Package g1001_1100.s1021_remove_outermost_parentheses

Class Solution

java.lang.Object
g1001_1100.s1021_remove_outermost_parentheses.Solution
public class Solution extends Object
1021 - Remove Outermost Parentheses\. Easy 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.

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • removeOuterParentheses

      public String removeOuterParentheses(String s)