k3s/vendor/github.com/google/btree/btree.go
Darren Shepherd fa08d6076c Update vendor
2019-01-11 21:58:27 -07:00

650 lines
18 KiB
Go

// Copyright 2014 Google Inc.
//
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
// Package btree implements in-memory B-Trees of arbitrary degree.
//
// btree implements an in-memory B-Tree for use as an ordered data structure.
// It is not meant for persistent storage solutions.
//
// It has a flatter structure than an equivalent red-black or other binary tree,
// which in some cases yields better memory usage and/or performance.
// See some discussion on the matter here:
// http://google-opensource.blogspot.com/2013/01/c-containers-that-save-memory-and-time.html
// Note, though, that this project is in no way related to the C++ B-Tree
// implmentation written about there.
//
// Within this tree, each node contains a slice of items and a (possibly nil)
// slice of children. For basic numeric values or raw structs, this can cause
// efficiency differences when compared to equivalent C++ template code that
// stores values in arrays within the node:
// * Due to the overhead of storing values as interfaces (each
// value needs to be stored as the value itself, then 2 words for the
// interface pointing to that value and its type), resulting in higher
// memory use.
// * Since interfaces can point to values anywhere in memory, values are
// most likely not stored in contiguous blocks, resulting in a higher
// number of cache misses.
// These issues don't tend to matter, though, when working with strings or other
// heap-allocated structures, since C++-equivalent structures also must store
// pointers and also distribute their values across the heap.
//
// This implementation is designed to be a drop-in replacement to gollrb.LLRB
// trees, (http://github.com/petar/gollrb), an excellent and probably the most
// widely used ordered tree implementation in the Go ecosystem currently.
// Its functions, therefore, exactly mirror those of
// llrb.LLRB where possible. Unlike gollrb, though, we currently don't
// support storing multiple equivalent values or backwards iteration.
package btree
import (
"fmt"
"io"
"sort"
"strings"
)
// Item represents a single object in the tree.
type Item interface {
// Less tests whether the current item is less than the given argument.
//
// This must provide a strict weak ordering.
// If !a.Less(b) && !b.Less(a), we treat this to mean a == b (i.e. we can only
// hold one of either a or b in the tree).
Less(than Item) bool
}
const (
DefaultFreeListSize = 32
)
// FreeList represents a free list of btree nodes. By default each
// BTree has its own FreeList, but multiple BTrees can share the same
// FreeList.
// Two Btrees using the same freelist are not safe for concurrent write access.
type FreeList struct {
freelist []*node
}
// NewFreeList creates a new free list.
// size is the maximum size of the returned free list.
func NewFreeList(size int) *FreeList {
return &FreeList{freelist: make([]*node, 0, size)}
}
func (f *FreeList) newNode() (n *node) {
index := len(f.freelist) - 1
if index < 0 {
return new(node)
}
f.freelist, n = f.freelist[:index], f.freelist[index]
return
}
func (f *FreeList) freeNode(n *node) {
if len(f.freelist) < cap(f.freelist) {
f.freelist = append(f.freelist, n)
}
}
// ItemIterator allows callers of Ascend* to iterate in-order over portions of
// the tree. When this function returns false, iteration will stop and the
// associated Ascend* function will immediately return.
type ItemIterator func(i Item) bool
// New creates a new B-Tree with the given degree.
//
// New(2), for example, will create a 2-3-4 tree (each node contains 1-3 items
// and 2-4 children).
func New(degree int) *BTree {
return NewWithFreeList(degree, NewFreeList(DefaultFreeListSize))
}
// NewWithFreeList creates a new B-Tree that uses the given node free list.
func NewWithFreeList(degree int, f *FreeList) *BTree {
if degree <= 1 {
panic("bad degree")
}
return &BTree{
degree: degree,
freelist: f,
}
}
// items stores items in a node.
type items []Item
// insertAt inserts a value into the given index, pushing all subsequent values
// forward.
func (s *items) insertAt(index int, item Item) {
*s = append(*s, nil)
if index < len(*s) {
copy((*s)[index+1:], (*s)[index:])
}
(*s)[index] = item
}
// removeAt removes a value at a given index, pulling all subsequent values
// back.
func (s *items) removeAt(index int) Item {
item := (*s)[index]
(*s)[index] = nil
copy((*s)[index:], (*s)[index+1:])
*s = (*s)[:len(*s)-1]
return item
}
// pop removes and returns the last element in the list.
func (s *items) pop() (out Item) {
index := len(*s) - 1
out = (*s)[index]
(*s)[index] = nil
*s = (*s)[:index]
return
}
// find returns the index where the given item should be inserted into this
// list. 'found' is true if the item already exists in the list at the given
// index.
func (s items) find(item Item) (index int, found bool) {
i := sort.Search(len(s), func(i int) bool {
return item.Less(s[i])
})
if i > 0 && !s[i-1].Less(item) {
return i - 1, true
}
return i, false
}
// children stores child nodes in a node.
type children []*node
// insertAt inserts a value into the given index, pushing all subsequent values
// forward.
func (s *children) insertAt(index int, n *node) {
*s = append(*s, nil)
if index < len(*s) {
copy((*s)[index+1:], (*s)[index:])
}
(*s)[index] = n
}
// removeAt removes a value at a given index, pulling all subsequent values
// back.
func (s *children) removeAt(index int) *node {
n := (*s)[index]
(*s)[index] = nil
copy((*s)[index:], (*s)[index+1:])
*s = (*s)[:len(*s)-1]
return n
}
// pop removes and returns the last element in the list.
func (s *children) pop() (out *node) {
index := len(*s) - 1
out = (*s)[index]
(*s)[index] = nil
*s = (*s)[:index]
return
}
// node is an internal node in a tree.
//
// It must at all times maintain the invariant that either
// * len(children) == 0, len(items) unconstrained
// * len(children) == len(items) + 1
type node struct {
items items
children children
t *BTree
}
// split splits the given node at the given index. The current node shrinks,
// and this function returns the item that existed at that index and a new node
// containing all items/children after it.
func (n *node) split(i int) (Item, *node) {
item := n.items[i]
next := n.t.newNode()
next.items = append(next.items, n.items[i+1:]...)
n.items = n.items[:i]
if len(n.children) > 0 {
next.children = append(next.children, n.children[i+1:]...)
n.children = n.children[:i+1]
}
return item, next
}
// maybeSplitChild checks if a child should be split, and if so splits it.
// Returns whether or not a split occurred.
func (n *node) maybeSplitChild(i, maxItems int) bool {
if len(n.children[i].items) < maxItems {
return false
}
first := n.children[i]
item, second := first.split(maxItems / 2)
n.items.insertAt(i, item)
n.children.insertAt(i+1, second)
return true
}
// insert inserts an item into the subtree rooted at this node, making sure
// no nodes in the subtree exceed maxItems items. Should an equivalent item be
// be found/replaced by insert, it will be returned.
func (n *node) insert(item Item, maxItems int) Item {
i, found := n.items.find(item)
if found {
out := n.items[i]
n.items[i] = item
return out
}
if len(n.children) == 0 {
n.items.insertAt(i, item)
return nil
}
if n.maybeSplitChild(i, maxItems) {
inTree := n.items[i]
switch {
case item.Less(inTree):
// no change, we want first split node
case inTree.Less(item):
i++ // we want second split node
default:
out := n.items[i]
n.items[i] = item
return out
}
}
return n.children[i].insert(item, maxItems)
}
// get finds the given key in the subtree and returns it.
func (n *node) get(key Item) Item {
i, found := n.items.find(key)
if found {
return n.items[i]
} else if len(n.children) > 0 {
return n.children[i].get(key)
}
return nil
}
// min returns the first item in the subtree.
func min(n *node) Item {
if n == nil {
return nil
}
for len(n.children) > 0 {
n = n.children[0]
}
if len(n.items) == 0 {
return nil
}
return n.items[0]
}
// max returns the last item in the subtree.
func max(n *node) Item {
if n == nil {
return nil
}
for len(n.children) > 0 {
n = n.children[len(n.children)-1]
}
if len(n.items) == 0 {
return nil
}
return n.items[len(n.items)-1]
}
// toRemove details what item to remove in a node.remove call.
type toRemove int
const (
removeItem toRemove = iota // removes the given item
removeMin // removes smallest item in the subtree
removeMax // removes largest item in the subtree
)
// remove removes an item from the subtree rooted at this node.
func (n *node) remove(item Item, minItems int, typ toRemove) Item {
var i int
var found bool
switch typ {
case removeMax:
if len(n.children) == 0 {
return n.items.pop()
}
i = len(n.items)
case removeMin:
if len(n.children) == 0 {
return n.items.removeAt(0)
}
i = 0
case removeItem:
i, found = n.items.find(item)
if len(n.children) == 0 {
if found {
return n.items.removeAt(i)
}
return nil
}
default:
panic("invalid type")
}
// If we get to here, we have children.
child := n.children[i]
if len(child.items) <= minItems {
return n.growChildAndRemove(i, item, minItems, typ)
}
// Either we had enough items to begin with, or we've done some
// merging/stealing, because we've got enough now and we're ready to return
// stuff.
if found {
// The item exists at index 'i', and the child we've selected can give us a
// predecessor, since if we've gotten here it's got > minItems items in it.
out := n.items[i]
// We use our special-case 'remove' call with typ=maxItem to pull the
// predecessor of item i (the rightmost leaf of our immediate left child)
// and set it into where we pulled the item from.
n.items[i] = child.remove(nil, minItems, removeMax)
return out
}
// Final recursive call. Once we're here, we know that the item isn't in this
// node and that the child is big enough to remove from.
return child.remove(item, minItems, typ)
}
// growChildAndRemove grows child 'i' to make sure it's possible to remove an
// item from it while keeping it at minItems, then calls remove to actually
// remove it.
//
// Most documentation says we have to do two sets of special casing:
// 1) item is in this node
// 2) item is in child
// In both cases, we need to handle the two subcases:
// A) node has enough values that it can spare one
// B) node doesn't have enough values
// For the latter, we have to check:
// a) left sibling has node to spare
// b) right sibling has node to spare
// c) we must merge
// To simplify our code here, we handle cases #1 and #2 the same:
// If a node doesn't have enough items, we make sure it does (using a,b,c).
// We then simply redo our remove call, and the second time (regardless of
// whether we're in case 1 or 2), we'll have enough items and can guarantee
// that we hit case A.
func (n *node) growChildAndRemove(i int, item Item, minItems int, typ toRemove) Item {
child := n.children[i]
if i > 0 && len(n.children[i-1].items) > minItems {
// Steal from left child
stealFrom := n.children[i-1]
stolenItem := stealFrom.items.pop()
child.items.insertAt(0, n.items[i-1])
n.items[i-1] = stolenItem
if len(stealFrom.children) > 0 {
child.children.insertAt(0, stealFrom.children.pop())
}
} else if i < len(n.items) && len(n.children[i+1].items) > minItems {
// steal from right child
stealFrom := n.children[i+1]
stolenItem := stealFrom.items.removeAt(0)
child.items = append(child.items, n.items[i])
n.items[i] = stolenItem
if len(stealFrom.children) > 0 {
child.children = append(child.children, stealFrom.children.removeAt(0))
}
} else {
if i >= len(n.items) {
i--
child = n.children[i]
}
// merge with right child
mergeItem := n.items.removeAt(i)
mergeChild := n.children.removeAt(i + 1)
child.items = append(child.items, mergeItem)
child.items = append(child.items, mergeChild.items...)
child.children = append(child.children, mergeChild.children...)
n.t.freeNode(mergeChild)
}
return n.remove(item, minItems, typ)
}
// iterate provides a simple method for iterating over elements in the tree.
// It could probably use some work to be extra-efficient (it calls from() a
// little more than it should), but it works pretty well for now.
//
// It requires that 'from' and 'to' both return true for values we should hit
// with the iterator. It should also be the case that 'from' returns true for
// values less than or equal to values 'to' returns true for, and 'to'
// returns true for values greater than or equal to those that 'from'
// does.
func (n *node) iterate(from, to func(Item) bool, iter ItemIterator) bool {
for i, item := range n.items {
if !from(item) {
continue
}
if len(n.children) > 0 && !n.children[i].iterate(from, to, iter) {
return false
}
if !to(item) {
return false
}
if !iter(item) {
return false
}
}
if len(n.children) > 0 {
return n.children[len(n.children)-1].iterate(from, to, iter)
}
return true
}
// Used for testing/debugging purposes.
func (n *node) print(w io.Writer, level int) {
fmt.Fprintf(w, "%sNODE:%v\n", strings.Repeat(" ", level), n.items)
for _, c := range n.children {
c.print(w, level+1)
}
}
// BTree is an implementation of a B-Tree.
//
// BTree stores Item instances in an ordered structure, allowing easy insertion,
// removal, and iteration.
//
// Write operations are not safe for concurrent mutation by multiple
// goroutines, but Read operations are.
type BTree struct {
degree int
length int
root *node
freelist *FreeList
}
// maxItems returns the max number of items to allow per node.
func (t *BTree) maxItems() int {
return t.degree*2 - 1
}
// minItems returns the min number of items to allow per node (ignored for the
// root node).
func (t *BTree) minItems() int {
return t.degree - 1
}
func (t *BTree) newNode() (n *node) {
n = t.freelist.newNode()
n.t = t
return
}
func (t *BTree) freeNode(n *node) {
for i := range n.items {
n.items[i] = nil // clear to allow GC
}
n.items = n.items[:0]
for i := range n.children {
n.children[i] = nil // clear to allow GC
}
n.children = n.children[:0]
n.t = nil // clear to allow GC
t.freelist.freeNode(n)
}
// ReplaceOrInsert adds the given item to the tree. If an item in the tree
// already equals the given one, it is removed from the tree and returned.
// Otherwise, nil is returned.
//
// nil cannot be added to the tree (will panic).
func (t *BTree) ReplaceOrInsert(item Item) Item {
if item == nil {
panic("nil item being added to BTree")
}
if t.root == nil {
t.root = t.newNode()
t.root.items = append(t.root.items, item)
t.length++
return nil
} else if len(t.root.items) >= t.maxItems() {
item2, second := t.root.split(t.maxItems() / 2)
oldroot := t.root
t.root = t.newNode()
t.root.items = append(t.root.items, item2)
t.root.children = append(t.root.children, oldroot, second)
}
out := t.root.insert(item, t.maxItems())
if out == nil {
t.length++
}
return out
}
// Delete removes an item equal to the passed in item from the tree, returning
// it. If no such item exists, returns nil.
func (t *BTree) Delete(item Item) Item {
return t.deleteItem(item, removeItem)
}
// DeleteMin removes the smallest item in the tree and returns it.
// If no such item exists, returns nil.
func (t *BTree) DeleteMin() Item {
return t.deleteItem(nil, removeMin)
}
// DeleteMax removes the largest item in the tree and returns it.
// If no such item exists, returns nil.
func (t *BTree) DeleteMax() Item {
return t.deleteItem(nil, removeMax)
}
func (t *BTree) deleteItem(item Item, typ toRemove) Item {
if t.root == nil || len(t.root.items) == 0 {
return nil
}
out := t.root.remove(item, t.minItems(), typ)
if len(t.root.items) == 0 && len(t.root.children) > 0 {
oldroot := t.root
t.root = t.root.children[0]
t.freeNode(oldroot)
}
if out != nil {
t.length--
}
return out
}
// AscendRange calls the iterator for every value in the tree within the range
// [greaterOrEqual, lessThan), until iterator returns false.
func (t *BTree) AscendRange(greaterOrEqual, lessThan Item, iterator ItemIterator) {
if t.root == nil {
return
}
t.root.iterate(
func(a Item) bool { return !a.Less(greaterOrEqual) },
func(a Item) bool { return a.Less(lessThan) },
iterator)
}
// AscendLessThan calls the iterator for every value in the tree within the range
// [first, pivot), until iterator returns false.
func (t *BTree) AscendLessThan(pivot Item, iterator ItemIterator) {
if t.root == nil {
return
}
t.root.iterate(
func(a Item) bool { return true },
func(a Item) bool { return a.Less(pivot) },
iterator)
}
// AscendGreaterOrEqual calls the iterator for every value in the tree within
// the range [pivot, last], until iterator returns false.
func (t *BTree) AscendGreaterOrEqual(pivot Item, iterator ItemIterator) {
if t.root == nil {
return
}
t.root.iterate(
func(a Item) bool { return !a.Less(pivot) },
func(a Item) bool { return true },
iterator)
}
// Ascend calls the iterator for every value in the tree within the range
// [first, last], until iterator returns false.
func (t *BTree) Ascend(iterator ItemIterator) {
if t.root == nil {
return
}
t.root.iterate(
func(a Item) bool { return true },
func(a Item) bool { return true },
iterator)
}
// Get looks for the key item in the tree, returning it. It returns nil if
// unable to find that item.
func (t *BTree) Get(key Item) Item {
if t.root == nil {
return nil
}
return t.root.get(key)
}
// Min returns the smallest item in the tree, or nil if the tree is empty.
func (t *BTree) Min() Item {
return min(t.root)
}
// Max returns the largest item in the tree, or nil if the tree is empty.
func (t *BTree) Max() Item {
return max(t.root)
}
// Has returns true if the given key is in the tree.
func (t *BTree) Has(key Item) bool {
return t.Get(key) != nil
}
// Len returns the number of items currently in the tree.
func (t *BTree) Len() int {
return t.length
}
// Int implements the Item interface for integers.
type Int int
// Less returns true if int(a) < int(b).
func (a Int) Less(b Item) bool {
return a < b.(Int)
}