Just use flatten method:
scala> val m = List(List(1, 2), List(3, 4))
res2: List[List[Int]] = List(List(1, 2), List(3, 4))
scala> m.flatten
res3: List[Int] = List(1, 2, 3, 4)
Just out of sheer curiosity, here another implementation with much better performace:
scala> import scala.collection.mutable.ListBuffer
import scala.collection.mutable.ListBuffer
scala> def myFlatten[T](nl:List[List[T]]):List[T] = {
| val b = new ListBuffer[T]
| for(list<-nl){
| var tlist = list
| while(!tlist.isEmpty){
| b+=tlist.head
| tlist = tlist.tail
| }
| }
| b.toList
| }
myFlatten: [T](nl: List[List[T]])List[T]
scala> myFlatten(m)
res1: List[Int] = List(1, 2, 3, 4)
My thoughts on tackling IT complexity by breaking it into atomic components. and I term it lego oriented solution, LOS for short.
Wednesday, October 7, 2009
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