Functional Programming: Does a list only contain unique items?

Answer 1

I would simply use distinct method:

scala> val l = List(1,2,3,4,3)
l: List[Int] = List(1, 2, 3, 4, 3)

scala> l.distinct.size == l.size
res2: Boolean = false

ADD: Standard distinct implementation (from scala.collection.SeqLike) uses mutable HashSet, to find duplicate elements:

  def distinct: Repr = {
    val b = newBuilder
    val seen = mutable.HashSet[A]()
    for (x <- this) {
      if (!seen(x)) {
        b += x
        seen += x
      }
    }
    b.result
  }

Answer 2

Here is the fastest purely functional solution I can think of:

def isUniqueList(l: List[T]) = isUniqueList1(l, new HashSet[T])

@tailrec
def isUniqueList1(l: List[T], s: Set[T]) = l match {
  case Nil => true
  case (h :: t) => if (s(h)) false else isUniqueList1(t, s + h)
}

This should be faster, but uses mutable data structure (based on the distinct implementation given by Vasil Remeniuk):

def isUniqueList(l: List[T]): Boolean = {
  val seen = mutable.HashSet[A]()
  for (x <- this) {
    if (seen(x)) {
      return false
    }
    else {
      seen += x
    }
  }
  true
}

And here is the simplest (equivalent to yours):

def isUniqueList(l: List[T]) = l.toSet.size == l.size

Answer 3

A more efficient method would be to attempt to find a dupe; this would return more quickly if one were found:

var dupes : Set[A] = Set.empty

def isDupe(a : A) = if (dupes(a)) true else { dupes += a; false }

//then
l exists isDupe