Find anagrams for words using dictionary

gistfile1.scala

import scala.collection.JavaConversions._
    
object Solution {
  /**
   * Read dictionary
   */
  val dictionary = scala.io.Source.fromFile("/tmp/wl.txt").mkString.split("\n");
	
  def annagramm(word: List[String], dictionary: Array[String]): List[String] = {
    /**
     * Take all same size words
     */
    val mappedDictionary = dictionary.toList.filter { e =>
      e.length == word.length && word.exists( c => c == e(0).toString )
    } map { e =>
      (e toString) :: word
    }
		
    def removeIndex(list: List[String], ix: Int) = if (list.size < ix) list
      else list.take(ix) ++ list.drop(ix+1);
		
    def isAnnagramm(word: List[String], n: Int = 0): Boolean = {
      word match {
        case word :: Nil => true
        case word :: tail => {
          val indexOf = tail.indexOf(word(n).toString);
          if (indexOf >= 0) {
            isAnnagramm( word :: removeIndex(tail, indexOf), n + 1 )
          } else false
        }
        case Nil => false
      }
    }

    /**
     * Reduce dictionary
     */
    mappedDictionary filter { e =>
      isAnnagramm(e)
    } map { e =>
      e(0)
    }
  }
	
	
    println( annagramm("horse".toList.map { e => e.toString }, dictionary) )
	  // > List(heros, horse, shore)
    println( annagramm("abundance".toList.map { e => e.toString }, dictionary) )
	  // > List(abundance)
    println( annagramm("acadia".toList.map { e => e.toString }, dictionary) )
	  // > List(acadia)
    println( annagramm("asdfkalskdjflaksjdf".toList.map { e => e.toString }, dictionary) ) // If there is no options, then return empty list
	  // > List()
    println( annagramm("shore".toList.map { e => e.toString }, dictionary) )
	  // > List(heros, horse, shore)

}