斯卡拉的懒人快艇

Lazy Quicksort in Scala

在scala中可以做这种事情吗？

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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)
}
}

它也可以用视图来完成，尽管速度肯定会慢得多：

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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)
}