Scala flatten List

Question

I want to write a function that flats a List.

object Flat {
  def flatten[T](list: List[T]): List[T] = list match {
    case Nil => Nil
    case head :: N         


        

Answer 1

If someone does not understand this line of the accepted solution, or did not know that you can annotate a pattern with a type:

case (head: List[_]) :: tail => flatten(head) ++ flatten(tail)

Then look at an equivalent without the type annotation:

case (y :: ys) :: tail => flatten3(y :: ys) ::: flatten3(tail)
case Nil :: tail => flatten3(tail)

So, just for better understanding some alternatives:

def flatten2(xs: List[Any]): List[Any] = xs match {
  case x :: xs => x match {
    case y :: ys => flatten2(y :: ys) ::: flatten2(xs)
    case Nil => flatten2(xs)
    case _ => x :: flatten2(xs)
  }
  case x => x
}

def flatten3(xs: List[Any]): List[Any] = xs match {
  case Nil => Nil
  case (y :: ys) :: zs => flatten3(y :: ys) ::: flatten3(zs)
  case Nil :: ys => flatten3(ys)
  case y :: ys => y :: flatten3(ys)
}

val yss = List(List(1,2,3), List(), List(List(1,2,3), List(List(4,5,6))))
flatten2(yss) // res2: List[Any] = List(1, 2, 3, 1, 2, 3, 4, 5, 6) 
flatten3(yss) // res2: List[Any] = List(1, 2, 3, 1, 2, 3, 4, 5, 6) 

By the way, the second posted answer will do the following, which you probably don't want.

val yss = List(List(1,2,3), List(), List(List(1,2,3), List(List(4,5,6))))
flatten(yss) // res1: List[Any] = List(1, 2, 3, List(), 1, 2, 3, 4, 5, 6)