How to group a variable-length, repeating sequence in Scala

Question

I have a collection of ints that repeat themselves in a pattern:

val repeatingSequence = List(1,2,3,1,2,3,4,1,2,1,2,3,4,5)

I\'d like to sec

Answer 1

import scala.collection.mutable.ListBuffer
import scala.collection.breakOut

val repeatingSequence = List(1,2,3,1,2,3,4,1,2,1,2,3,4,5)
val groupedBySequence: List[List[Int]] = repeatingSequence.foldLeft(ListBuffer[ListBuffer[Int]]()) {
  case (acc, 1) => acc += ListBuffer(1)
  case (acc, n) => acc.last += n; acc
}.map(_.toList)(breakOut)

Answer 2

Here's a not-exactly-elegant solution I bashed out using span:

def groupWhen[A](fn: A => Boolean)(xs: List[A]): List[List[A]] = {
  xs.span(!fn(_)) match {
    case (Nil, Nil) => Nil
    case (Nil, z::zs) => groupWhen(fn)(zs) match {
      case ys::yss => (z::ys) :: yss
      case Nil => List(List(z))
    }
    case (ys, zs) => ys :: groupWhen(fn)(zs)
  }
}

scala> groupWhen[Int](_==1)(List(1,2,3,1,2,3,4,1,2,3,4,5))
res39: List[List[Int]] = List(List(1, 2, 3), List(1, 2, 3, 4), List(1, 2, 3, 4, 5))

scala> groupWhen[Int](_==1)(List(5,4,3,2,1,2,3,1,2,3,4,1,2,3,4,5))
res40: List[List[Int]] = List(List(5, 4, 3, 2), List(1, 2, 3), List(1, 2, 3, 4), List(1, 2, 3, 4, 5))

Answer 3

Given an iterator itr, this will do the trick:

val head = iter.next()
val out = (
  Iterator continually {iter takeWhile (_ != head)}
  takeWhile {!_.isEmpty}
  map {head :: _.toList}
).toList

Answer 4

As well all know, fold can do everything... ;)

  val rs = List(1,2,3,1,2,3,4,1,2,1,2,3,4,5)
  val res = (rs++List(1)).foldLeft((List[List[Int]](),List[Int]()))((acc,e) => acc match {
    case (res,subl) => {
      if (e == 1) ((subl.reverse)::res,1::Nil) else (res, e::subl)
    }
  })
  println(res._1.reverse.tail)

Please regard this as an entry for the obfuscated Scala contest rather than as a real answer.