假设我想展平相同类型的嵌套列表......例如
ListA(Element(A), Element(B), ListA(Element(C), Element(D)), ListB(Element(E),Element(F)))
ListA
包含相同类型的嵌套列表 ( ListA(Element(C), Element(D))
) 所以我想用它包含的值替换它,所以上面示例的结果应该如下所示:
ListA(Element(A), Element(B), Element(C), Element(D), ListB(Element(E),Element(F)))
当前类层次结构:
abstract class SpecialList() extends Exp {
val elements: List[Exp]
}
case class Element(name: String) extends Exp
case class ListA(elements: List[Exp]) extends SpecialList {
override def toString(): String = "ListA("+elements.mkString(",")+")"
}
case class ListB(elements: List[Exp]) extends SpecialList {
override def toString(): String = "ListB("+elements.mkString(",")+")"
}
object ListA{def apply(elements: Exp*):ListA = ListA(elements.toList)}
object ListB{def apply(elements: Exp*):ListB = ListB(elements.toList)}
我已经提出了三种可行的解决方案,但我认为必须有更好的方法来实现这一点:
第一个解决方案:
def flatten[T <: SpecialList](parentList: T): List[Exp] = {
val buf = new ListBuffer[Exp]
for (feature <- parentList.elements) feature match {
case listA:ListA if parentList.isInstanceOf[ListA] => buf ++= listA.elements
case listB:ListB if parentList.isInstanceOf[ListB] => buf ++= listB.elements
case _ => buf += feature
}
buf.toList
}
第二种解决方案:
def flatten[T <: SpecialList](parentList: T): List[Exp] = {
val buf = new ListBuffer[Exp]
parentList match {
case listA:ListA => for (elem <- listA.elements) elem match {
case listOfTypeA:ListA => buf ++= listOfTypeA.elements
case _ => buf += elem
}
case listB:ListB => for (elem <- listB.elements) elem match {
case listOfTypeB:ListB => buf ++= listOfTypeB.elements
case _ => buf += elem
}
}
buf.toList
}
第三个解决方案
def flatten[T <: SpecialList](parentList: T): List[Exp] = parentList.elements flatMap {
case listA:ListA if parentList.isInstanceOf[ListA] => listA.elements
case listB:ListB if parentList.isInstanceOf[ListB] => listB.elements
case other => List(other)
}
我的问题是,是否有更好、更通用的方法来实现与所有上面三个解决方案相同的功能是否存在重复代码?