algorithm - Go 中的优先级队列实现

Question

我刚刚看到了优先级队列的一种通用实现方式，其中任何 满足接口(interface)的类型可以放入队列中。这是 go 的方法还是这会带来任何问题？

// Copyright 2012 Stefan Nilsson
//
// 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 prio provides a priority queue.
// The queue can hold elements that implement the two methods of prio.Interface.
package prio

/*
A type that implements prio.Interface can be inserted into a priority queue.

The simplest use case looks like this:

        type myInt int

        func (x myInt) Less(y prio.Interface) bool { return x < y.(myInt) }
        func (x myInt) Index(i int)                {}

To use the Remove method you need to keep track of the index of elements
in the heap, e.g. like this:

        type myType struct {
                value int
                index int // index in heap
        }

        func (x *myType) Less(y prio.Interface) bool { return x.value < y.(*myType).value }
        func (x *myType) Index(i int)                { x.index = i }
*/
type Interface interface {
        // Less returns whether this element should sort before element x.
        Less(x Interface) bool
        // Index is called by the priority queue when this element is moved to index i.
        Index(i int)
}

// Queue represents a priority queue.
// The zero value for Queue is an empty queue ready to use.
type Queue struct {
        h []Interface
}

// New returns an initialized priority queue with the given elements.
// A call of the form New(x...) uses the underlying array of x to implement
// the queue and hence might change the elements of x.
// The complexity is O(n), where n = len(x).
func New(x ...Interface) Queue {
        q := Queue{x}
        heapify(q.h)
        return q
}

// Push pushes the element x onto the queue.
// The complexity is O(log(n)) where n = q.Len().
func (q *Queue) Push(x Interface) {
        n := len(q.h)
        q.h = append(q.h, x)
        up(q.h, n) // x.Index(n) is done by up.
}

// Pop removes a minimum element (according to Less) from the queue and returns it.
// The complexity is O(log(n)), where n = q.Len().
func (q *Queue) Pop() Interface {
        h := q.h
        n := len(h) - 1
        x := h[0]
        h[0], h[n] = h[n], nil
        h = h[:n]
        if n > 0 {
                down(h, 0) // h[0].Index(0) is done by down.
        }
        q.h = h
        x.Index(-1) // for safety
        return x
}

// Peek returns, but does not remove, a minimum element (according to Less) of the queue.
func (q *Queue) Peek() Interface {
        return q.h[0]
}

// Remove removes the element at index i from the queue and returns it.
// The complexity is O(log(n)), where n = q.Len().
func (q *Queue) Remove(i int) Interface {
        h := q.h
        n := len(h) - 1
        x := h[i]
        h[i], h[n] = h[n], nil
        h = h[:n]
        if i < n {
                down(h, i) // h[i].Index(i) is done by down.
                up(h, i)
        }
        q.h = h
        x.Index(-1) // for safety
        return x
}

// Len returns the number of elements in the queue.
func (q *Queue) Len() int {
        return len(q.h)
}

// Establishes the heap invariant in O(n) time.
func heapify(h []Interface) {
        n := len(h)
        for i := n - 1; i >= n/2; i-- {
                h[i].Index(i)
        }
        for i := n/2 - 1; i >= 0; i-- { // h[i].Index(i) is done by down.
                down(h, i)
        }
}

// Moves element at position i towards top of heap to restore invariant.
func up(h []Interface, i int) {
        for {
                parent := (i - 1) / 2
                if i == 0 || h[parent].Less(h[i]) {
                        h[i].Index(i)
                        break
                }
                h[parent], h[i] = h[i], h[parent]
                h[i].Index(i)
                i = parent
        }
}

// Moves element at position i towards bottom of heap to restore invariant.
func down(h []Interface, i int) {
        for {
                n := len(h)
                left := 2*i + 1
                if left >= n {
                        h[i].Index(i)
                        break
                }
                j := left
                if right := left + 1; right < n && h[right].Less(h[left]) {
                        j = right
                }
                if h[i].Less(h[j]) {
                        h[i].Index(i)
                        break
                }
                h[i], h[j] = h[j], h[i]
                h[i].Index(i)
                i = j
        }
}

Answer 1

这个包不是一般的选择，但是如果你喜欢它并且它满足你的需求，请使用它。我没有看到任何重大问题。

这个包相对于容器/堆的理念是把接口(interface)放在节点上而不是容器上。容器/堆通过使用容器提供了更大的灵活性。 (您可能已经在容器中拥有节点，并且该容器甚至可能不是 slice 。它只需要可索引即可。)另一方面，您不关心容器并且会被索引，这可能是一种常见的情况很高兴让软件包为您管理它。与容器/堆相比，该包的索引管理是一个很好的功能，尽管即使不需要索引管理，它也会增加方法调用的开销。

总有一些权衡。容器/堆非常通用。这个包使用了较小的方法集(2 个而不是 5 个)，并在顶部添加了索引管理，但只是牺牲了一点通用性，在某些情况下可能还牺牲了一点性能。 (如果您真的关心的话，您需要进行基准测试。还有其他差异可能会使索引调用的开销相形见绌。)