Efficient way to convert Scala Array to Unique Sorted List

Question

Can anybody optimize following statement in Scala:

// maybe large
val someArray = Array(9, 1, 6, 2, 1, 9, 4, 5, 1, 6, 5, 0, 6) 

// output a sorted list whic         


        

Answer 1

I haven't measured, but I'm with Duncan, sort in place then use something like:

util.Sorting.quickSort(array)
array.foldRight(List.empty[Int]){ 
  case (a, b) => 
    if (!b.isEmpty && b(0) == a) 
      b 
    else 
      a :: b 
}

In theory this should be pretty efficient.