1021. Remove Outermost Parentheses

QUESTION:

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 Bnonempty 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:

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

EXPLANATION:

其实这个问题就很简单，只要有一个思路就可以：

怎么样才能算一个primitive的string呢。那就是括号需要对应。怎么才能对应呢？

那就是n个”(“对上n个”)”，那结果就显而易见了。

遇到”(“的时候，标记+1

遇到”)”的时候，标记-1

当标记为0的时候，要不然是开始，要不然就是结束。

SOLUTION:

class Solution {
    public String removeOuterParentheses(String S) {
        int index = 0,start=0,end=0;
        StringBuilder sb = new StringBuilder();
        for(int i = 0;i<S.length();i++){
            if(index==0) start = i;
            char tmp = S.charAt(i);
            if(tmp=='(') index++;
            if(tmp==')') index--;
            if(index==0) {
                end = i;
                sb.append(S.substring(start+1,end));
            }
        }
        return sb.toString();
    }
}
--> return <--