Code to enumerate permutations in Scala

Question

I coded a function to enumerate all permutations of a given list. What do you think of the code below?

def interleave(x:Int, l:List[Int]):List[List[Int]] = {         


        

Answer 1

I guess you are practicing your Scala programming skill. Here is another one, where the idea is to take different elements as head of sequence and repeats are removed through filter. Complexity of the code shall be fine, since O(n)+O(n or maybe n^2)+O(n)*P(n-1) is dominated by O(n)*P(n-1), where P(n) is the permutation number and cannot be improved, .

def permute(xs:List[Int]):List[List[Int]] = xs match {
  case Nil => List(List())
  case head::tail => {
    val len = xs.length
    val tps = (0 to len-1).map(xs.splitAt(_)).toList.filter(tp => !tp._1.contains(tp._2.head))
    tps.map(tp => permute(tp._1:::tp._2.tail).map(tp._2.head :: _)).flatten
  }
}