scala - 来自 "Programming Scala"的合并排序导致堆栈溢出

Question

以下算法的直接剪切和粘贴:

def msort[T](less: (T, T) => Boolean)
            (xs: List[T]): List[T] = {
  def merge(xs: List[T], ys: List[T]): List[T] =
    (xs, ys) match {
      case (Nil, _) => ys
      case (_, Nil) => xs
      case (x :: xs1, y :: ys1) =>
        if (less(x, y)) x :: merge(xs1, ys)
        else y :: merge(xs, ys1)
    }
  val n = xs.length / 2
  if (n == 0) xs
  else {
    val (ys, zs) = xs splitAt n
     merge(msort(less)(ys), msort(less)(zs))
  }
}

在 5000 个长列表上导致 StackOverflowError。

有没有什么办法可以优化这个，这样就不会发生？

Answer 1

这样做是因为它不是尾递归的。您可以通过使用非严格集合或使其成为尾递归来解决此问题。

后一个解决方案是这样的:

def msort[T](less: (T, T) => Boolean) 
            (xs: List[T]): List[T] = { 
  def merge(xs: List[T], ys: List[T], acc: List[T]): List[T] = 
    (xs, ys) match { 
      case (Nil, _) => ys.reverse ::: acc 
      case (_, Nil) => xs.reverse ::: acc
      case (x :: xs1, y :: ys1) => 
        if (less(x, y)) merge(xs1, ys, x :: acc) 
        else merge(xs, ys1, y :: acc) 
    } 
  val n = xs.length / 2 
  if (n == 0) xs 
  else { 
    val (ys, zs) = xs splitAt n 
    merge(msort(less)(ys), msort(less)(zs), Nil).reverse
  } 
} 

使用非严格性涉及按名称传递参数，或使用非严格性集合，例如 Stream .以下代码使用 Stream只是为了防止堆栈溢出，还有 List别处:

def msort[T](less: (T, T) => Boolean) 
            (xs: List[T]): List[T] = { 
  def merge(left: List[T], right: List[T]): Stream[T] = (left, right) match {
    case (x :: xs, y :: ys) if less(x, y) => Stream.cons(x, merge(xs, right))
    case (x :: xs, y :: ys) => Stream.cons(y, merge(left, ys))
    case _ => if (left.isEmpty) right.toStream else left.toStream
  }
  val n = xs.length / 2 
  if (n == 0) xs 
  else { 
    val (ys, zs) = xs splitAt n 
    merge(msort(less)(ys), msort(less)(zs)).toList
  } 
}