private Map<Integer, Set<Integer>> nobreaks = new HashMap();
nobreaks.put(1, new HashSet(Arrays.asList(new int[]{2, 4, 5})));
System.out.println("expected size is 3: " +nobreaks.get(1).size());

输出为

expected size is 3: 1

将其更改为

nobreaks.put(1, new HashSet(Arrays.asList( 2, 4, 5 )));

输出为

expected size is 3: 3

在Java 10中：

String[] strs = {"A", "B"};
Set<String> set = Set.copyOf(Arrays.asList(strs));

Set.copyOf返回包含给定集合元素的不可修改集合。

给定的集合不能为null，并且不能包含任何null元素。

有时使用一些标准库会有很大帮助。尝试查看Apache Commons集合。在这种情况下，您的问题简单地转化为以下内容

String[] keys = {"blah", "blahblah"}
Set<String> myEmptySet = new HashSet<String>();
CollectionUtils.addAll(pythonKeywordSet, keys);

这里是CollectionsUtils javadoc

使用stanford-stagger-3.0.jar中的CollectionUtils或ArrayUtils

import static edu.stanford.nlp.util.ArrayUtils.asSet;
or 
import static edu.stanford.nlp.util.CollectionUtils.asSet;

  ...
String [] array = {"1", "q"};
Set<String> trackIds = asSet(array);

快速：您可以做到：

// Fixed-size list
List list = Arrays.asList(array);

// Growable list
list = new LinkedList(Arrays.asList(array));

// Duplicate elements are discarded
Set set = new HashSet(Arrays.asList(array));

并反转

// Create an array containing the elements in a list
Object[] objectArray = list.toArray();
MyClass[] array = (MyClass[])list.toArray(new MyClass[list.size()]);

// Create an array containing the elements in a set
objectArray = set.toArray();
array = (MyClass[])set.toArray(new MyClass[set.size()]);

new HashSet＜Object＞（Arrays.asList（Object[]a））；

但我认为这会更有效率：

final Set s = new HashSet<Object>();    
for (Object o : a) { s.add(o); }