scala - Scala 中的惰性快速排序

Question

有没有可能做this Scala 中的那种东西？

Answer 1

def quicksort[A](xs: Stream[A])(implicit o: Ordering[A]): Stream[A] = {
  import o._ 
  if (xs.isEmpty) xs else {
      val (smaller, bigger) = xs.tail.partition(_ < xs.head)
      quicksort(smaller) #::: xs.head #:: quicksort(bigger)
  }
}

它也可以用 View 来完成，尽管它肯定会慢得多:

def quicksort[A](xs: List[A])(implicit o: Ordering[A]) = {
  import o._
  def qs(xs: SeqView[A, List[A]]): SeqView[A, Seq[_]] = if (xs.isEmpty) xs else {
    val (smaller, bigger) = xs.tail.partition(_ < xs.head)
    qs(smaller) ++ (xs.head +: qs(bigger))
  }
  qs(xs.view)
}