如何在Scala中编写“通用”mergesort？

Question

Here is what I have so far :

  def mergesort[T <: Ordered[T]](elements : List[T]) : List[T] = {
    def merge(first : List[T], second : List[T]) : List[T] = (first, second) match {
      case (Nil, _) => second
      case (_, Nil) => first
      case (x :: xs, y :: ys) => if (x < y) x :: merge(xs, second) else y :: merge(first, ys)
    }

    if (elements.isEmpty) Nil
    else {
      val length = elements.length
      val (firstHalf, secondHalf) = elements.splitAt(length/2)

      merge(mergesort(firstHalf), mergesort(secondHalf))
    }
  }

The problem I'm having is that this fails

mergesort(List(1, 3, 6, 3, 1, 0))

error: inferred type arguments [Int] do not conform to method mergesort's type parameter bounds [T <: Ordered[T]]
       mergesort(List(1, 3, 6, 3, 1, 0))
       ^

Is there any way to make this work for all types which are ordered? I though Scala would have some sort of implicit conversion to a 'rich' integer type, which I assumed would have the Ordered trait.

Answer 1

What you need is a view bound def mergesort[T <% Ordered[T]] . See answers to this question: Generic method to return first of two values .

This will now complie but you have some bugs in your algorithm giving you a stackoverflow error in the splitAt line.