你如何将类型 Node 转换为不可变树？
这个类实现了一个范围树，它不允许重叠或相邻的范围，而是将它们连接起来。例如，如果根节点是 {min = 10; max = 20} 则它是右 child ，并且它的所有孙子节点的最小值和最大值必须大于 21。范围的最大值必须大于或等于最小值。我包含了一个测试函数，所以你可以按原样运行它，它会转储任何失败的案例。
我建议从 Insert 方法开始阅读此代码。
module StackOverflowQuestion
open System
type Range =
{ min : int64; max : int64 }
with
override this.ToString() =
sprintf "(%d, %d)" this.min this.max
[<AllowNullLiteralAttribute>]
type Node(left:Node, right:Node, range:Range) =
let mutable left = left
let mutable right = right
let mutable range = range
// Symmetric to clean right
let rec cleanLeft(node : Node) =
if node.Left = null then
()
elif range.max < node.Left.Range.min - 1L then
cleanLeft(node.Left)
elif range.max <= node.Left.Range.max then
range <- {min = range.min; max = node.Left.Range.max}
node.Left <- node.Left.Right
else
node.Left <- node.Left.Right
cleanLeft(node)
// Clean right deals with merging when the node to merge with is not on the
// left outside of the tree. It travels right inside the tree looking for an
// overlapping node. If it finds one it merges the range and replaces the
// node with its left child thereby deleting it. If it finds a subset node
// it replaces it with its left child, checks it and continues looking right.
let rec cleanRight(node : Node) =
if node.Right = null then
()
elif range.min > node.Right.Range.max + 1L then
cleanRight(node.Right)
elif range.min >= node.Right.Range.min then
range <- {min = node.Right.Range.min; max = range.max}
node.Right <- node.Right.Left
else
node.Right <- node.Right.Left
cleanRight(node)
// Merger left is called whenever the min value of a node decreases.
// It handles the case of left node overlap/subsets and merging/deleting them.
// When no more overlaps are found on the left nodes it calls clean right.
let rec mergeLeft(node : Node) =
if node.Left = null then
()
elif range.min <= node.Left.Range.min - 1L then
node.Left <- node.Left.Left
mergeLeft(node)
elif range.min <= node.Left.Range.max + 1L then
range <- {min = node.Left.Range.min; max = range.max}
node.Left <- node.Left.Left
else
cleanRight(node.Left)
// Symmetric to merge left
let rec mergeRight(node : Node) =
if node.Right = null then
()
elif range.max >= node.Right.Range.max + 1L then
node.Right <- node.Right.Right
mergeRight(node)
elif range.max >= node.Right.Range.min - 1L then
range <- {min = range.min; max = node.Right.Range.max}
node.Right <- node.Right.Right
else
cleanLeft(node.Right)
let (|Before|After|BeforeOverlap|AfterOverlap|Superset|Subset|) r =
if r.min > range.max + 1L then After
elif r.min >= range.min then
if r.max <= range.max then Subset
else AfterOverlap
elif r.max < range.min - 1L then Before
elif r.max <= range.max then
if r.min >= range.min then Subset
else BeforeOverlap
else Superset
member this.Insert r =
match r with
| After ->
if right = null then
right <- Node(null, null, r)
else
right.Insert(r)
| AfterOverlap ->
range <- {min = range.min; max = r.max}
mergeRight(this)
| Before ->
if left = null then
left <- Node(null, null, r)
else
left.Insert(r)
| BeforeOverlap ->
range <- {min = r.min; max = range.max}
mergeLeft(this)
| Superset ->
range <- r
mergeLeft(this)
mergeRight(this)
| Subset -> ()
member this.Left with get() : Node = left and set(x) = left <- x
member this.Right with get() : Node = right and set(x) = right <- x
member this.Range with get() : Range = range and set(x) = range <- x
static member op_Equality (a : Node, b : Node) =
a.Range = b.Range
override this.ToString() =
sprintf "%A" this.Range
type RangeTree() =
let mutable root = null
member this.Add(range) =
if root = null then
root <- Node(null, null, range)
else
root.Insert(range)
static member fromArray(values : Range seq) =
let tree = new RangeTree()
values |> Seq.iter (fun value -> tree.Add(value))
tree
member this.Seq
with get() =
let rec inOrder(node : Node) =
seq {
if node <> null then
yield! inOrder node.Left
yield {min = node.Range.min; max = node.Range.max}
yield! inOrder node.Right
}
inOrder root
let TestRange() =
printf "\n"
let source(n) =
let rnd = new Random(n)
let rand x = rnd.NextDouble() * float x |> int64
let rangeRnd() =
let a = rand 1500
{min = a; max = a + rand 15}
[|for n in 1 .. 50 do yield rangeRnd()|]
let shuffle n (array:_[]) =
let rnd = new Random(n)
for i in 0 .. array.Length - 1 do
let n = rnd.Next(i)
let temp = array.[i]
array.[i] <- array.[n]
array.[n] <- temp
array
let testRangeAdd n i =
let dataSet1 = source (n+0)
let dataSet2 = source (n+1)
let dataSet3 = source (n+2)
let result1 = Array.concat [dataSet1; dataSet2; dataSet3] |> shuffle (i+3) |> RangeTree.fromArray
let result2 = Array.concat [dataSet2; dataSet3; dataSet1] |> shuffle (i+4) |> RangeTree.fromArray
let result3 = Array.concat [dataSet3; dataSet1; dataSet2] |> shuffle (i+5) |> RangeTree.fromArray
let test1 = (result1.Seq, result2.Seq) ||> Seq.forall2 (fun a b -> a.min = b.min && a.max = b.max)
let test2 = (result2.Seq, result3.Seq) ||> Seq.forall2 (fun a b -> a.min = b.min && a.max = b.max)
let test3 = (result3.Seq, result1.Seq) ||> Seq.forall2 (fun a b -> a.min = b.min && a.max = b.max)
let print dataSet =
dataSet |> Seq.iter (fun r -> printf "%s " <| string r)
if not (test1 && test2 && test3) then
printf "\n\nTest# %A: " n
printf "\nSource 1: %A: " (n+0)
dataSet1 |> print
printf "\nSource 2: %A: " (n+1)
dataSet2 |> print
printf "\nSource 3: %A: " (n+2)
dataSet3 |> print
printf "\nResult 1: %A: " (n+0)
result1.Seq |> print
printf "\nResult 2: %A: " (n+1)
result2.Seq |> print
printf "\nResult 3: %A: " (n+2)
result3.Seq |> print
()
for i in 1 .. 10 do
for n in 1 .. 1000 do
testRangeAdd n i
printf "\n%d" (i * 1000)
printf "\nDone"
TestRange()
System.Console.ReadLine() |> ignore
范围的测试用例
After (11, 14) | | <-->
AfterOverlap (10, 14) | |<--->
AfterOverlap ( 9, 14) | +---->
AfterOverlap ( 6, 14) |<--+---->
"Test Case" ( 5, 9) +---+
BeforeOverlap ( 0, 8) <----+-->|
BeforeOverlap ( 0, 5) <----+ |
BeforeOverlap ( 0, 4) <--->| |
Before ( 0, 3) <--> | |
Superset ( 4, 10) <+---+>
Subset ( 5, 9) +---+
Subset ( 6, 8) |<->|
这不是一个答案。
我修改了我的测试用例来运行 Juliet 的代码。它在许多情况下都失败了，但是我确实看到它通过了一些测试。
type Range =
{ min : int64; max : int64 }
with
override this.ToString() =
sprintf "(%d, %d)" this.min this.max
let rangeSeqToJTree ranges =
ranges |> Seq.fold (fun tree range -> tree |> insert (range.min, range.max)) Nil
let JTreeToRangeSeq node =
let rec inOrder node =
seq {
match node with
| JNode(left, min, max, right) ->
yield! inOrder left
yield {min = min; max = max}
yield! inOrder right
| Nil -> ()
}
inOrder node
let TestJTree() =
printf "\n"
let source(n) =
let rnd = new Random(n)
let rand x = rnd.NextDouble() * float x |> int64
let rangeRnd() =
let a = rand 15
{min = a; max = a + rand 5}
[|for n in 1 .. 5 do yield rangeRnd()|]
let shuffle n (array:_[]) =
let rnd = new Random(n)
for i in 0 .. array.Length - 1 do
let n = rnd.Next(i)
let temp = array.[i]
array.[i] <- array.[n]
array.[n] <- temp
array
let testRangeAdd n i =
let dataSet1 = source (n+0)
let dataSet2 = source (n+1)
let dataSet3 = source (n+2)
let result1 = Array.concat [dataSet1; dataSet2; dataSet3] |> shuffle (i+3) |> rangeSeqToJTree
let result2 = Array.concat [dataSet2; dataSet3; dataSet1] |> shuffle (i+4) |> rangeSeqToJTree
let result3 = Array.concat [dataSet3; dataSet1; dataSet2] |> shuffle (i+5) |> rangeSeqToJTree
let test1 = (result1 |> JTreeToRangeSeq, result2 |> JTreeToRangeSeq) ||> Seq.forall2 (fun a b -> a.min = b.min && a.max = b.max)
let test2 = (result2 |> JTreeToRangeSeq, result3 |> JTreeToRangeSeq) ||> Seq.forall2 (fun a b -> a.min = b.min && a.max = b.max)
let test3 = (result3 |> JTreeToRangeSeq, result1 |> JTreeToRangeSeq) ||> Seq.forall2 (fun a b -> a.min = b.min && a.max = b.max)
let print dataSet =
dataSet |> Seq.iter (fun r -> printf "%s " <| string r)
if not (test1 && test2 && test3) then
printf "\n\nTest# %A: " n
printf "\nSource 1: %A: " (n+0)
dataSet1 |> print
printf "\nSource 2: %A: " (n+1)
dataSet2 |> print
printf "\nSource 3: %A: " (n+2)
dataSet3 |> print
printf "\n\nResult 1: %A: " (n+0)
result1 |> JTreeToRangeSeq |> print
printf "\nResult 2: %A: " (n+1)
result2 |> JTreeToRangeSeq |> print
printf "\nResult 3: %A: " (n+2)
result3 |> JTreeToRangeSeq |> print
()
for i in 1 .. 1 do
for n in 1 .. 10 do
testRangeAdd n i
printf "\n%d" (i * 10)
printf "\nDone"
TestJTree()
最佳答案
搞定了!我认为最难的部分是弄清楚如何在将状态传回堆栈的同时对子项进行递归调用。
表演比较有趣。当主要插入冲突并合并在一起的范围时，可变版本更快，而如果您主要插入不重叠的节点并填充树，则不可变版本更快。我已经看到性能在两种方式下都达到了 100% 的最大值。
这是完整的代码。
module StackOverflowQuestion
open System
type Range =
{ min : int64; max : int64 }
with
override this.ToString() =
sprintf "(%d, %d)" this.min this.max
type RangeTree =
| Node of RangeTree * int64 * int64 * RangeTree
| Nil
// Clean right deals with merging when the node to merge with is not on the
// left outside of the tree. It travels right inside the tree looking for an
// overlapping node. If it finds one it merges the range and replaces the
// node with its left child thereby deleting it. If it finds a subset node
// it replaces it with its left child, checks it and continues looking right.
let rec cleanRight n node =
match node with
| Node(left, min, max, (Node(left', min', max', right') as right)) ->
if n > max' + 1L then
let node, n' = right |> cleanRight n
Node(left, min, max, node), n'
elif n >= min' then
Node(left, min, max, left'), min'
else
Node(left, min, max, left') |> cleanRight n
| _ -> node, n
// Symmetric to clean right
let rec cleanLeft x node =
match node with
| Node(Node(left', min', max', right') as left, min, max, right) ->
if x < min' - 1L then
let node, x' = left |> cleanLeft x
Node(node, min, max, right), x'
elif x <= max' then
Node(right', min, max, right), max'
else
Node(right', min, max, right) |> cleanLeft x
| Nil -> node, x
| _ -> node, x
// Merger left is called whenever the min value of a node decreases.
// It handles the case of left node overlap/subsets and merging/deleting them.
// When no more overlaps are found on the left nodes it calls clean right.
let rec mergeLeft n node =
match node with
| Node(Node(left', min', max', right') as left, min, max, right) ->
if n <= min' - 1L then
Node(left', min, max, right) |> mergeLeft n
elif n <= max' + 1L then
Node(left', min', max, right)
else
let node, min' = left |> cleanRight n
Node(node, min', max, right)
| _ -> node
// Symmetric to merge left
let rec mergeRight x node =
match node with
| Node(left, min, max, (Node(left', min', max', right') as right)) ->
if x >= max' + 1L then
Node(left, min, max, right') |> mergeRight x
elif x >= min' - 1L then
Node(left, min, max', right')
else
let node, max' = right |> cleanLeft x
Node(left, min, max', node)
| node -> node
let (|Before|After|BeforeOverlap|AfterOverlap|Superset|Subset|) (min, max, min', max') =
if min > max' + 1L then After
elif min >= min' then
if max <= max' then Subset
else AfterOverlap
elif max < min' - 1L then Before
elif max <= max' then
if min >= min' then Subset
else BeforeOverlap
else Superset
let rec insert min' max' this =
match this with
| Node(left, min, max, right) ->
match (min', max', min, max) with
| After -> Node(left, min, max, right |> insert min' max')
| AfterOverlap -> Node(left, min, max', right) |> mergeRight max'
| Before -> Node(left |> insert min' max', min, max, right)
| BeforeOverlap -> Node(left, min', max, right) |> mergeLeft min'
| Superset -> Node(left, min', max', right) |> mergeLeft min' |> mergeRight max'
| Subset -> this
| Nil -> Node(Nil, min', max', Nil)
let rangeSeqToRangeTree ranges =
ranges |> Seq.fold (fun tree range -> tree |> insert range.min range.max) Nil
let rangeTreeToRangeSeq node =
let rec inOrder node =
seq {
match node with
| Node(left, min, max, right) ->
yield! inOrder left
yield {min = min; max = max}
yield! inOrder right
| Nil -> ()
}
inOrder node
let TestImmutableRangeTree() =
printf "\n"
let source(n) =
let rnd = new Random(n)
let rand x = rnd.NextDouble() * float x |> int64
let rangeRnd() =
let a = rand 15000
{min = a; max = a + rand 150}
[|for n in 1 .. 200 do yield rangeRnd()|]
let shuffle n (array:_[]) =
let rnd = new Random(n)
for i in 0 .. array.Length - 1 do
let n = rnd.Next(i)
let temp = array.[i]
array.[i] <- array.[n]
array.[n] <- temp
array
let print dataSet =
dataSet |> Seq.iter (fun r -> printf "%s " <| string r)
let testRangeAdd n i =
let dataSet1 = source (n+0)
let dataSet2 = source (n+1)
let dataSet3 = source (n+2)
let result1 = Array.concat [dataSet1; dataSet2; dataSet3] |> shuffle (i+3) |> rangeSeqToRangeTree
let result2 = Array.concat [dataSet2; dataSet3; dataSet1] |> shuffle (i+4) |> rangeSeqToRangeTree
let result3 = Array.concat [dataSet3; dataSet1; dataSet2] |> shuffle (i+5) |> rangeSeqToRangeTree
let test1 = (result1 |> rangeTreeToRangeSeq, result2 |> rangeTreeToRangeSeq) ||> Seq.forall2 (fun a b -> a.min = b.min && a.max = b.max)
let test2 = (result2 |> rangeTreeToRangeSeq, result3 |> rangeTreeToRangeSeq) ||> Seq.forall2 (fun a b -> a.min = b.min && a.max = b.max)
let test3 = (result3 |> rangeTreeToRangeSeq, result1 |> rangeTreeToRangeSeq) ||> Seq.forall2 (fun a b -> a.min = b.min && a.max = b.max)
if not (test1 && test2 && test3) then
printf "\n\nTest# %A: " n
printf "\nSource 1: %A: " (n+0)
dataSet1 |> print
printf "\nSource 2: %A: " (n+1)
dataSet2 |> print
printf "\nSource 3: %A: " (n+2)
dataSet3 |> print
printf "\n\nResult 1: %A: " (n+0)
result1 |> rangeTreeToRangeSeq |> print
printf "\nResult 2: %A: " (n+1)
result2 |> rangeTreeToRangeSeq |> print
printf "\nResult 3: %A: " (n+2)
result3 |> rangeTreeToRangeSeq |> print
()
for i in 1 .. 10 do
for n in 1 .. 100 do
testRangeAdd n i
printf "\n%d" (i * 10)
printf "\nDone"
TestImmutableRangeTree()
System.Console.ReadLine() |> ignore
关于f# - 你如何将这棵可变树转换成一棵不可变树？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/1856794/