Class Solution
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public final class Solution1021 - Remove Outermost Parentheses.
Easy
A valid parentheses string is either empty
"","(" + A + ")", orA + B, whereAandBare valid parentheses strings, and+represents string concatenation.For example,
"","()","(())()", and"(()(()))"are all valid parentheses strings.
A valid parentheses string
sis primitive if it is nonempty, and there does not exist a way to split it intos = A + B, withAandBnonempty valid parentheses strings.Given a valid parentheses string
s, consider its primitive decomposition: <code>s = P<sub>1</sub> + P<sub>2</sub> + ... + P<sub>k</sub></code>, where <code>P<sub>i</sub></code> are primitive valid parentheses strings.Return
safter removing the outermost parentheses of every primitive string in the primitive decomposition ofs.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:
<code>1 <= s.length <= 10<sup>5</sup></code>
s[i]is either'('or')'.sis a valid parentheses string.
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final StringremoveOuterParentheses(String s)-
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Method Detail
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removeOuterParentheses
final String removeOuterParentheses(String s)
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