Question

/**
(**) Pack consecutive duplicates of list elements into sublists.
If a list contains repeated elements they should be placed in separate sublists.
Example:

scala> pack(List('a, 'a, 'a, 'a, 'b, 'c, 'c, 'a, 'a, 'd, 'e, 'e, 'e, 'e))
res0: List[List[Symbol]] = List(List('a, 'a, 'a, 'a), List('b), List('c, 'c), List('a, 'a), List('d), List('e, 'e, 'e, 'e))
  */

正常情况如下：

object MatchTest1 extends Application {
  def matchTest(x: Int): String = x match {
    case 1 => "one"
    case 2 => "two"
    case _ => "many"
  }
  println(matchTest(3))
}

我知道尝试将x与模式匹配。

但是如下所示，它匹配什么？

// 1- My solution
    /**
     * Similar solution to recusive P08 solution
     */
    def pack(input: List[Any]): List[List[Any]] = {
      def comp(l: List[Any], lastElem: Any, lastList: List[Any]): List[List[Any]] = {
        l match {
          case Nil => List[List[Any]](lastList)
          case head :: tail if head == lastElem => comp(tail, head, head :: lastList)
          case head :: tail if lastList == Nil => comp(tail, head, head :: Nil)
          case head :: tail => lastList :: comp(tail, head, head :: Nil)
        }
      }
      comp(input, Nil, Nil)
    }

由于

Answer 1

检查List[Any] l以下内容：

// Is it an empty list?
case Nil => ...

// Is it not empty and does the first item (the head) equal `lastElem`?
case head :: tail if head == lastElem => ...

// Is it not emptybut lastList is empty?
case head :: tail if lastList == Nil => ...

// Every other case
case head :: tail => ...

如果列表中包含 head （第一项）和 tail （列表的其余部分），则head :: tail匹配。 head 必须是商品， tail 可以是空列表。这意味着[1]和[1, 2]都匹配head :: tail，而[]仅匹配Nil。

斯卡拉模式匹配。这个匹配现在是什么

1 个答案: