Scala:使用子列表的谓词拆分列表

我刚刚有一个用例,我需要将一个列表拆分为n个子列表,这样元素就可以从原始列表中按顺序提取并分组,同时子列表的谓词为真(当它为假时,会启动一个新的子列表)。我在标准库中没有找到这个功能,我认为尝试以函数式风格解决它是一个很好的练习(因为我远非函数大师)。

下面是我想出的代码。但我怀疑它可以改进很多。你能帮我找到一个更好的编码方法吗?

class ListWithSplitter[A](val theList:List[A])
{
  private def sublistWhile(list:List[A], pred:(List[A] => Boolean)):(List[A],List[A]) =
  {
    def combine(okList:List[A], remaining:List[A], pred:(List[A] => Boolean)):(List[A],List[A]) =
    {
      if(pred(okList ::: remaining.head :: Nil))
        combine(okList ::: remaining.head :: Nil, remaining.tail, pred)
      else
        (okList, remaining)
    }

    list match {
      case Nil => (Nil, Nil)
      case x :: Nil => (list, Nil)
      case x :: xs => combine(List(x), xs, pred)
    }
  }

  private def combinedSplit(list:List[A], pred:(List[A] => Boolean)):List[List[A]] =
  {
    val r = sublistWhile(list, pred)
    r match {
      case (Nil, Nil) => List(Nil)
      case (x, Nil) => List(x)
      case (x, y) => x :: combinedSplit(y, pred)
    }
  }

  def combinedSplit(pred:(List[A] => Boolean)):List[List[A]] =
  {
    combinedSplit(theList, pred)
  }
}

trait ListCombinedSplit
{
  implicit def list2combSplitter[A](x:List[A]) : ListWithSplitter[A] = new ListWithSplitter(x)
}

object ListSplitter extends ListCombinedSplit {

  def main(args:Array[String])
  {
    // sample usage: sum of each sublist is less than 100
    val a = List(4, 59, 10, 24, 42, 9, 2, 44, 44, 44, 44)
    val b = a combinedSplit { list:List[Int] => ((0 /: list)(_ + _)) < 100 }

    b foreach println
  }
}

样本结果为:

List(4, 59, 10, 24)
List(42, 9, 2, 44)
List(44, 44)
List(44)