算法：移除最外层的括号1021. Remove Outermost Parentheses

Posted 2021-10-18 架构师易筋

1021. Remove Outermost Parentheses

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 = P1 + P2 + … + Pk, where Pi 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: s = "(()())(())"
Output: "()()()"
Explanation: 
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

Example 2:

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

Example 3:

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

Constraints:

• 1 <= s.length <= 105
• s[i] is either ‘(’ or ‘)’.
• s is a valid parentheses string.

超级优雅的解法，不用stack

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