识别Scala序列中的连续间隔

2 回答

• 4

  我相信这是foldLeft的候选人 . 随着时间的推移迭代 Build 所需结构的时间间隔 . 我发现将新项添加到seq的头部更容易，因此最后使用reverse：

  case class Interval(a: Int, b: Int)
  
  val intervals = Seq(
      Interval(0, 1),
      Interval(1, 2),
      Interval(2, 4),
      Interval(4, 6),
      Interval(7, 9), // continuous sequence is broken
      Interval(9, 10),
      Interval(11, 14), // continuous sequence is broken
      Interval(15, 20), // continuous sequence is broken
      Interval(20, 24)
  )
  
  intervals.foldLeft(Seq[Seq[Interval]]()){(acc, i) => 
      if(acc.isEmpty){
          Seq(Seq(i))
      }else {
          if(acc.head.head.b == i.a) {
              (i +: acc.head) +: acc.tail
          } else {
              Seq(i) +: acc
          }
      }
  }.map(_.reverse).reverse.foreach(println)
  

  生产环境 ：

  List(Interval(0,1), Interval(1,2), Interval(2,4), Interval(4,6))
  List(Interval(7,9), Interval(9,10))
  List(Interval(11,14))
  List(Interval(15,20), Interval(20,24))
  

  编辑：

  替代使用模式匹配：

  intervals.foldLeft(Seq[Seq[Interval]]()){
      case (Seq(Seq()), i) => Seq(Seq(i))
      case (h :: tail, i) if h.head.b == i.a => (i +: h) +: tail
      case (acc, i) => Seq(i) +: acc
  }.map(_.reverse).reverse.foreach(println)
  

  回复于 2024-05-06T02:33:46+08:00
• 2

  我建议使用foldLeft和Vectors来累积结果 . 因为我发现它更具可读性，附加值而不是前置然后反转所有内容：

  val testData = Seq(
    new Interval(0L, 1L),
    new Interval(1L, 2L),
    new Interval(2L, 4L),
    new Interval(4L, 6L),
    new Interval(7L, 9L), // continuous sequence is broken
    new Interval(9L, 10L),
    new Interval(11L, 14L), // continuous sequence is broken
    new Interval(15L, 20L), // continuous sequence is broken
    new Interval(20L, 24L)
  )
  
  val resultVector = testData.foldLeft(Vector.empty[Vector[Interval]]) {
    case (Vector(), current) => 
        Vector(Vector(current))
      case (init :+ last, current) if last.last.getEndMillis == current.getStartMillis =>
        init :+ (last :+ current)
      case (result, current) =>
        result :+ Vector(current)
  }
  

  为了使结果更具可读性，我添加了以下内容：

  val readable = resultVector.map(_.map(interv => s"(${interv.getStartMillis} -> ${interv.getEndMillis})")).mkString("\n")
  

  输出是：

  readable: String = Vector((0 -> 1), (1 -> 2), (2 -> 4), (4 -> 6))
    Vector((7 -> 9), (9 -> 10))
    Vector((11 -> 14))
    Vector((15 -> 20), (20 -> 24))
  

  回复于 2024-05-06T02:33:46+08:00