Here is one way to solve this.

import org.apache.spark.sql.Row
import org.apache.spark.sql.functions._
import scala.collection.mutable.WrappedArray

val data = Seq((Seq(1,2,3),Seq(4,5,6),Seq(7,8,9)))
val df = sqlContext.createDataFrame(data)
val first = df.first

// use a pattern match to deferral the type
val mapped = first.getAs[WrappedArray[Int]](0)

// now we can use it like normal collection
mapped.mkString("\n")

// get rows where has array
val rows = df.collect.map {
    case Row(a: Seq[Any], b: Seq[Any], c: Seq[Any]) => 
        (a, b, c)
}
rows.mkString("\n")

If you have a WrappedArray there are several ways to convert it to a Java collection. You can leverage scala.collection.JavaConverters helper methods:

WrappedArray<Long> wrappedArray = row.getAs("ids");
List<Long> list = scala.collection.JavaConverters.seqAsJavaList(wrappedArray.seq());

Alternatively, you can use the Row.getList method to obtain directly a Java List:

List<Long> list = row.getList(row.fieldIndex("ids"));