1078. Remove Outermost Parentheses

难度: Easy

题目描述

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.

解题思路

代码实现

解决方案

java

class Solution {
    public String removeOuterParentheses(String s) {

        char[] cArray = s.toCharArray();
        ArrayList<Character> list = new ArrayList<>();
        StringBuilder sb = new StringBuilder();
        int leftCount=0;
        int rightCount=0;
        for (char c : cArray) {
            switch (c) {
                case '(':
                    list.add('(');
                    leftCount++;
                    break;
                case ')':
                    list.add(')');
                    rightCount++;
                    if(leftCount==rightCount){
                        if(list.size()>2){
                            for(int i=1;i<list.size()-1;i++){
                                sb.append(list.get(i));
                            }
                        }
                        leftCount=0;
                        rightCount=0;
                        list.clear();
                    }


                    break;
            }

        }

        return sb.toString();

    }
}