Question

我的树对象：

    private[package] object Tree {

      /** A node in a  Tree. */
      class Node[T](val parent: Node[T]) extends Serializable {
        var item: T = _
        var count: Long = 0L
        val children: mutable.Map[T, Node[T]] = mutable.Map.empty

        def isRoot: Boolean = parent == null
      }

      /** Attribute in a tree */
      private class Attribute [T] extends Serializable {
        var count: Long = 0L
        val nodes: ListBuffer[Node[T]] = ListBuffer.empty
      }
    }

上课：

    private[package] class Tree[T] extends Serializable {
      import Tree._
      val root: Node[T] = new Node(null)
      private val attributes: mutable.Map[T, Attribute[T]] = mutable.Map.empty

    def extract(
       minCount: Long,
       validateSuffix: T => Boolean = _ => true): Iterator[(List[T], Long)] = {
          //missing code
      }

函数 extract 必须生成包含路径root-to-leaf的Iterator [List [T]]。如果每个节点的计数大于minCount，则该路径有效。

编辑：这是我的尝试：

 def extract(minCount: Long, validateSuffix: T => Boolean = _ => true): Iterator[(List[T], Long)] = {

    def traverse(node: Node[T], path: List[T]): Iterator[(List[T], Long)] = {
      path.::(node.item)
      node.children.iterator.flatMap { case (item, child) =>
        traverse(child, path).map { case (t, c) =>
          (item :: t, c)
        }
      } ++ {
          if (node.children.isEmpty && node.count >= minCount) {
            Iterator.single((path, node.count))
          } else {
            Iterator.empty
          }
        }
    }
    traverse(root, List.empty)

编辑：这就是我如何建造一棵树：

    val tree = new Tree[String]
      .add(Seq("a", "b", "c"))
      .add(Seq("a", "b", "y"))
      .add(Seq("a", "b"))
      .add(Seq("a"))
      .add(Seq("b"))
      .add(Seq("b", "n"))

    val paths = tree.extract(3L).map { case (items, count) =>
      (items.toSet, count)
    }.toSet

Answer 1

这样做：

paths

private void SaveCustomer(TestModel ctx, Customer customer) { //save back ctx.Customers.Attach(customer); //GET ERROR HERE ctx.Entry(customer).State = EntityState.Modified; ctx.SaveChanges(); }方法返回一个路径数组，每个路径都是一个值数组。

在Scala中的常规树中提取所有根到叶路径

1 个答案: