[Swift]LeetCode87. 扰乱字符串 | Scramble String

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Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

    great
   /    
  gr    eat
 /     /  
g   r  e   at
           / 
          a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

    rgeat
   /    
  rg    eat
 /     /  
r   g  e   at
           / 
          a   t

We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

    rgtae
   /    
  rg    tae
 /     /  
r   g  ta  e
       / 
      t   a

We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true

Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

---

给定一个字符串 s1，我们可以把它递归地分割成两个非空子字符串，从而将其表示为二叉树。

下图是字符串 s1 = "great" 的一种可能的表示形式。

    great
   /    
  gr    eat
 /     /  
g   r  e   at
           / 
          a   t

在扰乱这个字符串的过程中，我们可以挑选任何一个非叶节点，然后交换它的两个子节点。

例如，如果我们挑选非叶节点 "gr" ，交换它的两个子节点，将会产生扰乱字符串 "rgeat" 。

    rgeat
   /    
  rg    eat
 /     /  
r   g  e   at
           / 
          a   t

我们将 "rgeat” 称作 "great" 的一个扰乱字符串。

同样地，如果我们继续将其节点 "eat" 和 "at" 进行交换，将会产生另一个新的扰乱字符串 "rgtae" 。

    rgtae
   /    
  rg    tae
 /     /  
r   g  ta  e
       / 
      t   a

我们将 "rgtae” 称作 "great" 的一个扰乱字符串。

给出两个长度相等的字符串 s1 和 s2，判断 s2 是否是 s1 的扰乱字符串。

示例 1:

输入: s1 = "great", s2 = "rgeat"
输出: true

示例 2:

输入: s1 = "abcde", s2 = "caebd"
输出: false

---

56ms

class Solution {
    func isScramble(_ s1: String, _ s2: String) -> Bool {
        if s1 == s2 {
            return true
        }

        var letters = [Int](repeating: 0, count: 26)

        for i in 0..<s1.count {
            let aASCII = Character("a").ascii
            letters[s1[i].ascii - aASCII] += 1
            letters[s2[i].ascii - aASCII] -= 1
        }
        for i in 0..<26 {
            if letters[i] != 0 {
                return false
            }
        }
        for i in 1..<s1.count {
            if isScramble(s1[0..<i], s2[0..<i]) &&
                isScramble(s1[i..<s1.count], s2[i..<s2.count]) {
                return true
            }
            if isScramble(s1[0..<i], s2[(s2.count - i)..<s2.count]) &&
                isScramble(s1[i..<s1.count], s2[0..<(s2.count - i)]) {
                return true
            }
        }

        return false
    }
}

extension String {
    subscript (i: Int) -> Character {
        return self[index(startIndex, offsetBy: i)]
    }
    
    subscript(_ range: CountableRange<Int>) -> String {
        let idx1 = index(startIndex, offsetBy: max(0, range.lowerBound))
        let idx2 = index(startIndex, offsetBy: min(self.count, range.upperBound))
        return String(self[idx1..<idx2])
    }
}

extension Character {
    var ascii: Int {
        get {
            let s = String(self).unicodeScalars
            return Int(s[s.startIndex].value)
        }
    }
    
    func isDigit() -> Bool {
        return self >= "0" && self <= "9"
    }
}

---

56ms

class Solution {
    func isScramble(_ s1: String, _ s2: String) -> Bool {
        return isScramble_recursion(s1: s1, s2: s2)
    }
    
    func isScramble_recursion(s1: String?, s2: String?) -> Bool {
        if s1 == nil || s2 == nil || s1?.count != s2?.count {
            return false
        }
        if s1 == s2 {
            return true
        }
        if (s1?.sorted())! != (s2?.sorted())! {
            return false
        }
        let count: Int = (s1?.count)!
        for i in 1 ..< count {
            if isScramble_recursion(s1: s1![0..<i], s2: s2![0..<i]) && isScramble_recursion(s1: s1![i..<Int.max], s2: s2![i..<Int.max]) {
                return true
            }
            if isScramble_recursion(s1: s1![0..<i], s2: s2![(s2?.count)! - i..<Int.max]) && isScramble_recursion(s1: s1![i..<Int.max], s2: s2![0..<(s2?.count)! - i]) {
                return true
            }
        }
        return false
    }
    static func isScramble_iteration(s1: String?, s2: String?) -> Bool {
        if s1 == nil || s2 == nil || s1?.count != s2?.count {
            return false
        }
        let len = s1?.count
        var dp: Array<Array<Array<Bool>>> = Array<Array<Array<Bool>>>(repeating: Array<Array<Bool>>(repeating: Array<Bool>(repeating: false, count: 100), count: 100), count: 100)
        for i in (0...len! - 1).reversed() {
            for j in (0...len!-1).reversed() {
                dp[i][j][1] = (s1![i] == s2![j])
                var l = 2
                while i + l <= len! && j + l <= len! {
                    for n in 1 ..< l {
                        dp[i][j][l] = dp[i][j][l] || ( dp[i][j][n] && dp[i+n][j+n][l-n] )
                        dp[i][j][l] = dp[i][j][l] || ( dp[i][j+l-n][n] && dp[i+n][j][l-n] )
                    }
                    l += 1
                }
            }
        }
        return dp[0][0][len!]
    }
}
private extension String {
    subscript (range: Range<Int>) -> String {
        guard let localEndIndex = self.index(self.startIndex, offsetBy: range.upperBound, limitedBy: self.endIndex) else {
            return String(self[self.index(self.startIndex, offsetBy: range.lowerBound)..<self.endIndex])
        }
        return String(self[self.index(self.startIndex, offsetBy: range.lowerBound)..<localEndIndex])
    }
    subscript (index: Int) -> Character {
        return self[self.index(self.startIndex, offsetBy: index)]
    }
}

---

180ms

class Solution {
    
    func isScramble(_ s1: String, _ s2: String) -> Bool {
        guard s1.count == s2.count else { return false }
        guard s1.count > 0 else { return false }
        
        var dp = [[[Bool]]](repeating: [[Bool]](repeating: [Bool](repeating: false, count: s1.count + 1), count: s1.count), count: s1.count)
        
        for i in 0..<s1.count {
            for j in 0..<s2.count {
                dp[i][j][1] = s1.charAt(i) == s2.charAt(j)
            }
        }
        
        if s1.count > 1 {
            for len in 2...s1.count {
                for i in 0...s1.count - len {
                    for j in 0...s2.count - len {
                        for w in 1..<len {
                            if dp[i][j][w] && dp[i + w][j + w][len - w] {
                                dp[i][j][len] = true
                                break
                            }
                            if dp[i][j + len - w][w] && dp[i + w][j][len - w] {
                                dp[i][j][len] = true
                                break
                            }
                        }
                    }
                }
            }
        }
        
        return dp[0][0][s1.count]
    }
}

extension String {
    
    func charAt(_ i: Int) -> Character {
        let index = self.index(self.startIndex, offsetBy: i)
        return self[index]
    }
}