scala - 预置 Scala 向量的复杂性

Question

我指的是官方documentation

它将 Vector 的复杂性显示为“有效常数”(eC)。但我的理解是，对于向量，前置意味着还需要调整所有其他索引，这将使操作 O(n) 或 L(线性)。任何人都可以解释如何在向量 eC 中添加(有效地恒定)。

Answer 1

找到以下对 prepend 操作的可视化解释，其中每一步都添加一个字符。为了便于说明，图片仅显示每个块 2 个插槽，但在向量的情况下，每个块将是 32 个插槽。 Vector 维护一个开始索引(或图片中的偏移量)以跟踪空白位置。



以下是来自 Vector.scala 的源代码。由于它不会移动所有元素，因此它不是 O(n)。

override def prepended[B >: A](value: B): Vector[B] = {
    if (endIndex != startIndex) {
      val blockIndex = (startIndex - 1) & ~31
      val lo = (startIndex - 1) & 31

      if (startIndex != blockIndex + 32) {
        val s = new Vector(startIndex - 1, endIndex, blockIndex)
        s.initFrom(this)
        s.dirty = dirty
        s.gotoPosWritable(focus, blockIndex, focus ^ blockIndex)
        s.display0(lo) = value.asInstanceOf[AnyRef]
        s
      } else {

        val freeSpace = (1 << (5 * depth)) - endIndex           // free space at the right given the current tree-structure depth
        val shift = freeSpace & ~((1 << (5 * (depth - 1))) - 1) // number of elements by which we'll shift right (only move at top level)
        val shiftBlocks = freeSpace >>> (5 * (depth - 1))       // number of top-level blocks

        if (shift != 0) {
          // case A: we can shift right on the top level
          if (depth > 1) {
            val newBlockIndex = blockIndex + shift
            val newFocus = focus + shift

            val s = new Vector(startIndex - 1 + shift, endIndex + shift, newBlockIndex)
            s.initFrom(this)
            s.dirty = dirty
            s.shiftTopLevel(0, shiftBlocks) // shift right by n blocks
            s.gotoFreshPosWritable(newFocus, newBlockIndex, newFocus ^ newBlockIndex) // maybe create pos; prepare for writing
            s.display0(lo) = value.asInstanceOf[AnyRef]
            s
          } else {
            val newBlockIndex = blockIndex + 32
            val newFocus = focus

            val s = new Vector(startIndex - 1 + shift, endIndex + shift, newBlockIndex)
            s.initFrom(this)
            s.dirty = dirty
            s.shiftTopLevel(0, shiftBlocks) // shift right by n elements
            s.gotoPosWritable(newFocus, newBlockIndex, newFocus ^ newBlockIndex) // prepare for writing
            s.display0(shift - 1) = value.asInstanceOf[AnyRef]
            s
          }
        } else if (blockIndex < 0) {
          // case B: we need to move the whole structure
          val move = (1 << (5 * (depth + 1))) - (1 << (5 * depth))
          val newBlockIndex = blockIndex + move
          val newFocus = focus + move

          val s = new Vector(startIndex - 1 + move, endIndex + move, newBlockIndex)
          s.initFrom(this)
          s.dirty = dirty
          s.gotoFreshPosWritable(newFocus, newBlockIndex, newFocus ^ newBlockIndex) // could optimize: we know it will create a whole branch
          s.display0(lo) = value.asInstanceOf[AnyRef]
          s
        } else {
          val newBlockIndex = blockIndex
          val newFocus = focus

          val s = new Vector(startIndex - 1, endIndex, newBlockIndex)
          s.initFrom(this)
          s.dirty = dirty
          s.gotoFreshPosWritable(newFocus, newBlockIndex, newFocus ^ newBlockIndex)
          s.display0(lo) = value.asInstanceOf[AnyRef]
          s
        }
      }
    } else {
      // empty vector, just insert single element at the back
      val elems = new Array[AnyRef](32)
      elems(31) = value.asInstanceOf[AnyRef]
      val s = new Vector(31, 32, 0)
      s.depth = 1
      s.display0 = elems
      s
    }
  }