您如何旋转（循环移位）Scala集合

问题描述

我 可以使用for循环轻松，干净地完成此操作。例如，如果我想将每个元素中的 Seq 遍历回自身，我将执行以下操作：

I can do this quite easily, and cleanly, using a for loop. For instance, if I wanted to traverse a Seq from every element back to itself I would do the following:

val seq = Seq(1,2,3,4,5)

for (i <- seq.indices) {
    for (j <- seq.indices) {
        print(seq(i + j % seq.length))
    }
}

但是当我想对整个收藏集进行折叠时，我想知道是否还有一种更惯用的方法。递归方法将使我避免使用任何 var s。但基本上，我想知道是否可能出现以下情况：

But as I'm looking to fold over the collection, I'm wondering if there is a more idiomatic approach. A recursive approach would allow me to avoid any vars. But basically, I'm wondering if something like the following is possible:

seq.rotatedView(i)

这将创建旋转视图，例如旋转位（或循环移位）。

Which would create a rotated view, like rotating bits (or circular shift).

推荐答案

在OP希望将其折叠的注释之后，这是一种略有不同的用法，它避免了先计算序列的长度。

Following the OP's comment that they want to fold over it, here's a slightly different take on it that avoids calculating the length of the sequence first.

定义将在旋转序列上进行迭代的迭代器

Define an iterator that will iterate over the rotated sequence

class RotatedIterator[A](seq: Seq[A], start: Int) extends Iterator[A] {
  var (before, after) = seq.splitAt(start)
  def next = after match {
    case Seq()  =>
      val (h :: t) = before; before = t; h
    case h :: t => after = t; h
  }
  def hasNext = after.nonEmpty || before.nonEmpty
}

并像这样使用它：

val seq = List(1, 2, 3, 4, 5)
val xs = new RotatedIterator(seq, 2)
println(xs.toList)         //> List(3, 4, 5, 1, 2)

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