LeetCode 1021. Remove Outermost Parentheses

原题链接在这里：https://leetcode.com/problems/remove-outermost-parentheses/

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 "()()" + "()" = "()()()".

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

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

题解：

Count open parentheses, when it is (, count++. If it is outermost (, count == 1 now and skip.

When it is ), count--. If it is outermost ), count == 0 now and skip.

Otherwise append it to StringBuilder.

Time Complexity: O(n). n = S.length().

Space: O(n).

AC Java:

class Solution {
    public String removeOuterParentheses(String S) {
        int count = 0;
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i<S.length(); i++){
            char c = S.charAt(i);
            if(c == ‘(‘){
                count++;
                if(count == 1){
                    continue;
                }
            }else{
                count--;
                if(count == 0){
                    continue;
                }
            }
            
            sb.append(c);
        }
        
        return sb.toString();
    }
}