如何将选项添加到列表

Question

什么是补充的最佳途径Options到一个List.

这是我的第一次尝试:

def append[A](as: List[A], maybeA1 : Option[A], maybeA2: Option[A]) : List[A] = as  ++ maybeA1.toList ++ maybeA2.toList

Con:它创造了2个tmp List

(我知道.toList()是可选的,因为存在从Option [A]到Iterable [A]的隐式转换)

另一种尝试是

def append2[A](ls: List[A], maybeA : Option[A]) : List[A] = maybeA.map(_ :: ls).getOrElse(ls)
def append[A](as: List[A], maybeA1 : Option[A], maybeA2: Option[A]) : List[A] = append2(append2(as, maybeA1), maybeA2)

更好的性能但可读性更低......

还有另外一种方法吗？

Answer 1

def combine[A](s: Seq[A], o: Option[A]) = (s /: o)(_ :+ _)
def combineAll[A](s: Seq[A], os: Option[A]*) = (s /: os)(combine)

combineAll(List(1), Some(2), None, Some(3))
//res0: Seq[Int] = List(1, 2, 3)

Answer 2

使用List.++或List.++:：

def append[A](o:Option[A],l:List[A]):List[A] = l ++ o   
def prepend[A](o:Option[A],l:List[A]):List[A] = o ++: l