我一直很喜欢树,O(n*log(n))和它们的整洁。然而,我所认识的每个软件工程师都尖锐地问过我为什么要使用TreeSet。从CS的背景来看,我不认为你使用什么很重要,我也不关心在哈希函数和桶(在Java的情况下)上搞得一团糟。
在哪些情况下,我应该在树集上使用HashSet ?
我一直很喜欢树,O(n*log(n))和它们的整洁。然而,我所认识的每个软件工程师都尖锐地问过我为什么要使用TreeSet。从CS的背景来看,我不认为你使用什么很重要,我也不关心在哈希函数和桶(在Java的情况下)上搞得一团糟。
在哪些情况下,我应该在树集上使用HashSet ?
当前回答
A lot of answers have been given, based on technical considerations, especially around performance. According to me, choice between TreeSet and HashSet matters. But I would rather say the choice should be driven by conceptual considerations first. If, for the objects your need to manipulate, a natural ordering does not make sense, then do not use TreeSet. It is a sorted set, since it implements SortedSet. So it means you need to override function compareTo, which should be consistent with what returns function equals. For example if you have a set of objects of a class called Student, then I do not think a TreeSet would make sense, since there is no natural ordering between students. You can order them by their average grade, okay, but this is not a "natural ordering". Function compareTo would return 0 not only when two objects represent the same student, but also when two different students have the same grade. For the second case, equals would return false (unless you decide to make the latter return true when two different students have the same grade, which would make equals function have a misleading meaning, not to say a wrong meaning.) Please note this consistency between equals and compareTo is optional, but strongly recommended. Otherwise the contract of interface Set is broken, making your code misleading to other people, thus also possibly leading to unexpected behavior.
这个链接可能是关于这个问题的一个很好的信息来源。
其他回答
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 ║
╚══════════════╩═══════════════════════════════════════════════════════════════╝
大多数人使用HashSet的原因是操作(平均)是O(1)而不是O(log n)。如果集合包含标准项,你就不会像以前那样“乱搞哈希函数”。如果集合包含自定义类,则必须实现hashCode才能使用HashSet(尽管Effective Java演示了如何使用),但如果使用TreeSet,则必须使其具有可比性或提供比较器。如果类没有特定的顺序,这可能是一个问题。
我有时会使用TreeSet(或者实际上是TreeMap)用于非常小的集合/映射(< 10项),尽管我没有检查这样做是否有任何真正的好处。对于大型机组,差异可能相当大。
现在,如果您需要排序,那么TreeSet是合适的,尽管即使如此,如果更新频繁,对排序结果的需求并不频繁,有时将内容复制到列表或数组中并对它们排序会更快。
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
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));
}
}
HashSet比TreeSet快得多(对于添加、删除和包含等大多数操作,HashSet是常量时间,而不是日志时间),但不像TreeSet那样提供排序保证。
HashSet
该类为基本操作(添加、删除、包含和大小)提供恒定的时间性能。 它不能保证元素的顺序随时间保持不变 迭代性能取决于初始容量和HashSet的负载因子。 接受默认的负载因子是相当安全的,但您可能希望指定的初始容量大约是您期望该集增长的两倍。
TreeSet
保证基本操作(添加、删除和包含)的时间成本为log(n) 确保set的元素将被排序(升序,自然或由你通过它的构造函数指定)(实现SortedSet) 不为迭代性能提供任何调优参数 提供了一些方便的方法来处理有序集,如first(), last(), headSet()和tailSet()等
重要的几点:
Both guarantee duplicate-free collection of elements It is generally faster to add elements to the HashSet and then convert the collection to a TreeSet for a duplicate-free sorted traversal. None of these implementations are synchronized. That is if multiple threads access a set concurrently, and at least one of the threads modifies the set, it must be synchronized externally. LinkedHashSet is in some sense intermediate between HashSet and TreeSet. Implemented as a hash table with a linked list running through it, however,it provides insertion-ordered iteration which is not same as sorted traversal guaranteed by TreeSet.
因此,使用方法的选择完全取决于您的需要,但我认为,即使您需要一个有序的集合,那么您仍然应该使用HashSet来创建Set,然后将其转换为TreeSet。
例如:SortedSet<String> s = new TreeSet<String>(hashSet);