如何用第二个序列的元素填充第一个序列

非常简单的一行解决方案。创建第二个列表的第一个元素和第二个列表中小于第一个列表的当前映射元素的所有元素的序列,并获取该序列中最大的元素。

val first = 1 to 7
val second = List(3, 5, 6)

val result = (first.map(x => (second.head +: second.filter(_ <= x)).last))

简单的方法:

scala> val first = 1 to 7 toList
first: List[Int] = List(1, 2, 3, 4, 5, 6, 7)

scala> val second = List(3,5)
second: List[Int] = List(3, 5)

scala> val result = first.map(x => if(x < second.tail.head) second.head else second.tail.head)
result: List[Int] = List(3, 3, 3, 3, 5, 5, 5)

然而,这不是很灵活,但它满足了您的问题。

我自己找到了解决办法。问题中的给定序列

first.foldLeft((second, Seq.empty[Int])) { case ((sec, acc), f) =>
    if (sec.tail.isEmpty) (sec, sec.head +: acc)
    else if (f >= sec.tail.head) (sec.tail, sec.tail.head +: acc)
    else (sec, sec.head +: acc)
}._2.reverse

提供正确的结果

使用递归解决方案 span

@annotation.tailrec
def recurse(first: Seq[Int], second: Seq[Int], acc: Seq[Int] = List.empty): Seq[Int] = {
  second match {
    case a :: b :: tail =>
      val (f1, f2) = first.span(_ != b)
      recurse(f2, b :: tail, List.fill(f1.length)(a) ++ acc)
    case a :: Nil =>
      (List.fill(first.length)(a) ++ acc).reverse
  }
}

val first = 1 to 7

println(recurse(first, Seq(3, 5)))
// List(3, 3, 3, 3, 5, 5, 5)

println(recurse(first, Seq(3, 5, 6)))
// List(3, 3, 3, 3, 5, 6, 6)