![https://leetcode.com/problems/remove-outermost-parentheses/]
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
class Solution {
public String removeOuterParentheses(String S) {
StringBuilder sb = new StringBuilder();
int opened = 0;
for (char c : S.toCharArray()) {
if (c == '(') {
// skip first opened
if (opened != 0) {
sb.append(c);
}
opened++;
} else {
opened--;
// skip last closed
if (opened != 0) {
sb.append(c);
}
}
}
return sb.toString();
}
}