2

I wrote a recursive mergeSort function:

func mergeSort<T: Comparable>(inout array: [T]) {
    if array.count <= 1 {
        return
    }

    var leftSlice = [T](array[0..<array.count / 2])
    var rightSlice = [T](array[array.count / 2...array.endIndex - 1])

    mergeSort(&leftSlice)
    mergeSort(&rightSlice)
    array = merge(leftSlice, rightSlice)
}

func merge<T: Comparable>(var left: [T], var right: [T]) -> [T] {
    var mergedValues = [T]()

    while !left.isEmpty && !right.isEmpty {
        mergedValues.append(left.first! < right.first! ? left.removeAtIndex(0) : right.removeAtIndex(0))
    }

    if !left.isEmpty {
        mergedValues += left
    } else if !right.isEmpty {
        mergedValues += right
    }

    return mergedValues
}

Now, since merge() is only supposed to be used by mergeSort() I placed it inside of mergeSort(), therefore making merge() a nested function:

func mergeSort<T: Comparable>(inout array: [T]) {
    func merge<T: Comparable>(var left: [T], var right: [T]) -> [T] {
        var mergedValues = [T]()

        while !left.isEmpty && !right.isEmpty {
            mergedValues.append(left.first! < right.first! ? left.removeAtIndex(0) : right.removeAtIndex(0))
        }

        if !left.isEmpty {
            mergedValues += left
        } else if !right.isEmpty {
            mergedValues += right
        }

        return mergedValues
    }

    if array.count <= 1 {
        return
    }

    var leftSlice = [T](array[0..<array.count / 2])
    var rightSlice = [T](array[array.count / 2...array.endIndex - 1])

    mergeSort(&leftSlice)
    mergeSort(&rightSlice)
    array = merge(leftSlice, rightSlice)
}

Now the first version works fine, but the second one doesn't.
How can that be?

4

2 回答 2

5

看起来您在编译器中发现了与嵌套泛型函数相关的错误。这是一个也会导致 1.2 编译器崩溃的缩减:

func f<T>(t: T) {
    func g<U>(u: U) { }
}

但在这种情况下,您实际上并不需要通用版本的merge. 它的泛型参数与外部函数相同,因此只需使用它:

func mergeSort<T: Comparable>(inout array: [T]) {
    // no generic placeholder needed, T is the T for mergeSort
    func merge(var left: [T], var right: [T]) -> [T] {
      // etc.
    }
}

这似乎工作正常。

但是,还值得指出的是,在您的merge函数中,您是removeAtIndex在一个循环中调用的,这是一个O(n)函数。这意味着您的合并排序不会具有预期的复杂性。

这是要考虑的替代版本:

func mergeSort<T: Comparable>(inout array: [T], range: Range<Int>? = nil) {

    func merge(left: Range<Int>, right: Range<Int>) -> [T] {    
        var tmp: [T] = []
        tmp.reserveCapacity(count(left) + count(right))

        var l = left.startIndex, r = right.startIndex

        while l != left.endIndex && r != right.endIndex {
            if array[l] < array[r] {
                tmp.append(array[l++])
            }
            else {
                tmp.append(array[r++])
            }
        }
        // where left or right may be empty, this is a no-op
        tmp += source[l..<left.endIndex]
        tmp += source[r..<right.endIndex]

        return tmp
    }

    // this allows the original caller to omit the range,
    // the default being the full array
    let r = range ?? indices(array)
    if count(r) > 1 {
        let mid = r.startIndex.advancedBy(r.startIndex.distanceTo(r.endIndex)/2)
        let left = r.startIndex..<mid
        let right = mid..<r.endIndex

        mergeSort(&array, range: left)
        mergeSort(&array, range: right)
        let merged = merge(left, right)
        array.replaceRange(r, with: merged)
    }
}

我还要说,因为merge它本身可能是一个通用的有用函数,你不妨让它独立而不是嵌套它(类似地,partition在实现快速排序时)。嵌套不会给您带来任何好处(除了我上面使用的从其中引用外部参数的技巧之外,无论如何这可能是一种不好的做法,我这样做主要是为了向您展示:)

于 2015-02-23T18:20:02.637 回答
3

您不需要制作merge通用功能。GenericT已经定义了,mergeSort所以你只需[T]在内部函数中设置为参数:

func merge(var left: [T], var right: [T]) -> [T] {
    var mergedValues = [T]()
    ...
}
于 2015-02-23T18:07:41.000 回答