Swift34/90Days - objc.io 的 Swift 片斷 5

Swift90Days - objc.io 的 Swift 片斷 4

全排列

咱們想經過 Swift 簡單的實現全排列。

首先先寫個例子，列出全部把元素插入到數組中的狀況：

func between<T>(x: T, ys: [T]) -> [[T]] {
    if let (head, tail) = ys.decompose {
        return [[x] + ys] + between(x, tail).map { [head] + $0 }
    } else {
        return [[x]]
    }
}


between(0, [1, 2, 3])   // [[0, 1, 2, 3], [1, 0, 2, 3], [1, 2, 0, 3], [1, 2, 3, 0]]

接下來再經過 between 函數實現全排列：

func permutations<T>(xs: [T]) -> [[T]] {
    if let (head, tail) = xs.decompose {
        return permutations(tail) >>= { permTail in
            between(head, permTail)
        }
    } else {
        return [[]]
    }
}

其中 >>= 這個符號前面有提到過，返回全部的組合結果。完整的實現代碼以下：

extension Array {
    var decompose : (head: T, tail: [T])? {
        return (count > 0) ? (self[0], Array(self[1..<count])) : nil
    }
}

infix operator >>= {}
func >>=<A, B>(xs: [A], f: A -> [B]) -> [B] {
    return xs.map(f).reduce([], combine: +)
}

func between<T>(x: T, ys: [T]) -> [[T]] {
    if let (head, tail) = ys.decompose {
        return [[x] + ys] + between(x, tail).map { [head] + $0 }
    } else {
        return [[x]]
    }
}


func permutations<T>(xs: [T]) -> [[T]] {
    if let (head, tail) = xs.decompose {
        return permutations(tail) >>= { permTail in
            between(head, permTail)
        }
    } else {
        return [[]]
    }
}

輕量級解包

咱們能夠在原來的API上面封裝一層減小頻繁解包的麻煩：

typealias JSONDictionary = [String:AnyObject]

func decodeJSON(data: NSData) -> JSONDictionary? {
    return NSJSONSerialization.JSONObjectWithData(data, 
        options: .allZeros, error: nil) as? JSONDictionary
}

固然 encode 也同樣：

func encodeJSON(input: JSONDictionary) -> NSData? {
    return NSJSONSerialization.dataWithJSONObject(input, 
        options: .allZeros, error: nil)
}

---

References

• Permutations
• Lightweight API Wrappers