Class Solution
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public final class Solution
1021 - Remove Outermost Parentheses\.
Easy
A valid parentheses string is either empty
""
,"(" + A + ")"
, orA + B
, whereA
andB
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 intos = A + B
, withA
andB
nonempty 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
s
after 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')'
.s
is 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 String
removeOuterParentheses(String s)
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Method Detail
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removeOuterParentheses
final String removeOuterParentheses(String s)
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