- btree iterator implemented with tests
Emir Pasic
9 years ago
parent
76711f5b71
commit
53898058bb
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// Copyright (c) 2015, Emir Pasic. All rights reserved.
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// Use of this source code is governed by a BSD-style
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// license that can be found in the LICENSE file.
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package btree
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import "github.com/emirpasic/gods/containers"
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func assertIteratorImplementation() {
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var _ containers.ReverseIteratorWithKey = (*Iterator)(nil)
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}
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// Iterator holding the iterator's state
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type Iterator struct {
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tree *Tree
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node *Node
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entry *Entry
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position position
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}
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type position byte
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const (
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begin, between, end position = 0, 1, 2
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)
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// Iterator returns a stateful iterator whose elements are key/value pairs.
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func (tree *Tree) Iterator() Iterator {
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return Iterator{tree: tree, node: nil, position: begin}
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}
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// Next moves the iterator to the next element and returns true if there was a next element in the container.
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// If Next() returns true, then next element's key and value can be retrieved by Key() and Value().
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// If Next() was called for the first time, then it will point the iterator to the first element if it exists.
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// Modifies the state of the iterator.
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func (iterator *Iterator) Next() bool {
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// If already at end, go to end
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if iterator.position == end {
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goto end
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}
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// If at beginning, get the left-most entry in the tree
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if iterator.position == begin {
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left := iterator.tree.Left()
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if left == nil {
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goto end
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}
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iterator.node = left
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iterator.entry = left.Entries[0]
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goto between
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}
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{
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// Find current entry position in current node
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e, _ := iterator.tree.search(iterator.node, iterator.entry.Key)
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// Try to go down to the child right of the current entry
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if e+1 < len(iterator.node.Children) {
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iterator.node = iterator.node.Children[e+1]
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// Try to go down to the child left of the current node
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for len(iterator.node.Children) > 0 {
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iterator.node = iterator.node.Children[0]
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}
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// Return the left-most entry
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iterator.entry = iterator.node.Entries[0]
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goto between
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}
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// Above assures that we have reached a leaf node, so return the next entry in current node (if any)
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if e+1 < len(iterator.node.Entries) {
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iterator.entry = iterator.node.Entries[e+1]
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goto between
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}
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}
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// Reached leaf node and there are no entries to the right of the current entry, so go up to the parent
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for iterator.node.Parent != nil {
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iterator.node = iterator.node.Parent
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// Find next entry position in current node (note: search returns the first equal or bigger than entry)
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e, _ := iterator.tree.search(iterator.node, iterator.entry.Key)
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// Check that there is a next entry position in current node
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if e < len(iterator.node.Entries) {
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iterator.entry = iterator.node.Entries[e]
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goto between
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}
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}
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end:
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iterator.End()
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return false
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between:
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iterator.position = between
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return true
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}
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// Prev moves the iterator to the previous element and returns true if there was a previous element in the container.
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// If Prev() returns true, then previous element's key and value can be retrieved by Key() and Value().
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// Modifies the state of the iterator.
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func (iterator *Iterator) Prev() bool {
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// If already at beginning, go to begin
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if iterator.position == begin {
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goto begin
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}
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// If at end, get the right-most entry in the tree
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if iterator.position == end {
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right := iterator.tree.Right()
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if right == nil {
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goto begin
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}
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iterator.node = right
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iterator.entry = right.Entries[len(right.Entries)-1]
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goto between
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}
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{
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// Find current entry position in current node
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e, _ := iterator.tree.search(iterator.node, iterator.entry.Key)
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// Try to go down to the child left of the current entry
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if e < len(iterator.node.Children) {
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iterator.node = iterator.node.Children[e]
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// Try to go down to the child right of the current node
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for len(iterator.node.Children) > 0 {
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iterator.node = iterator.node.Children[len(iterator.node.Children)-1]
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}
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// Return the right-most entry
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iterator.entry = iterator.node.Entries[len(iterator.node.Entries)-1]
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goto between
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}
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// Above assures that we have reached a leaf node, so return the previous entry in current node (if any)
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if e-1 >= 0 {
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iterator.entry = iterator.node.Entries[e-1]
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goto between
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}
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}
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// Reached leaf node and there are no entries to the left of the current entry, so go up to the parent
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for iterator.node.Parent != nil {
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iterator.node = iterator.node.Parent
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// Find previous entry position in current node (note: search returns the first equal or bigger than entry)
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e, _ := iterator.tree.search(iterator.node, iterator.entry.Key)
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// Check that there is a previous entry position in current node
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if e-1 >= 0 {
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iterator.entry = iterator.node.Entries[e-1]
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goto between
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}
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}
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begin:
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iterator.Begin()
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return false
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between:
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iterator.position = between
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return true
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}
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// Value returns the current element's value.
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// Does not modify the state of the iterator.
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func (iterator *Iterator) Value() interface{} {
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return iterator.entry.Value
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}
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// Key returns the current element's key.
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// Does not modify the state of the iterator.
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func (iterator *Iterator) Key() interface{} {
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return iterator.entry.Key
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}
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// Begin resets the iterator to its initial state (one-before-first)
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// Call Next() to fetch the first element if any.
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func (iterator *Iterator) Begin() {
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iterator.node = nil
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iterator.position = begin
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iterator.entry = nil
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}
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// End moves the iterator past the last element (one-past-the-end).
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// Call Prev() to fetch the last element if any.
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func (iterator *Iterator) End() {
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iterator.node = nil
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iterator.position = end
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iterator.entry = nil
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}
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// First moves the iterator to the first element and returns true if there was a first element in the container.
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// If First() returns true, then first element's key and value can be retrieved by Key() and Value().
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// Modifies the state of the iterator
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func (iterator *Iterator) First() bool {
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iterator.Begin()
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return iterator.Next()
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}
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// Last moves the iterator to the last element and returns true if there was a last element in the container.
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// If Last() returns true, then last element's key and value can be retrieved by Key() and Value().
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// Modifies the state of the iterator.
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func (iterator *Iterator) Last() bool {
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iterator.End()
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return iterator.Prev()
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}
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