明明可以吃橘子，为什么要吃苹果?

Seriously guys and gals - if your collection is large, read and written to gazillions of times, and you're paying for CPU cycles, then the choice of the collection is relevant ONLY if you NEED it to perform better. However, in most cases, this doesn't really matter - a few milliseconds here and there go unnoticed in human terms. If it really mattered that much, why aren't you writing code in assembler or C? [cue another discussion]. So the point is if you're happy using whatever collection you chose, and it solves your problem [even if it's not specifically the best type of collection for the task] knock yourself out. The software is malleable. Optimise your code where necessary. Uncle Bob says Premature Optimisation is the root of all evil. Uncle Bob says so

HashSet是O(1)来访问元素，所以这当然很重要。但是保持集合中对象的顺序是不可能的。

如果维护顺序(根据值而不是插入顺序)对您很重要，TreeSet是有用的。但是，正如您所注意到的，您正在以顺序换取访问元素的更慢时间:基本操作为O(log n)。

来自TreeSet的javadocs:

该实现为基本操作(添加、删除和包含)提供了log(n)的时间成本。

基于@shevchyk在地图上可爱的视觉回答，以下是我的看法:

╔══════════════╦═════════════════════╦═══════════════════╦═════════════════════╗
║   Property   ║       HashSet       ║      TreeSet      ║     LinkedHashSet   ║
╠══════════════╬═════════════════════╬═══════════════════╬═════════════════════╣
║              ║  no guarantee order ║ sorted according  ║                     ║
║   Order      ║ will remain constant║ to the natural    ║    insertion-order  ║
║              ║      over time      ║    ordering       ║                     ║
╠══════════════╬═════════════════════╬═══════════════════╬═════════════════════╣
║ Add/remove   ║        O(1)         ║     O(log(n))     ║        O(1)         ║
╠══════════════╬═════════════════════╬═══════════════════╬═════════════════════╣
║              ║                     ║   NavigableSet    ║                     ║
║  Interfaces  ║         Set         ║       Set         ║         Set         ║
║              ║                     ║    SortedSet      ║                     ║
╠══════════════╬═════════════════════╬═══════════════════╬═════════════════════╣
║              ║                     ║    not allowed    ║                     ║
║  Null values ║       allowed       ║ 1st element only  ║      allowed        ║
║              ║                     ║     in Java 7     ║                     ║
╠══════════════╬═════════════════════╩═══════════════════╩═════════════════════╣
║              ║   Fail-fast behavior of an iterator cannot be guaranteed      ║
║   Fail-fast  ║ impossible to make any hard guarantees in the presence of     ║
║   behavior   ║           unsynchronized concurrent modification              ║
╠══════════════╬═══════════════════════════════════════════════════════════════╣
║      Is      ║                                                               ║
║ synchronized ║              implementation is not synchronized               ║
╚══════════════╩═══════════════════════════════════════════════════════════════╝

1.HashSet允许空对象。

2.树集不允许空对象。如果你试图添加空值，它将抛出一个NullPointerException。

3.HashSet比TreeSet快得多。

e.g.

 TreeSet<String> ts = new TreeSet<String>();
 ts.add(null); // throws NullPointerException

 HashSet<String> hs = new HashSet<String>();
 hs.add(null); // runs fine

如果您没有插入足够多的元素导致频繁重散列(或冲突，如果您的HashSet不能调整大小)，那么HashSet当然可以为您提供常量时间访问的好处。但是对于有大量增长或收缩的集合，使用Treesets实际上可能会获得更好的性能，这取决于实现。

如果我没记错的话，平摊时间可以接近于一个功能性红黑树的O(1)。冈崎的书会有比我更好的解释。(或参阅他的出版物列表)

即使在11年后，也没有人想到提到一个非常重要的区别。

你认为如果HashSet等于TreeSet，那么反过来也成立吗?看看这段代码:

TreeSet<String> treeSet = new TreeSet<>(String.CASE_INSENSITIVE_ORDER);
HashSet<String> hashSet = new HashSet<>();
treeSet.add("a");
hashSet.add("A");
System.out.println(hashSet.equals(treeSet));
System.out.println(treeSet.equals(hashSet));

尝试猜测输出，然后徘徊在代码片段下面，看看真正的输出是什么。准备好了吗?给你:

假 真正的

没错，如果比较器与等号不一致，它们就不具有等价关系。原因是TreeSet使用比较器来确定等价性，而HashSet使用等号。在内部，它们使用HashMap和TreeMap，所以你应该预料到上述map也会有这种行为。

最初的回答