Question

此问题末尾的代码将可能的数字替换为1到9且不重复。对于给定的数字序列，List（0,0,1,5,0,0,0,0,0），它将返回以下结果。总共有720种排列。

List(2, 3, 1, 5, 4, 6, 8, 7, 9)
List(2, 3, 1, 5, 4, 6, 8, 9, 7)
List(2, 3, 1, 5, 4, 7, 8, 6, 9)
List(2, 3, 1, 5, 4, 7, 8, 9, 6)
List(2, 3, 1, 5, 4, 9, 8, 6, 7)
List(2, 3, 1, 5, 4, 9, 8, 7, 6)
List(2, 3, 1, 5, 6, 4, 8, 7, 9)
...

我的问题是如何将我的代码转换为NOT而不是使用ArrayBuffer（coll）作为我的临时存储，而最终结果是从函数（search0）返回的？

由于

/ LIM /

import collection.mutable.ArrayBuffer

object ScratchPad extends App {
  def search(l : List[Int]) : ArrayBuffer[List[Int]] = {
    def search0(la : List[Int], pos : Int, occur : List[Int], coll : ArrayBuffer[List[Int]]) : Unit = {
    if (pos == l.length) {println(la); coll += la }
    val bal = (1 to 9) diff occur
    if (!bal.isEmpty) {
        la(pos) match {
        case 0 => bal map { x => search0(la.updated(pos, x), pos + 1, x :: occur, coll)}
        case n => if  (occur contains n) Nil else search0(la, pos + 1, n :: occur, coll)
        }
    }
    }

    val coll = ArrayBuffer[List[Int]]()

    search0(l, 0, Nil, coll)
    coll
  }

  println(search(List(0, 0, 1, 5, 0, 0, 8, 0, 0)).size)
}

Answer 1

这是一个使用不可变集合的简短解决方案：

scala> def search(xs: Seq[Int])(implicit ys: Seq[Int] = (1 to 9).diff(xs)): Seq[Seq[Int]] = ys match {
     |   case Seq() => Seq(xs)
     |   case _ => ys.flatten(y => search(xs.updated(xs.indexOf(0), y))(ys.diff(Seq(y))))
     | }
search: (xs: Seq[Int])(implicit ys: Seq[Int])Seq[Seq[Int]]

scala> search(List(0, 0, 1, 5, 0, 0, 8, 0, 0)).size
res0: Int = 720

scala> search(List(0, 0, 1, 5, 0, 0, 8, 0, 0)) take 10 foreach println
List(2, 3, 1, 5, 4, 6, 8, 7, 9)
List(2, 3, 1, 5, 4, 6, 8, 9, 7)
List(2, 3, 1, 5, 4, 7, 8, 6, 9)
List(2, 3, 1, 5, 4, 7, 8, 9, 6)
List(2, 3, 1, 5, 4, 9, 8, 6, 7)
List(2, 3, 1, 5, 4, 9, 8, 7, 6)
List(2, 3, 1, 5, 6, 4, 8, 7, 9)
List(2, 3, 1, 5, 6, 4, 8, 9, 7)
List(2, 3, 1, 5, 6, 7, 8, 4, 9)
List(2, 3, 1, 5, 6, 7, 8, 9, 4)

更短的解决方案：

scala> :paste
// Entering paste mode (ctrl-D to finish)

def search(xs: Seq[Int], ys: Seq[Int] = 1 to 9): Seq[Seq[Int]] = ys.diff(xs) match {
  case Seq() => Seq(xs)
  case zs => zs.flatten(z => search(xs.updated(xs.indexOf(0),z),zs))
}

// Exiting paste mode, now interpreting.

search: (xs: Seq[Int], ys: Seq[Int])Seq[Seq[Int]]

scala> search(List(0, 0, 1, 5, 0, 0, 8, 0, 0)).size
res1: Int = 720

scala> search(List(0, 0, 1, 5, 0, 0, 8, 0, 0)) take 10 foreach println
List(2, 3, 1, 5, 4, 6, 8, 7, 9)
List(2, 3, 1, 5, 4, 6, 8, 9, 7)
List(2, 3, 1, 5, 4, 7, 8, 6, 9)
List(2, 3, 1, 5, 4, 7, 8, 9, 6)
List(2, 3, 1, 5, 4, 9, 8, 6, 7)
List(2, 3, 1, 5, 4, 9, 8, 7, 6)
List(2, 3, 1, 5, 6, 4, 8, 7, 9)
List(2, 3, 1, 5, 6, 4, 8, 9, 7)
List(2, 3, 1, 5, 6, 7, 8, 4, 9)
List(2, 3, 1, 5, 6, 7, 8, 9, 4)

Answer 2

仅使用不可变数据结构对代码进行简单的重构：

object ScratchPad extends App {
    def search(l: List[Int]): List[List[Int]] = {
        def search0(la: List[Int], pos: Int, occur: List[Int]): List[List[Int]] =
            if (pos == l.length)
                List(la)
            else {
                val bal = (1 to 9) diff occur
                if (pos < l.length && !bal.isEmpty)
                    la(pos) match {
                        case 0 => bal.toList flatMap {x => search0(la.updated(pos, x), pos + 1, x :: occur)}
                        case n => if (occur contains n) List.empty[List[Int]] else search0(la, pos + 1, n :: occur)
                    }
                else
                    List.empty[List[Int]]
            }

        search0(l, 0, Nil)
    }

    val result = search(List(0, 0, 1, 5, 0, 0, 8, 0, 0))
    result foreach println
    println(result.size)
}

如何在不使用可变ArrayBuffer的情况下累积结果？

2 个答案: