Swift34/90Days - objc.io 的 Swift 片断 5

Swift90Days - objc.io 的 Swift 片断 4

全排列

咱们想经过 Swift 简单的实现全排列。

首先先写个例子，列出全部把元素插入到数组中的状况：

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]]

接下来再经过 between 函数实现全排列：

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

其中 >>= 这个符号前面有提到过，返回全部的组合结果。完整的实现代码以下：

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 [[]]
    }
}

轻量级解包

咱们能够在原来的API上面封装一层减小频繁解包的麻烦：

typealias JSONDictionary = [String:AnyObject]

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

固然 encode 也同样：

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

---

References

• Permutations
• Lightweight API Wrappers