
            if (c == ')' && opened-- > 1) s.append(c);
        }
        return s.toString();
    }
}

//C++
class Solution {
public:
    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;
    }
};