[Swift]LeetCode856. 括号的分数 | Score of Parentheses

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Given a balanced parentheses string S, compute the score of the string based on the following rule:

• () has score 1
• AB has score A + B, where A and B are balanced parentheses strings.
• (A) has score 2 * A, where A is a balanced parentheses string. 

Example 1:

Input: "()"
Output: 1

Example 2:

Input: "(())"
Output: 2

Example 3:

Input: "()()"
Output: 2

Example 4:

Input: "(()(()))"
Output: 6 

Note:

1. S is a balanced parentheses string, containing only (and ).
2. 2 <= S.length <= 50

---

给定一个平衡括号字符串 S，按下述规则计算该字符串的分数：

• () 得 1 分。
• AB 得 A + B 分，其中 A 和 B 是平衡括号字符串。
• (A) 得 2 * A 分，其中 A 是平衡括号字符串。 

示例 1：

输入： "()"
输出： 1

示例 2：

输入： "(())"
输出： 2

示例 3：

输入： "()()"
输出： 2

示例 4：

输入： "(()(()))"
输出： 6 

提示：

1. S 是平衡括号字符串，且只含有 ( 和 ) 。
2. 2 <= S.length <= 50

---

Runtime: 4 ms

Memory Usage: 19.5 MB

class Solution {
    func scoreOfParentheses(_ S: String) -> Int {
        var stack:[Int] = [Int]()
        var cur:Int = 0
        for c in S
        {
            if c == "("
            {
                stack.append(cur)
                cur = 0
            }
            else
            {
                cur = stack.removeLast() + max(cur * 2, 1)
            }
        }
        return cur
    }
}

---

4ms

class Solution {
    func scoreOfParentheses(_ S: String) -> Int {
        guard S.count > 1 && S.count%2 == 0 else {
            return 0
        }
        
        var mulArr = [Int]()
        mulArr.append(0)
        var mulInx = 0
        let strArr = Array(S)
        for i in 0..<strArr.count-1 {
            let phrase = "\(strArr[i])\(strArr[i+1])"
            switch phrase{
                case "((":
                    mulInx += 1
                    if mulInx == mulArr.count  {
                        mulArr.append(0)
                    } 
                case "))":                
                    mulArr[mulInx-1] += mulArr[mulInx]*2
                    mulArr[mulInx] = 0
                    mulInx -= 1
                    
                case "()":
                    mulArr[mulInx] += 1
                case ")(":
                    continue
                default:
                    continue
            }
        }
        return mulArr[0]
    }
}

---

8ms

class Solution {
    func scoreOfParentheses(_ S: String) -> Int {
        guard S.count % 2 == 0 else {
            // this means that it is an invalid parentheses combination
            return -1
        }
        
        var sum: Int = 0 
        var stack: [Character] = []
        var shouldCount: Bool = false 
        for character in S {
            if character == "(" {
                stack.append("(")
                shouldCount = true 
            } else {
                // if character == ")"
                stack.removeLast()
                if shouldCount {
                    sum += Int(pow(Double(2), Double(stack.count)))
                    shouldCount = false 
                }
            }
        }
        return sum
    }
}

---

12ms

class Solution {
    func scoreOfParentheses(_ S: String) -> Int {
        if S.count == 0 {
            return 0
        }
        let chars = Array(S)
        
        // O(n)方法，因为每一次括号的嵌套都会乘以2，所以对于一个基础"()"，
        // 它前面有多少个开"("，就要乘以多少个2，2^(d-1),d是"()"自己的"("加上前面发现的开"("
        // 把(()(()))转化为 -> (()) + ((())) = 2^1 + 2^2 = 6
        var sum = 0, leftCount = 0
        var stack = [Character]()
        
        for char in chars {
            if char == "(" {
                // add to stack
                leftCount += 1
            } else {
                if let previous = stack.last, previous == "(" {
                    sum += Int(pow(Double(2), Double(leftCount - 1)))
                }
                leftCount -= 1
            }
            stack.append(char)
        }
        
        return sum
    }
}

---

12ms

class Solution {
    func scoreOfParentheses(_ S: String) -> Int {
    var arr = [0]    
    for c in Array(S) {
        if c == "(" {
            arr.append(0)
        } else {
            let last = arr.removeLast()
            if last == 0 {
                arr[arr.count-1] += 1
            } else {
                arr[arr.count-1] += (2 * last)
            }
        }
    }    
    return arr.last!
  }
}

---

16ms

class Solution {        
    func scoreOfParentheses(_ S: String) -> Int {
        if let result = Int(S) {
            return result
        }
        
        var array = Array(S).map({ String($0) })
        var i = array.count - 1
        while i > 0 {
            if array[i - 1] == "(" && array[i] == ")" {
                array.remove(at: i)
                array[i - 1] = "1"
                i -= 1
            }
            
            i -= 1
        }
        
        return Int(scoreOfParentheses(array)) ?? 0
    }
    
    func scoreOfParentheses(_ string: [String]) -> String {
        if string.count == 1 {
            return string[0]
        }
        var string = string
        
        var i = string.count - 1
        while i > 0 {
            let index = i
            
            if let num1 = Int(String(string[index])), 
                let num2 = Int(String(string[index - 1])){
                string.remove(at: index)
                string[index - 1] = String(num1 + num2)
                i -= 1
            }
            i -= 1
        }
        
        i = string.count - 1
        while i > 1 {
                let index = i
            
                if string[index] == ")", string[index - 2] == "(", 
                    let num = Int(string[index - 1]) {
                    string.remove(at: index)
                    string.remove(at: index - 1)
                    string[index - 2] = String(num * 2)
                    i -= 2
                }
            i -= 1
        }
        
        return scoreOfParentheses(string)
    }
}