[Swift]LeetCode756. 金字塔转换矩阵 | Pyramid Transition Matrix

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We are stacking blocks to form a pyramid. Each block has a color which is a one letter string, like `'Z'`.

For every block of color `C` we place not in the bottom row, we are placing it on top of a left block of color `A` and right block of color `B`. We are allowed to place the block there only if `(A, B, C)` is an allowed triple.

We start with a bottom row of bottom, represented as a single string. We also start with a list of allowed triples allowed. Each allowed triple is represented as a string of length 3.

Return true if we can build the pyramid all the way to the top, otherwise false.

Example 1:

Input: bottom = "XYZ", allowed = ["XYD", "YZE", "DEA", "FFF"]
Output: true
Explanation:
We can stack the pyramid like this:
    A
   / \
  D   E
 / \ / \
X   Y   Z

This works because ('X', 'Y', 'D'), ('Y', 'Z', 'E'), and ('D', 'E', 'A') are allowed triples. 

Example 2:

Input: bottom = "XXYX", allowed = ["XXX", "XXY", "XYX", "XYY", "YXZ"]
Output: false
Explanation:
We can't stack the pyramid to the top.
Note that there could be allowed triples (A, B, C) and (A, B, D) with C != D. 

Note:

1. bottom will be a string with length in range [2, 8].
2. allowed will have length in range [0, 200].
3. Letters in all strings will be chosen from the set {'A', 'B', 'C', 'D', 'E', 'F', 'G'}.

---

现在，我们用一些方块来堆砌一个金字塔。 每个方块用仅包含一个字母的字符串表示，例如 “Z”。

使用三元组表示金字塔的堆砌规则如下：

(A, B, C) 表示，“C”为顶层方块，方块“A”、“B”分别作为方块“C”下一层的的左、右子块。当且仅当(A, B, C)是被允许的三元组，我们才可以将其堆砌上。

初始时，给定金字塔的基层 bottom，用一个字符串表示。一个允许的三元组列表 allowed，每个三元组用一个长度为 3 的字符串表示。

如果可以由基层一直堆到塔尖返回true，否则返回false。

示例 1:

输入: bottom = "XYZ", allowed = ["XYD", "YZE", "DEA", "FFF"]
输出: true
解析:
可以堆砌成这样的金字塔:
    A
   / \
  D   E
 / \ / \
X   Y   Z

因为符合('X', 'Y', 'D'), ('Y', 'Z', 'E') 和 ('D', 'E', 'A') 三种规则。

示例 2:

输入: bottom = "XXYX", allowed = ["XXX", "XXY", "XYX", "XYY", "YXZ"]
输出: false
解析:
无法一直堆到塔尖。
注意, 允许存在三元组(A, B, C)和 (A, B, D) ，其中 C != D.

注意：

1. bottom 的长度范围在 [2, 8]。
2. allowed 的长度范围在[0, 200]。
3. 方块的标记字母范围为{'A', 'B', 'C', 'D', 'E', 'F', 'G'}。

---

24ms

class Solution {
    func pyramidTransition(_ bottom: String, _ allowed: [String]) -> Bool {
        var patterDict = [String: [String]]()
        for w in allowed {
           let key = String(w.prefix(2))
            let val = String(w.suffix(1))
            if patterDict[key] == nil {
                patterDict[key] = [String]()
            }
            patterDict[key]?.append(val)
        }
        
        func buildNext(_ A: String, _ ans: String, _ i: Int) -> Bool {
            //print("A = \(A), ans = \(ans), A = \(i)")
            if A.count == 0 {  //找到可行解
                return true
            }
            if i + 1 == A.count {  //某一排全部摆放完成
                return buildNext(ans, "", 0)
            }
            let w = A.subString(from: i, length: 2)
            if let dic = patterDict[w] {
                for x in dic {
                    if buildNext(A, ans + x, i + 1) {
                        return true
                    }
                }
            } else {
                return false
            }

            return false
        }

        return buildNext(bottom, "", 0)
    }
}

extension String {
    func subString(from: Int, length: Int) -> String {
        let indexStartOfText = self.index(self.startIndex, offsetBy: from)
        let indexEndOfText = self.index(self.startIndex, offsetBy: from + length)
        let substring = self[indexStartOfText..<indexEndOfText]
        return String(substring)
    }
}

---

40ms

class Solution {
    func pyramidTransition(_ bottom: String, _ allowed: [String]) -> Bool {
        var map = [String: [Character]]()
        for s in allowed {
            let key = String(s.prefix(2))
            var arr = map[key, default: [Character]()]
            arr.append(s.last!)
            map[key] = arr
        }
        var curr = ""
        var pre = bottom
        return DFS(&pre, &curr, 0, map)
    }

    func DFS(_ prev: inout String, _ curr: inout String, _ idx: Int, _ map: [String: [Character]]) -> Bool {
        if prev.count == 1 {                   // the total pyramid has been constructed
            return true
        }
        if idx + 1 == prev.count {               // the current layer has been constructed
            var next = ""
            return DFS(&curr, &next, 0, map);
        }
        else {
            let chars = Array(prev)
            let key = String(chars[idx..<idx+2])
            if map[key] == nil {  // no candidate
                return false
            }
            else {                                  // try each candidate
                let chars = map[key]!
                for i in 0..<chars.count {
                    curr.append(chars[i])
                    if DFS(&prev, &curr, idx+1, map) { return true }
                    curr.popLast() // backtracking
                }
                return false
            }
        }
    }
}

---

Runtime: 44 ms

Memory Usage: 20 MB

class Solution {
    let arrChar:[Character] = ["A","B","C","D","E","F","G"]
    func pyramidTransition(_ bottom: String, _ allowed: [String]) -> Bool {
        var n:Int = bottom.count
        var dp:[[[Bool]]] = [[[Bool]]](repeating:[[Bool]](repeating:[Bool](repeating:false,count:7),count:n),count:n)
        var m:[Character:Set<String>] = [Character:Set<String>]()
        for str in allowed
        {
            m[str[2],default:Set<String>()].insert(str.subString(0, 2))
        }
        for i in 0..<n
        {
            dp[n - 1][i][bottom[i].ascii - 65] = true
        }
        for i in stride(from:n - 2,through:0,by:-1)
        {
            for j in 0...i
            {
                for ch in arrChar
                {
                    if m[ch] == nil {continue}
                    for str in m[ch]!
                    {
                        if dp[i + 1][j][str[0].ascii - 65] && dp[i + 1][j + 1][str[1].ascii - 65]
                        {
                            dp[i][j][ch.ascii - 65] = true
                        }
                    }
                }
            }
        }
        for i in 0..<7
        {
            if dp[0][0][i] {return true}
        }
        return false
    }
}

extension String {
    //subscript函数可以检索数组中的值
    //直接按照索引方式截取指定索引的字符
    subscript (_ i: Int) -> Character {
        //读取字符
        get {return self[index(startIndex, offsetBy: i)]}
    }
    
    // 截取字符串：指定索引和字符数
    // - begin: 开始截取处索引
    // - count: 截取的字符数量
    func subString(_ begin:Int,_ count:Int) -> String {
        let start = self.index(self.startIndex, offsetBy: max(0, begin))
        let end = self.index(self.startIndex, offsetBy:  min(self.count, begin + count))
        return String(self[start..<end]) 
    }
    
}

//Character扩展 
extension Character  
{  
  //Character转ASCII整数值(定义小写为整数值)
   var ascii: Int {
       get {
           return Int(self.unicodeScalars.first?.value ?? 0)
       }       
    }    
}