我正在自学,它要求推理并实现一个偏向权重的左堆。这是我的基本实现:
(* 3.4 (b) *)
functor WeightBiasedLeftistHeap (Element : Ordered) : Heap =
struct
structure Elem = Element
datatype Heap = E | T of int * Elem.T * Heap * Heap
fun size E = 0
| size (T (s, _, _, _)) = s
fun makeT (x, a, b) =
let
val sizet = size a + size b + 1
in
if size a >= size b then T (sizet, x, a, b)
else T (sizet, x, b, a)
end
val empty = E
fun isEmpty E = true | isEmpty _ = false
fun merge (h, E) = h
| merge (E, h) = h
| merge (h1 as T (_, x, a1, b1), h2 as T (_, y, a2, b2)) =
if Elem.leq (x, y) then makeT (x, a1, merge (b1, h2))
else makeT (y, a2, merge (h1, b2))
fun insert (x, h) = merge (T (1, x, E, E), h)
fun findMin E = raise Empty
| findMin (T (_, x, a, b)) = x
fun deleteMin E = raise Empty
| deleteMin (T (_, x, a, b)) = merge (a, b)
end
现在,在 3.4 (c) 和 (d) 中,它要求:
现在,merge
分两个运作
传递:自上而下的传递,包括
打电话给merge
,以及自下而上的传递
包括对助手的调用
功能,makeT
。调整merge
到
以单一、自上而下的方式进行操作。
自上而下的方式有什么好处
的版本merge
有一个懒惰的
环境?在并发的
环境?
我改变了merge
通过简单的内联函数makeT
,但我没有看到任何好处,所以我认为我还没有掌握这部分练习的精神。我缺少什么?
fun merge (h, E) = h
| merge (E, h) = h
| merge (h1 as T (s1, x, a1, b1), h2 as T (s2, y, a2, b2)) =
let
val st = s1 + s2
val (v, a, b) =
if Elem.leq (x, y) then (x, a1, merge (b1, h2))
else (y, a2, merge (h1, b2))
in
if size a >= size b then T (st, v, a, b)
else T (st, v, b, a)
end
我想我已经找到了关于惰性评估的一点。如果我不使用递归合并来计算大小,则在需要子级之前不需要评估递归调用:
fun merge (h, E) = h
| merge (E, h) = h
| merge (h1 as T (s1, x, a1, b1), h2 as T (s2, y, a2, b2)) =
let
val st = s1 + s2
val (v, ma, mb1, mb2) =
if Elem.leq (x, y) then (x, a1, b1, h2)
else (y, a2, h1, b2)
in
if size ma >= size mb1 + size mb2
then T (st, v, ma, merge (mb1, mb2))
else T (st, v, merge (mb1, mb2), ma)
end
这就是全部?但我不确定并发性。