Flattening nested lists of the same type

pattern-matching, review, scala

Solution

A true functional way. Without using a variable.

def flatten[A](list: List[A]): List[A] = list match {
   case Nil => Nil
   case (ls: List[A]) :: tail => flatten(ls) ::: flatten(tail)
   case h :: tail => h :: flatten(tail)
}

Problem

Let's say I want to flatten nested lists of the same type... For example ``` ListA(Element(A), Element(B), ListA(Element(C), Element(D)), ListB(Element(E),Element(F))) ``` `ListA` contains nested list of the same type (`ListA(Element(C), Element(D))`) so I want to substitute it with the values it contains, so the result of the upper example should look like this: ``` ListA(Element(A), Element(B), Element(C), Element(D), ListB(Element(E),Element(F))) ``` Current class hierarchy: ``` abstract class SpecialList() extends Exp { val elements: List[Exp] } case class Element(name: String) extends Exp case class ListA(elements: List[Exp]) extends SpecialList { override def toString(): String = "ListA("+elements.mkString(",")+")" } case class ListB(elements: List[Exp]) extends SpecialList { override def toString(): String = "ListB("+elements.mkString(",")+")" } object ListA{def apply(elements: Exp*):ListA = ListA(elements.toList)} object ListB{def apply(elements: Exp*):ListB = ListB(elements.toList)} ``` I have made three solutions that works, but I think there have to be better way to achieve this: First solution: ``` def flatten[T <: SpecialList](parentList: T): List[Exp] = { val buf = new ListBuffer[Exp] for (feature <- parentList.elements) feature match { case listA:ListA if parentList.isInstanceOf[ListA] => buf ++= listA.elements case listB:ListB if parentList.isInstanceOf[ListB] => buf ++= listB.elements case _ => buf += feature } buf.toList } ``` Second solution: ``` def flatten[T <: SpecialList](parentList: T): List[Exp] = { val buf = new ListBuffer[Exp] parentList match { case listA:ListA => for (elem <- listA.elements) elem match { case listOfTypeA:ListA => buf ++= listOfTypeA.elements case _ => buf += elem } case listB:ListB => for (elem <- listB.elements) elem match { case listOfTypeB:ListB => buf ++= listOfTypeB.elements case _ => buf += elem } } buf.toList } ``` Third solution ``` def flatten[T <: SpecialList](parentList: T): List[Exp] = parentList.elements flatMap { case listA:ListA if parentList.isInstanceOf[ListA] => listA.elements case listB:ListB if parentList.isInstanceOf[ListB] => listB.elements case other => List(other) } ``` My question is whether there is any better, more generic way to achieve same functionality as in all of upper three solutions there is repetition of code?

Original source