Question

我想要一组并行的字符串值，按长度排序最长 - ＆gt;最短的。

这是我的代码（JAVA 8）：

private ConcurrentSkipListSet<String> sortedSourceTypeNames = new ConcurrentSkipListSet<>(Comparator.comparing(String::length).reversed());

以下是java 8文档：

    /**
     * Constructs a new, empty set that orders its elements according to
     * the specified comparator.
     *
     * @param comparator the comparator that will be used to order this set.
     *        If {@code null}, the {@linkplain Comparable natural
     *        ordering} of the elements will be used.
     */
    public ConcurrentSkipListSet(Comparator<? super E> comparator) {
        m = new ConcurrentSkipListMap<E,Object>(comparator);
    }

现在这是有线的事情：

添加新值＆＃34; some_str＆＃34; - ＆GT;确定
添加新值＆＃34; some_els＆＃34; - ＆GT; 未添加
添加新值＆＃34;一些＆＃34; - ＆GT;确定

调试这种现象我看到ConcurrentSkipListSet拒绝新的唯一字符串，这些字符串与集合中已存在的字符串具有相同的长度。

我就像Waaaattt？！？！？

这是一种意外行为，未在任何文档中提及。

这是JAVA ConcurrentSkipListSet实现中的错误吗？或者它是我做过的事情？

修改

感谢大家的快速回复！

我只想指出这种行为记录在JAVA SortedSet接口中（但不在ConcurrentSkipListSet中）：

* <p>Note that the ordering maintained by a sorted set (whether or not an
 * explicit comparator is provided) must be <i>consistent with equals</i> if
 * the sorted set is to correctly implement the <tt>Set</tt> interface.  (See
 * the <tt>Comparable</tt> interface or <tt>Comparator</tt> interface for a
 * precise definition of <i>consistent with equals</i>.)  This is so because
 * the <tt>Set</tt> interface is defined in terms of the <tt>equals</tt>
 * operation, but a sorted set performs all element comparisons using its
 * <tt>compareTo</tt> (or <tt>compare</tt>) method, so two elements that are
 * deemed equal by this method are, from the standpoint of the sorted set,
 * equal.  The behavior of a sorted set <i>is</i> well-defined even if its
 * ordering is inconsistent with equals; it just fails to obey the general
 * contract of the <tt>Set</tt> interface.

Answer 1

你有提供者的比较器返回相同长度的字符串相等，因此重复被忽略。

默认使用ConcurrentSkipListSet就像

Set<String> set = new ConcurrentSkipListSet<>(
        Comparator.comparing(s -> s));

或

Set<String> set = new ConcurrentSkipListSet<>(
        Comparator.naturalOrder());

设置比较器时，如果比较器指出两个对象相等，则替换默认比较器并且它不会默认返回默认行为。

解决这个问题的一种方法是按长度排序，然后按相同长度的内容排序。

Set<String> set = new ConcurrentSkipListSet<>(
        Comparator.comparing(String::length).reversed()
        .thenComparing(s -> s));

set.add("aa");
set.add("bb");
set.add("aaa");
set.add("ccc");
System.out.println(set);

打印

[aaa, ccc, aa, bb]

使用Comparator的ConcurrentSkipListSet不会添加新的唯一值

1 个答案: