leetcode刷題記錄 easy(3) 1021.Remove Outermost Parentheses

英文題目：

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_iare 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:

1. S.length <= 10000
2. S[i] is "(" or ")"
3. S is a valid parentheses string

中文題目解釋：

注意：

1. S.length <= 10000
2. S[i]是"("或")"
3. S 是一個有效的括號字符串

解析：

定義變量count=0，start=0,

遍歷字符串S，字符的索引下標爲i：

遇到一個“(”就count+1，此時若count==1，則statrt置爲該“(”的索引下標i；(表明一個有效括號字符串的開始)

遇到一個“)”就count-1，此時若count==0證明是一個有效括號字符串，則截取S的start+1到i的字符串放入新字符串中；（去掉這個有效括號字符串最外層的“()”，並放入新字符串中）

提交結果：

class Solution {
    public String removeOuterParentheses(String S) {
        char[] chars=S.toCharArray();
        StringBuilder sb=new StringBuilder();
        int count=0,start=0;
        for (int i=0;i<chars.length;i++) {
            if(chars[i]=='('){
                count++;
                if(count==1){
                    start=i;
                }
            }else if(chars[i]==')'){
                count--;
                if(count==0){
                    sb.append(S.substring(start+1,i));
                }
            }
        }
        return sb.toString();
    }
}