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

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

消息编辑(完全重写)当顺序无关紧要时，就是这样。两者都应该给出Log(n) -看看其中一个是否比另一个快5%以上是有用的。HashSet可以在循环中给出O(1)测试，应该可以揭示它是否正确。

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

import java.util.HashSet;
import java.util.Set;
import java.util.TreeSet;

public class HashTreeSetCompare {

    //It is generally faster to add elements to the HashSet and then
    //convert the collection to a TreeSet for a duplicate-free sorted
    //Traversal.

    //really? 
    O(Hash + tree set) > O(tree set) ??
    Really???? Why?



    public static void main(String args[]) {

        int size = 80000;
        useHashThenTreeSet(size);
        useTreeSetOnly(size);

    }

    private static void useTreeSetOnly(int size) {

        System.out.println("useTreeSetOnly: ");
        long start = System.currentTimeMillis();
        Set<String> sortedSet = new TreeSet<String>();

        for (int i = 0; i < size; i++) {
            sortedSet.add(i + "");
        }

        //System.out.println(sortedSet);
        long end = System.currentTimeMillis();

        System.out.println("useTreeSetOnly: " + (end - start));
    }

    private static void useHashThenTreeSet(int size) {

        System.out.println("useHashThenTreeSet: ");
        long start = System.currentTimeMillis();
        Set<String> set = new HashSet<String>();

        for (int i = 0; i < size; i++) {
            set.add(i + "");
        }

        Set<String> sortedSet = new TreeSet<String>(set);
        //System.out.println(sortedSet);
        long end = System.currentTimeMillis();

        System.out.println("useHashThenTreeSet: " + (end - start));
    }
}