Return the corresponding integer - scala

Question

I need to write a recursive function that would return a corresponding integer from the given list. For example, list1 = List(2,3,1,7,5)
fromList(list1) would return 57132
This is what I have.
def fromList(list1: List[Int]): Int = {
val num1 = list1.head
val num2 = fromList(list1.tail)
if (list1.isEmpty) num1
else num1 + num2
}
I'm able to solve this if I use power operation but this operation is limited.

Answer 1

Something like this?
def fromList(list1: List[Int]): Int =
list1.reverse.fold(0)(10 * _ + _)
fromList(List(2,3,6)) //res0: Int = 632
Can also be done with foldRight().
def fromList(list1: List[Int]): Int =
list1.foldRight(0)(_ + _ * 10)
If recursion is required...
#annotation.tailrec
def fromList(list1: List[Int], acc:Int = 0): Int =
if (list1.isEmpty) acc
else fromList(list1.init, 10*acc + list1.last)
fromList(List(2,3,6,1)) //res0: Int = 1632
Or the slightly more verbose but likely more efficient...
def fromList(list1: List[Int]): Int = {
#annotation.tailrec
def loop(lst: List[Int], acc:Int = 0):Int = lst match {
case hd::tl => loop(tl, 10*acc + hd)
case _ => acc
}
loop(list1.reverse, 0)
}
fromList(List(7,3,6,4)) //res0: Int = 4637
There are 2 major problems with your original design.
1st - Both list1.head and list1.tail are impossible if list1.isEmpty. So you need to test for that before decomposing the List.
2nd - You need to adjust the intermediate results before adding the current digit.
//recursive but not tail-recursive
def fromList(list1: List[Int]): Int =
if (list1.isEmpty) 0
else 10 * fromList(list1.tail) + list1.head

Answer 2

#jwvh solution is recursive(tail) if you might want a non tail recursive though with a good chance of a stack overflow for large values then
def solution (list : List[Int]): Int = {
def recursive(list1: List[Int]): Int= {
list1 match {
case head :: tail => head * scala.math.pow(10, list1.length-1).toInt + recursive(tail)
case Nil => 0
}
}
recursive(list.reverse)
}

Answer 3

Not recursive, but does the job :)
list1.reverse.mkString("").toInt