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王布衣/gox

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iterator.go 6.07 KB
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王布衣 提交于 2023-06-07 20:08 . 修订部分util工具库
// Copyright (c) 2015, Emir Pasic. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.
package btree
import (
"gitee.com/quant1x/gox/util/internal"
)
func assertIteratorImplementation() {
var _ internal.ReverseIteratorWithKey = (*Iterator)(nil)
}
// Iterator holding the iterator's state
type Iterator struct {
tree *Tree
node *Node
entry *Entry
position position
}
type position byte
const (
begin, between, end position = 0, 1, 2
)
// Iterator returns a stateful iterator whose elements are key/value pairs.
func (tree *Tree) Iterator() Iterator {
return Iterator{tree: tree, node: nil, position: begin}
}
// Next moves the iterator to the next element and returns true if there was a next element in the container.
// If Next() returns true, then next element's key and value can be retrieved by Key() and Value().
// If Next() was called for the first time, then it will point the iterator to the first element if it exists.
// Modifies the state of the iterator.
func (iterator *Iterator) Next() bool {
// If already at end, go to end
if iterator.position == end {
goto end
}
// If at beginning, get the left-most entry in the tree
if iterator.position == begin {
left := iterator.tree.Left()
if left == nil {
goto end
}
iterator.node = left
iterator.entry = left.Entries[0]
goto between
}
{
// Find current entry position in current node
e, _ := iterator.tree.search(iterator.node, iterator.entry.Key)
// Try to go down to the child right of the current entry
if e+1 < len(iterator.node.Children) {
iterator.node = iterator.node.Children[e+1]
// Try to go down to the child left of the current node
for len(iterator.node.Children) > 0 {
iterator.node = iterator.node.Children[0]
}
// Return the left-most entry
iterator.entry = iterator.node.Entries[0]
goto between
}
// Above assures that we have reached a leaf node, so return the next entry in current node (if any)
if e+1 < len(iterator.node.Entries) {
iterator.entry = iterator.node.Entries[e+1]
goto between
}
}
// Reached leaf node and there are no entries to the right of the current entry, so go up to the parent
for iterator.node.Parent != nil {
iterator.node = iterator.node.Parent
// Find next entry position in current node (note: search returns the first equal or bigger than entry)
e, _ := iterator.tree.search(iterator.node, iterator.entry.Key)
// Check that there is a next entry position in current node
if e < len(iterator.node.Entries) {
iterator.entry = iterator.node.Entries[e]
goto between
}
}
end:
iterator.End()
return false
between:
iterator.position = between
return true
}
// Prev moves the iterator to the previous element and returns true if there was a previous element in the container.
// If Prev() returns true, then previous element's key and value can be retrieved by Key() and Value().
// Modifies the state of the iterator.
func (iterator *Iterator) Prev() bool {
// If already at beginning, go to begin
if iterator.position == begin {
goto begin
}
// If at end, get the right-most entry in the tree
if iterator.position == end {
right := iterator.tree.Right()
if right == nil {
goto begin
}
iterator.node = right
iterator.entry = right.Entries[len(right.Entries)-1]
goto between
}
{
// Find current entry position in current node
e, _ := iterator.tree.search(iterator.node, iterator.entry.Key)
// Try to go down to the child left of the current entry
if e < len(iterator.node.Children) {
iterator.node = iterator.node.Children[e]
// Try to go down to the child right of the current node
for len(iterator.node.Children) > 0 {
iterator.node = iterator.node.Children[len(iterator.node.Children)-1]
}
// Return the right-most entry
iterator.entry = iterator.node.Entries[len(iterator.node.Entries)-1]
goto between
}
// Above assures that we have reached a leaf node, so return the previous entry in current node (if any)
if e-1 >= 0 {
iterator.entry = iterator.node.Entries[e-1]
goto between
}
}
// Reached leaf node and there are no entries to the left of the current entry, so go up to the parent
for iterator.node.Parent != nil {
iterator.node = iterator.node.Parent
// Find previous entry position in current node (note: search returns the first equal or bigger than entry)
e, _ := iterator.tree.search(iterator.node, iterator.entry.Key)
// Check that there is a previous entry position in current node
if e-1 >= 0 {
iterator.entry = iterator.node.Entries[e-1]
goto between
}
}
begin:
iterator.Begin()
return false
between:
iterator.position = between
return true
}
// Value returns the current element's value.
// Does not modify the state of the iterator.
func (iterator *Iterator) Value() interface{} {
return iterator.entry.Value
}
// Key returns the current element's key.
// Does not modify the state of the iterator.
func (iterator *Iterator) Key() interface{} {
return iterator.entry.Key
}
// Begin resets the iterator to its initial state (one-before-first)
// Call Next() to fetch the first element if any.
func (iterator *Iterator) Begin() {
iterator.node = nil
iterator.position = begin
iterator.entry = nil
}
// End moves the iterator past the last element (one-past-the-end).
// Call Prev() to fetch the last element if any.
func (iterator *Iterator) End() {
iterator.node = nil
iterator.position = end
iterator.entry = nil
}
// First moves the iterator to the first element and returns true if there was a first element in the container.
// If First() returns true, then first element's key and value can be retrieved by Key() and Value().
// Modifies the state of the iterator
func (iterator *Iterator) First() bool {
iterator.Begin()
return iterator.Next()
}
// Last moves the iterator to the last element and returns true if there was a last element in the container.
// If Last() returns true, then last element's key and value can be retrieved by Key() and Value().
// Modifies the state of the iterator.
func (iterator *Iterator) Last() bool {
iterator.End()
return iterator.Prev()
}
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