HashSetremoveAll方法非常慢-HashSetremoveAllmethodissurprisinglyslow

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The behaviour is (somewhat) documented in the javadoc:

行为（有些）记录在javadoc中：

This implementation determines which is the smaller of this set and the specified collection, by invoking the size method on each. If this set has fewer elements, then the implementation iterates over this set, checking each element returned by the iterator in turn to see if it is contained in the specified collection. If it is so contained, it is removed from this set with the iterator's remove method. If the specified collection has fewer elements, then the implementation iterates over the specified collection, removing from this set each element returned by the iterator, using this set's remove method.

此实现通过在每个集合上调用size方法来确定该集合和指定集合中较小的集合。如果此set具有较少的元素，则实现迭代该集合，依次检查迭代器返回的每个元素以查看它是否包含在指定的集合中。如果包含它，则使用迭代器的remove方法从该集合中删除它。如果指定的集合具有较少的元素，则实现将迭代指定的集合，使用此set的remove方法从此集合中删除迭代器返回的每个元素。

What this means in practice, when you call source.removeAll(removals);:

这在实践中意味着什么，当你调用source.removeAll（删除）;时：

• if the removals collection is of a smaller size than source, the remove method of HashSet is called, which is fast.

  如果删除集合的大小小于源，则调用HashSet的remove方法，这很快。

• if the removals collection is of equal or larger size than the source, then removals.contains is called, which is slow for an ArrayList.

  如果删除集合的大小等于或大于源，则调用removals.contains，这对于ArrayList来说很慢。

Quick fix:

快速解决：

Collection removals = new HashSet();

Note that there is an open bug that is very similar to what you describe. The bottom line seems to be that it is probably a poor choice but can't be changed because it is documented in the javadoc.

请注意，有一个与您描述的非常相似的开放式错误。底线似乎是它可能是一个糟糕的选择，但不能改变，因为它记录在javadoc中。

For reference, this is the code of removeAll (in Java 8 - haven't checked other versions):

作为参考，这是removeAll的代码（在Java 8中 - 尚未检查其他版本）：

public boolean removeAll(Collection c) {
    Objects.requireNonNull(c);
    boolean modified = false;

    if (size() > c.size()) {
        for (Iterator i = c.iterator(); i.hasNext(); )
            modified |= remove(i.next());
    } else {
        for (Iterator i = iterator(); i.hasNext(); ) {
            if (c.contains(i.next())) {
                i.remove();
                modified = true;
            }
        }
    }
    return modified;
}