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# Radix tree implementation. It isn't that much faster than a hash table,
# however it *is* faster.
import json
import options
import tables
type
RadixPair[T] = tuple[k: string, v: RadixNode[T]]
RadixNode*[T] = ref object
children*: seq[RadixPair[T]]
value*: Option[T]
func newRadixTree*[T](): RadixNode[T] =
new(result)
func toRadixTree*[T](table: Table[string, T]): RadixNode[T] =
result = newRadixTree[T]()
for k, v in table:
result[k] = v
# getOrDefault: we have to compare the entire string but if it doesn't match
# exactly we can just return default.
func getOrDefault[T](node: RadixNode[T], k: string, default: RadixNode[T]): RadixNode[T] =
var i = 0
while i < node.children.len:
if node.children[i].k[0] == k[0]:
if k.len != node.children[i].k.len:
return default
var j = 1
while j < k.len:
if node.children[i].k[j] != k[j]:
return default
inc j
return node.children[i].v
inc i
return default
iterator keys*[T](node: RadixNode[T]): string =
var i = 0
while i < node.children.len:
yield node.children[i].k
inc i
#func contains[T](node: RadixNode[T], k: string): bool =
# var i = 0
# while i < node.children.len:
# if node.children[i].k[0] == k[0]:
# if k.len != node.children[i].k.len:
# return false
# var j = 1
# while j < k.len:
# if node.children[i].k[j] != k[j]:
# return false
# inc j
# return true
# inc i
# return false
# O(1) add procedures for insert
proc add[T](node: RadixNode[T], k: string, v: T) =
node.children.add((k, RadixNode[T](value: v.some)))
proc add[T](node: RadixNode[T], k: string) =
node.children.add((k, RadixNode[T]()))
proc add[T](node: RadixNode[T], k: string, v: RadixNode[T]) =
node.children.add((k, v))
# insert
proc `[]=`*[T](tree: RadixNode[T], key: string, value: T) =
var n = tree
var p: RadixNode[T] = nil
var i = 0
var j = 0
var k = 0
var t = ""
# find last matching node
var conflict = false
while i < key.len:
let m = i
var o = 0
for pk in n.keys:
if pk[0] == key[i]:
var l = 0
while l < pk.len and i + l < key.len:
if pk[l] != key[i + l]:
conflict = true
#t = key.substr(0, i + l - 1) & pk.substr(l)
break
inc l
p = n
k = o
n = n.children[k].v
t &= pk
i += l
if not conflict and pk.len == l:
j = i
# t = key.substr(0, i - 1)
#elif not conflict and pk.len > l:
# t = key & pk.substr(l)
break
inc o
if i == m:
break
if conflict:
break
if n == tree:
# first node, just add normally
tree.add(key, value)
elif conflict:
# conflict somewhere, so:
# * add new non-leaf to parent
# * add old to non-leaf
# * add new to non-leaf
# * remove old from parent
p.add(key.substr(j, i - 1))
p.children[^1].v.add(t.substr(i), n)
p.children[^1].v.add(key.substr(i), value)
p.children.del(k)
elif key.len == t.len:
# new matches a node, so replace
p.children[k].v = RadixNode[T](value: value.some, children: n.children)
elif key.len > t.len:
# new is longer than the old, so add child to old
n.add(key.substr(i), value)
else:
# new is shorter than old, so:
# * add new to parent
# * add old to new
# * remove old from parent
p.add(key.substr(j, i - 1), value)
p.children[^1].v.add(t.substr(i), n)
p.children.del(k)
func `{}`*[T](node: RadixNode[T], key: string): RadixNode[T] =
return node.getOrDefault(key, node)
func hasPrefix*[T](node: RadixNode[T], prefix: string): bool =
var n = node
var i = 0
while i < prefix.len:
let m = i
var j = 0
for pk in n.keys:
if pk[0] == prefix[i]:
var l = 0
while l < pk.len and i + l < prefix.len:
if pk[l] != prefix[i + l]:
return false
inc l
n = n.children[j].v
i += l
break
inc j
if i == m:
return false
return true
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