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题目

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

分析

字符串S只包含左括号和右括号，左括号和右括号匹配成一对，这就是一个“原始对 primitive”。
如果字符串Si不能拆成A+B的形式（A和B都是匹配好的括号对），则Si是原始对。

比如:Input: “(()())(())(()(()))”
按括号匹配拆成(()())、(())、(()(()))
这三个不是原始对，继续拆
(()()) --> ()()
(()) --> ()
(()(())) -->()(())

就是去掉外层括号

思路：
找出一个最典型的符合单位
(()) --> ()

对于一个基本单位，去掉第一个’(’，去掉最后一个’)’
初始化标记flag=0；
遍历字符串，
如果当前字符为’(‘且flag++>0，则加入到结果集（说明当前字符不是第一个’(’）；
如果当前字符为’)‘且flag–>1，则加入到结果集（说明当前字符不是最后一个’)’）；

用(()) --> ()演绎一下；

解答

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();
    }
}

凡是“配对”的题目，都可以考虑利用一个数加减进行匹配操作。