blocks|key|223617|text|HashSet<String>+(或)任何Set实现都可以为您完成这项工作。Set不允许重复。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|223618|这是HashSet的javadoc。|223619|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/util/HashSet.html^0|0|F|L|3|12|3|0|A|7|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|H|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

<code>HashSet&lt;String&gt;</code> (or) any <code>Set</code> implementation may does the job for you. <code>Set</code> don't allow duplicates.

Here is <a href="http://docs.oracle.com/javase/6/docs/api/java/util/HashSet.html" rel="nofollow">javadoc</a> for HashSet.

blocks|key|223671|text|您可以只使用HashSet来维护唯一对象的集合。如果映射中的Integer值很重要，那么您可以改用映射的containsKey方法来测试您的键是否已经在映射中。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|223672|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/util/HashSet.html^0|6|7|U|7|1G|B|6|7|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@$9|U|A|V|1|W]]|E|$]]|$1|F|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

You could just use a <a href="http://docs.oracle.com/javase/6/docs/api/java/util/HashSet.html" rel="nofollow"><code>HashSet&lt;String&gt;</code></a> to maintain a collection of unique objects. If the <code>Integer</code> values in your map are important, then you can instead use the <code>containsKey</code> method of maps to test whether your key is already in the map.

blocks|key|441980|text|以new+HashSet<String>为例：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|441981|import+java.util.HashSet;
import+java.util.Set;

public+class+MainClass+{
++public+static+void+main(String+args[])+{
++++String[]+name1+=+{+"Amy",+"Jose",+"Jeremy",+"Alice",+"Patrick"+};

++++String[]+name2+=+{+"Alan",+"Amy",+"Jeremy",+"Helen",+"Alexi"+};

++++String[]+name3+=+{+"Adel",+"Aaron",+"Amy",+"James",+"Alice"+};

++++Set<String>+letter+=+new+HashSet<String>();

++++for+(int+i+=+0;+i+<+name1.length;+i%2B%2B)
++++++letter.add(name1[i]);

++++for+(int+j+=+0;+j+<+name2.length;+j%2B%2B)
++++++letter.add(name2[j]);

++++for+(int+k+=+0;+k+<+name3.length;+k%2B%2B)
++++++letter.add(name3[k]);

++++System.out.println(letter.size()+%2B+"+letters+must+be+sent+to:+"+%2B+letter);

++}
}|code-block|syntax|javascript|441982|entityMap^0|1|J|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Use <code>new HashSet&lt;String&gt;</code>
An example:

<pre><code>import java.util.HashSet;
import java.util.Set;

public class MainClass {
 public static void main(String args[]) {
 String[] name1 = { "Amy", "Jose", "Jeremy", "Alice", "Patrick" };

 String[] name2 = { "Alan", "Amy", "Jeremy", "Helen", "Alexi" };

 String[] name3 = { "Adel", "Aaron", "Amy", "James", "Alice" };

 Set&lt;String&gt; letter = new HashSet&lt;String&gt;();

 for (int i = 0; i &lt; name1.length; i++)
 letter.add(name1[i]);

 for (int j = 0; j &lt; name2.length; j++)
 letter.add(name2[j]);

 for (int k = 0; k &lt; name3.length; k++)
 letter.add(name3[k]);

 System.out.println(letter.size() + " letters must be sent to: " + letter);

 }
}
</code></pre>

blocks|key|224039|text|我想在这里为最初的海报澄清一些事情，其他人已经暗示了，但没有真正明确地说明。当你说你想要一个唯一的列表时，这就是有序集的定义。Set接口和List接口之间的其他一些关键区别在于，List允许您指定插入索引。所以，问题是，你真的需要列表界面(例如，为了与第三方库兼容，等等)，或者你能重新设计你的软件来使用Set界面吗？您还必须考虑如何处理接口。根据元素的索引查找元素很重要吗？你期望你的集合中有多少个元素？如果你要有很多元素，排序是重要的吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|224040|如果您确实需要一个只具有唯一约束的列表，可以使用Apache+Common+Utils类org.apache.commons.collections.list.SetUniqueList，它将为您提供列表接口和唯一约束。请注意，这会破坏List接口。但是，如果您需要按索引查找列表，您将从中获得更好的性能。如果您可以处理set接口，并且您有一个较小的数据集，那么LinkedHashSet可能是一个很好的选择。这取决于你的软件的设计和意图。|224041|同样，每个集合都有一定的优点和缺点。一些快速插入但缓慢读取，一些快速读取但缓慢插入，等等。花相当多的时间阅读集合文档以充分了解每个类和接口的详细信息是有意义的。|224042|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

I want to clarify some things here for the original poster which others have alluded to but haven't really explicitly stated. When you say that you want a Unique List, that is the very definition of an Ordered Set. Some other key differences between the Set Interface and the List interface are that List allows you to specify the insert index. So, the question is do you really need the List Interface (i.e. for compatibility with a 3rd party library, etc.), or can you redesign your software to use the Set interface? You also have to consider what you are doing with the interface. Is it important to find elements by their index? How many elements do you expect in your set? If you are going to have many elements, is ordering important? 

If you really need a List which just has a unique constraint, there is the Apache Common Utils class org.apache.commons.collections.list.SetUniqueList which will provide you with the List interface and the unique constraint. Mind you, this breaks the List interface though. You will, however, get better performance from this if you need to seek into the list by index. If you can deal with the Set interface, and you have a smaller data set, then LinkedHashSet might be a good way to go. It just depends on the design and intent of your software.

Again, there are certain advantages and disadvantages to each collection. Some fast inserts but slow reads, some have fast reads but slow inserts, etc. It makes sense to spend a fair amount of time with the collections documentation to fully learn about the finer details of each class and interface.

blocks|key|442231|text|我不知道这有多有效，但在一个简单的上下文中对我来说是有效的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|442232|List<int>+uniqueNumbers+=+new+ArrayList<>();

+++public+void+AddNumberToList(int+num)
++++{
++++++++if(!uniqueNumbers+.contains(num))+{
++++++++++++uniqueNumbers+.add(num);
++++++++}
++++}|code-block|syntax|javascript|442233|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I do not know how efficient this is, However worked for me in a simple context.

<pre><code>List&lt;int&gt; uniqueNumbers = new ArrayList&lt;&gt;();

 public void AddNumberToList(int num)
 {
 if(!uniqueNumbers .contains(num)) {
 uniqueNumbers .add(num);
 }
 }
</code></pre>

How to create a list of unique/distinct objects (no duplicates) in Java? 

Right now I am using <code>HashMap&lt;String, Integer&gt;</code> to do this as the key is overwritten and hence at the end we can get <code>HashMap.getKeySet()</code> which would be unique. But I am sure there should be a better way to do this as the value part is wasted here.

How to maintain a Unique List in Java?

Java

如何在Java中创建唯一/不同对象的列表(无重复)？现在我使用HashMap<String, Integer>来做这件事，因为键被覆盖了，因此在最后我们可以得到HashMap.getKeySet()，它将是唯一的。但我相信应该有更好的方法来做到这一点，因为价值部分在这里被浪费了。

问如何在Java中维护唯一列表？
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