并发设置队列

本文介绍了并发设置队列的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

也许这是一个愚蠢的问题，但我似乎找不到一个明显的答案。

Maybe this is a silly question, but I cannot seem to find an obvious answer.

我需要一个并发的FIFO队列，只包含唯一的值。尝试添加已存在于队列中的值将忽略该值。其中，如果不是为了线程安全将是微不足道的。在Java中有一个数据结构，或者是在具有这种行为的网络上的代码snipit？

I need a concurrent FIFO queue that contains only unique values. Attempting to add a value that already exists in the queue simply ignores that value. Which, if not for the thread safety would be trivial. Is there a data structure in Java or maybe a code snipit on the interwebs that exhibits this behavior?

推荐答案

如果你想要更好的并发比完全同步，有一种方式我知道做，使用ConcurrentHashMap作为背景图。以下是仅草图。

If you want better concurrency than full synchronization, there is one way I know of to do it, using a ConcurrentHashMap as the backing map. The following is a sketch only.

public final class ConcurrentHashSet<E> extends ForwardingSet<E> implements Set<E>, Queue<E> { private enum Dummy { VALUE } private final ConcurrentMap<E, Dummy> map; ConcurrentHashSet(ConcurrentMap<E, Dummy> map) { super(map.keySet()); this.map = Preconditions.checkNotNull(map); } @Override public boolean add(E element) { return map.put(element, Dummy.VALUE) == null; } @Override public boolean addAll(Collection<? extends E> newElements) { // just the standard implementation boolean modified = false; for (E element : newElements) { modified |= add(element); } return modified; } @Override public boolean offer(E element) { return add(element); } @Override public E remove() { E polled = poll(); if (polled == null) { throw new NoSuchElementException(); } return polled; } @Override public E poll() { for (E element : this) { // Not convinced that removing via iterator is viable (check this?) if (map.remove(element) != null) { return element; } } return null; } @Override public E element() { return iterator().next(); } @Override public E peek() { Iterator<E> iterator = iterator(); return iterator.hasNext() ? iterator.next() : null; } }

这种方法不是阳光。除了使用后台映射的 entrySet（）。iterator（）。next（），我们没有正确的方法来选择头元素，结果是映射获得更多更不平衡，随着时间的推移。

All is not sunshine with this approach. We have no decent way to select a head element other than using the backing map's entrySet().iterator().next(), the result being that the map gets more and more unbalanced as time goes on. This unbalancing is a problem both due to greater bucket collisions and greater segment contention.

注意：此代码使用几个地方的Guava 。