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:
S.length <= 10000S[i]is"("or")"Sis a valid parentheses string
有效括号字符串为空 ("")、"(" + A + ")" 或 A + B，其中 A 和 B 都是有效的括号字符串，+ 代表字符串的连接。例如，""，"()"，"(())()" 和 "(()(()))" 都是有效的括号字符串。
如果有效字符串 S 非空，且不存在将其拆分为 S = A+B 的方法，我们称其为原语（primitive），其中 A 和 B 都是非空有效括号字符串。
给出一个非空有效字符串 S，考虑将其进行原语化分解，使得：S = P_1 + P_2 + ... + P_k，其中 P_i 是有效括号字符串原语。
对 S 进行原语化分解，删除分解中每个原语字符串的最外层括号，返回 S 。
示例 1：
输入："(()())(())" 输出："()()()" 解释： 输入字符串为 "(()())(())"，原语化分解得到 "(()())" + "(())"， 删除每个部分中的最外层括号后得到 "()()" + "()" = "()()()"。
示例 2：
输入："(()())(())(()(()))" 输出："()()()()(())" 解释： 输入字符串为 "(()())(())(()(()))"，原语化分解得到 "(()())" + "(())" + "(()(()))"， 删除每隔部分中的最外层括号后得到 "()()" + "()" + "()(())" = "()()()()(())"。
示例 3：
输入："()()" 输出："" 解释： 输入字符串为 "()()"，原语化分解得到 "()" + "()"， 删除每个部分中的最外层括号后得到 "" + "" = ""。
提示：
S.length <= 10000S[i]为"("或")"S是一个有效括号字符串
24ms
1 class Solution { 2 func removeOuterParentheses(_ S: String) -> String { 3 guard !S.isEmpty else { return "" } 4 var result = [Character]() 5 var inSub = true, backCount = 0 6 let s = Array(S) 7 8 for i in 1..<s.count { 9 if s[i] == "(" { 10 if inSub { 11 backCount += 1 12 result.append("(") 13 } else { 14 inSub = true 15 backCount = 0 16 } 17 continue 18 } 19 20 if s[i] == ")" { 21 backCount -= 1 22 if backCount == -1 { 23 inSub = false 24 } else { 25 result.append(")") 26 } 27 continue 28 } 29 } 30 return String(result) 31 } 32 }
28ms
1 class Solution { 2 func removeOuterParentheses(_ S: String) -> String { 3 var chars = Array(S) 4 var count = 0 5 var previousCount = 0 6 var resultArray = [Character]() 7 8 for i in 0..<chars.count { 9 previousCount = count 10 if chars[i] == "(" { 11 count += 1 12 } else if chars[i] == ")" { 13 count -= 1 14 } 15 if count != 0 && previousCount != 0 { 16 resultArray.append(chars[i]) 17 } 18 } 19 let result = String(resultArray) 20 return result 21 } 22 }
32ms
1 class Solution { 2 func removeOuterParentheses(_ S: String) -> String 3 { 4 var oc = 0 5 var temp = "" 6 var ret = "" 7 for c in S 8 { 9 if c == "(" { oc += 1 } 10 else { oc -= 1 } 11 12 temp += String(c) 13 if oc == 0 { 14 ret += self.helper(temp) 15 temp = "" 16 } 17 } 18 return ret 19 } 20 21 private func helper(_ S: String) -> String 22 { 23 guard S.count > 0 else { return S } 24 25 var ret = S 26 if let idx = ret.index(of: "(") { ret.remove(at: idx) } 27 if let idx = ret.lastIndex(of: ")") { ret.remove(at: idx) } 28 return ret 29 } 30 }
36ms
1 class Solution { 2 func removeOuterParentheses(_ S: String) -> String { 3 var s = S.characters.map{String($0)} 4 var temp = [String]() 5 var hold = "" 6 var openCount = 0 7 var closeCount = 0 8 var i = 0 9 while i < s.count{ 10 if s[i] == "("{ 11 hold += s[i] 12 openCount += 1 13 }else if hold.count != 0 && s[i] == ")"{ 14 hold += ")" 15 closeCount += 1 16 } 17 if hold.count != 0 && closeCount == openCount{ 18 temp.append(hold) 19 openCount = 0 20 closeCount = 0 21 hold = "" 22 } 23 i += 1 24 } 25 26 var answer = "" 27 for pair in temp{ 28 let startIndex = pair.index(pair.startIndex, offsetBy: 1) 29 let endIndex = pair.index(pair.endIndex, offsetBy: -1) 30 answer += String(pair[startIndex..<endIndex]) 31 } 32 33 return answer 34 } 35 }
Runtime: 40 ms
Memory Usage: 19.5 MB
1 class Solution { 2 func removeOuterParentheses(_ S: String) -> String { 3 var ret:String = String() 4 var bal:Int = 0 5 var cur:String = String() 6 for c in S 7 { 8 if c == "(" 9 { 10 bal += 1 11 } 12 else 13 { 14 bal -= 1 15 } 16 cur.append(c) 17 if bal == 0 18 { 19 ret += cur.subString(1, cur.count - 2) 20 cur = String() 21 } 22 } 23 return ret 24 } 25 } 26 27 extension String { 28 // 截取字符串：指定索引和字符数 29 // - begin: 开始截取处索引 30 // - count: 截取的字符数量 31 func subString(_ begin:Int,_ count:Int) -> String { 32 let start = self.index(self.startIndex, offsetBy: max(0, begin)) 33 let end = self.index(self.startIndex, offsetBy: min(self.count, begin + count)) 34 return String(self[start..<end]) 35 } 36 }
48ms
1 class Solution { 2 func removeOuterParentheses(_ S: String) -> String { 3 if S.count == 0 { 4 return "" 5 } 6 var s = Array(S) 7 var sk = [Character]() 8 var otherSk = [Character]() 9 var anOtherSK = [Character]() 10 while s.isEmpty == false { 11 if sk.isEmpty { 12 if anOtherSK.isEmpty { 13 anOtherSK.append(s.removeLast()) 14 } else { 15 if s.last == "(" { 16 s.popLast() 17 anOtherSK.popLast() 18 continue 19 } 20 } 21 } 22 if sk.isEmpty && s.last == "(" { 23 continue 24 } 25 26 if let a = s.popLast() { 27 sk.append(a) 28 otherSk.append(a) 29 } 30 if sk.last == "(" { 31 sk.popLast() 32 sk.popLast() 33 } 34 } 35 return String(otherSk.reversed()) 36 } 37 }
52ms
1 class Solution { 2 func removeOuterParentheses(_ S: String) -> String { 3 let S = Array(S) 4 var stack: [Character] = [] 5 var res: [String] = [] 6 var prev = 0 7 for (i, ch) in S.enumerated() { 8 if ch == "(" || ch == "[" || ch == "{" { 9 stack.append(ch) 10 } 11 12 if ch == ")" || ch == "}" || ch == "]" { 13 stack.remove(at: stack.count - 1) 14 } 15 16 if stack.count == 0 { 17 res.append(String(S[prev...i])) 18 prev = i + 1 19 } 20 } 21 22 return res.reduce("", {res, str in return res + String(String(str.dropFirst()).dropLast())}) 23 } 24 }
72ms
1 class Solution { 2 func removeOuterParentheses(_ S: String) -> String { 3 let parentheses = Array(S) 4 var groups = [[String]]() 5 let stack = Stack<String>() 6 var startIndex = 0 7 var result = "" 8 for (index, parenthese) in parentheses.enumerated() { 9 if stack.isEmpty() && index == 0 { 10 stack.push(String(parenthese)) 11 } else { 12 if let element = stack.top?.element, element != String(parenthese) { 13 stack.pop() 14 } else { 15 stack.push(String(parenthese)) 16 } 17 } 18 if stack.isEmpty() { 19 groups.append(Array(parentheses[startIndex...index]).map{ String($0) }) 20 startIndex = index + 1 21 } 22 } 23 24 for group in groups { 25 let simplified = Array(group.dropFirst().dropLast()) 26 result += simplified.joined() 27 } 28 29 return result 30 } 31 } 32 33 final class Stack<Element> { 34 35 var top: Node<Element>? 36 37 @discardableResult 38 func pop() -> Element? { 39 guard let topNode = top else { return nil } 40 top = topNode.next 41 return topNode.element 42 } 43 44 func push(_ element: Element) { 45 if let topNode = top { 46 let newTop = Node(element) 47 newTop.next = topNode 48 top = newTop 49 } else { 50 top = Node(element) 51 } 52 } 53 54 func isEmpty() -> Bool { 55 return top == nil 56 } 57 } 58 59 extension Stack { 60 61 final class Node<Element> { 62 63 var element: Element? 64 var next: Node<Element>? 65 66 init(_ element: Element) { 67 self.element = element 68 } 69 } 70 }
