blocks|key|904898|text|可以对ArrayList对象使用设置交集操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|904899|如下所示：|904900|List<Integer>+l1+=+new+ArrayList<Integer>();

l1.add(1);
l1.add(2);
l1.add(3);

List<Integer>+l2=+new+ArrayList<Integer>();
l2.add(4);
l2.add(2);
l2.add(3);

System.out.println("l1+==+"%2Bl1);
System.out.println("l2+==+"%2Bl2);

List<Integer>+l3+=+new+ArrayList<Integer>(l2);
l3.retainAll(l1);

++++System.out.println("l3+==+"%2Bl3);|code-block|syntax|javascript|904901|现在，l3应该只有l1和l2之间的公共元素。|904902|CONSOLE+OUTPUT
l1+==+[1,+2,+3]
l2+==+[4,+2,+3]
l3+==+[2,+3]|904903|entityMap^0|3|9|0|0|0|3|2|9|2|C|2|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

You can use set intersection operations with your <code>ArrayList</code> objects.

Something like this:

<pre><code>List&lt;Integer&gt; l1 = new ArrayList&lt;Integer&gt;();

l1.add(1);
l1.add(2);
l1.add(3);

List&lt;Integer&gt; l2= new ArrayList&lt;Integer&gt;();
l2.add(4);
l2.add(2);
l2.add(3);

System.out.println("l1 == "+l1);
System.out.println("l2 == "+l2);

List&lt;Integer&gt; l3 = new ArrayList&lt;Integer&gt;(l2);
l3.retainAll(l1);

 System.out.println("l3 == "+l3);
</code></pre>

Now, <code>l3</code> should have only common elements between <code>l1</code> and <code>l2</code>.

<pre><code>CONSOLE OUTPUT
l1 == [1, 2, 3]
l2 == [4, 2, 3]
l3 == [2, 3]
</code></pre>

blocks|key|688640|text|使用Collection#retainAll()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|688641|listA.retainAll(listB);
//+listA+now+contains+only+the+elements+which+are+also+contained+in+listB.|code-block|syntax|javascript|688642|如果您希望避免更改在listA中受到影响，则需要创建一个新的更改。|688643|List<Integer>+common+=+new+ArrayList<Integer>(listA);
common.retainAll(listB);
//+common+now+contains+only+the+elements+which+are+contained+in+listA+and+listB.|688644|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/6/docs/api/java/util/Collection.html#retainAll%2528java.util.Collection%2529^0|2|M|2|M|0|0|0|A|5|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

Use <a href="http://download.oracle.com/javase/6/docs/api/java/util/Collection.html#retainAll%28java.util.Collection%29" rel="noreferrer"><code>Collection#retainAll()</code></a>.

<pre><code>listA.retainAll(listB);
// listA now contains only the elements which are also contained in listB.
</code></pre>

If you want to avoid that changes are being affected in <code>listA</code>, then you need to create a new one.

<pre><code>List&lt;Integer&gt; common = new ArrayList&lt;Integer&gt;(listA);
common.retainAll(listB);
// common now contains only the elements which are contained in listA and listB.
</code></pre>

blocks|key|904958|text|为什么要重新发明轮子呢？使用Commons+Collections|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|904959|CollectionUtils.intersection(java.util.Collection+a,+java.util.Collection+b)|code-block|syntax|javascript|904960|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-collections/javadocs/api-4.2/org/apache/commons/collections4/CollectionUtils.html^0|E|J|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Why reinvent the wheel? Use <a href="http://commons.apache.org/proper/commons-collections/javadocs/api-4.2/org/apache/commons/collections4/CollectionUtils.html" rel="noreferrer">Commons Collections</a>:

<pre><code>CollectionUtils.intersection(java.util.Collection a, java.util.Collection b)
</code></pre>

blocks|key|688697|text|如果你想自己做的话..|type|unstyled|depth|inlineStyleRanges|entityRanges|data|688698|List<Integer>+commons+=+new+ArrayList<Integer>();

for+(Integer+igr+:+group1)+{
++++if+(group2.contains(igr))+{
++++++++commons.add(igr);
++++}
}

System.out.println("Common+elements+are+::+-");
for+(Integer+igr+:+commons)+{
++++System.out.println("+"%2Bigr);
}|code-block|syntax|javascript|688699|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|$]]

In case you want to do it yourself..

<pre><code>List&lt;Integer&gt; commons = new ArrayList&lt;Integer&gt;();

for (Integer igr : group1) {
 if (group2.contains(igr)) {
 commons.add(igr);
 }
}

System.out.println("Common elements are :: -");
for (Integer igr : commons) {
 System.out.println(" "+igr);
}
</code></pre>

blocks|key|905019|text|结合List.contains()使用Java8的Stream.filter()方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|905020|import+static+java.util.Arrays.asList;
import+static+java.util.stream.Collectors.toList;

/*+...+*/

List<Integer>+list1+=+asList(1,+2,+3,+4,+5);
List<Integer>+list2+=+asList(1,+3,+5,+7,+9);
++++
List<Integer>+common+=+list1.stream().filter(list2::contains).collect(toList());|code-block|syntax|javascript|905021|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/util/List.html#contains-java.lang.Object-|1|https://docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html#filter-java.util.function.Predicate-^0|2|F|P|F|2|F|0|P|F|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]|$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]]]

Using Java 8's <a href="https://docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html#filter-java.util.function.Predicate-" rel="nofollow noreferrer"><code>Stream.filter()</code></a> method in combination with <a href="https://docs.oracle.com/javase/8/docs/api/java/util/List.html#contains-java.lang.Object-" rel="nofollow noreferrer"><code>List.contains()</code></a>:
<pre><code>import static java.util.Arrays.asList;
import static java.util.stream.Collectors.toList;

/* ... */

List&lt;Integer&gt; list1 = asList(1, 2, 3, 4, 5);
List&lt;Integer&gt; list2 = asList(1, 3, 5, 7, 9);
 
List&lt;Integer&gt; common = list1.stream().filter(list2::contains).collect(toList());
</code></pre>

blocks|key|688748|text|++++//+Create+two+collections:
++++LinkedList<String>+listA+=++new+LinkedList<String>();
++++ArrayList<String>+listB+=++new+ArrayList<String>();

++++//+Add+some+elements+to+listA:
++++listA.add("A");
++++listA.add("B");
++++listA.add("C");
++++listA.add("D");

++++//+Add+some+elements+to+listB:
++++listB.add("A");
++++listB.add("B");
++++listB.add("C");

++++//+use+

++++List<String>+common+=+new+ArrayList<String>(listA);
++++//+use+common.retainAll

++++common.retainAll(listB);

++++System.out.println("The+common+collection+is+:+"+%2B+common);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|688749|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> // Create two collections:
 LinkedList&lt;String&gt; listA = new LinkedList&lt;String&gt;();
 ArrayList&lt;String&gt; listB = new ArrayList&lt;String&gt;();

 // Add some elements to listA:
 listA.add("A");
 listA.add("B");
 listA.add("C");
 listA.add("D");

 // Add some elements to listB:
 listB.add("A");
 listB.add("B");
 listB.add("C");

 // use 

 List&lt;String&gt; common = new ArrayList&lt;String&gt;(listA);
 // use common.retainAll

 common.retainAll(listB);

 System.out.println("The common collection is : " + common);
</code></pre>

blocks|key|688809|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|688810|📷|atomic|offset|length|688811|688812|List<String>+lista+=new+ArrayList<String>();
List<String>+listb+=new+ArrayList<String>();

lista.add("Isabella");
lista.add("Angelina");
lista.add("Pille");
lista.add("Hazem");

listb.add("Isabella");
listb.add("Angelina");
listb.add("Bianca");

//+Create+an+aplusb+list+which+will+contain+both+list+(list1+and+list2)+in+which+common+element+will+occur+twice+
List<String>+listapluslistb+=new+ArrayList<String>(lista);++++
listapluslistb.addAll(listb);
++++++++++++++++
//+Create+an+aunionb+set+which+will+contain+both+list+(list1+and+list2)+in+which+common+element+will+occur+once
Set<String>+listaunionlistb+=new+HashSet<String>(lista);
listaunionlistb.addAll(listb);
++++++++++++++++
for(String+s:listaunionlistb)
{
++++listapluslistb.remove(s);
}
System.out.println(listapluslistb);|code-block|syntax|javascript|688813|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/42e20bf1d5bc4783ffdfaf46be1b9761.png|imageAlt^0|0|0|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@$E|X|F|Y|1|Z]]|A|$]]|$1|G|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|11|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T|U|-4]]]]

<a href="https://i.stack.imgur.com/XdBPu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XdBPu.png" alt="enter image description here" /></a>
<pre><code>List&lt;String&gt; lista =new ArrayList&lt;String&gt;();
List&lt;String&gt; listb =new ArrayList&lt;String&gt;();

lista.add(&quot;Isabella&quot;);
lista.add(&quot;Angelina&quot;);
lista.add(&quot;Pille&quot;);
lista.add(&quot;Hazem&quot;);

listb.add(&quot;Isabella&quot;);
listb.add(&quot;Angelina&quot;);
listb.add(&quot;Bianca&quot;);

// Create an aplusb list which will contain both list (list1 and list2) in which common element will occur twice 
List&lt;String&gt; listapluslistb =new ArrayList&lt;String&gt;(lista); 
listapluslistb.addAll(listb);
 
// Create an aunionb set which will contain both list (list1 and list2) in which common element will occur once
Set&lt;String&gt; listaunionlistb =new HashSet&lt;String&gt;(lista);
listaunionlistb.addAll(listb);
 
for(String s:listaunionlistb)
{
 listapluslistb.remove(s);
}
System.out.println(listapluslistb);
</code></pre>

blocks|key|688861|text|您可以使用"retainAll“方法获取两个列表之间的公共元素。此方法将从它所应用的列表中删除所有不匹配的元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|688862|Ex.:+list.retainAll(list1);|code-block|syntax|javascript|688863|在这种情况下，从列表中删除所有不在list1中的元素，只保留在list和list1之间通用的元素。|688864|List<Integer>+list+=+new+ArrayList<>();
list.add(10);
list.add(13);
list.add(12);
list.add(11);

List<Integer>+list1+=+new+ArrayList<>();
list1.add(10);
list1.add(113);
list1.add(112);
list1.add(111);
//before+retainAll
System.out.println(list);
System.out.println(list1);
//applying+retainAll+on+list
list.retainAll(list1);
//After+retainAll
System.out.println("list::"%2Blist);
System.out.println("list1::"%2Blist1);|688865|输出：|688866|[10,+13,+12,+11]
[10,+113,+112,+111]
list::[10]
list1::[10,+113,+112,+111]|688867|注意:在list上应用retainAll后，list中包含了list和list1之间的公共元素。|688868|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

You can get the common elements between two lists using the method 
"retainAll". This method will remove all unmatched elements from the list to
which it applies.

<pre><code>Ex.: list.retainAll(list1);
</code></pre>

In this case from the list, all the elements which are not in list1 will be 
removed and only those will be remaining which are common between list and 
list1.

<pre><code>List&lt;Integer&gt; list = new ArrayList&lt;&gt;();
list.add(10);
list.add(13);
list.add(12);
list.add(11);

List&lt;Integer&gt; list1 = new ArrayList&lt;&gt;();
list1.add(10);
list1.add(113);
list1.add(112);
list1.add(111);
//before retainAll
System.out.println(list);
System.out.println(list1);
//applying retainAll on list
list.retainAll(list1);
//After retainAll
System.out.println("list::"+list);
System.out.println("list1::"+list1);
</code></pre>

Output: 

<pre><code>[10, 13, 12, 11]
[10, 113, 112, 111]
list::[10]
list1::[10, 113, 112, 111]
</code></pre>

NOTE: After retainAll applied on the list, the list contains common element between 
list and list1.

blocks|key|688879|text|public+<T>+List<T>+getIntersectOfCollections(Collection<T>+first,+Collection<T>+second)+{
++++++++return+first.stream()
++++++++++++++++.filter(second::contains)
++++++++++++++++.collect(Collectors.toList());
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|688880|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public &lt;T&gt; List&lt;T&gt; getIntersectOfCollections(Collection&lt;T&gt; first, Collection&lt;T&gt; second) {
 return first.stream()
 .filter(second::contains)
 .collect(Collectors.toList());
 }
</code></pre>

blocks|key|688914|text|考虑两个list+L1+ans+L2|type|unstyled|depth|inlineStyleRanges|entityRanges|data|688915|使用Java8我们可以很容易地找到它|688916|L1.stream().filter(L2::contains).collect(Collectors.toList())|offset|length|style|CODE|688917|entityMap^0|0|0|0|1P|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@$F|O|G|P|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

consider two list L1 ans L2

Using Java8 we can easily find it out

<code>L1.stream().filter(L2::contains).collect(Collectors.toList())</code>

blocks|key|905215|text|下面的代码删除列表中的公共元素|type|unstyled|depth|inlineStyleRanges|entityRanges|data|905216|List<String>+result+=++list1.stream().filter(item->+!list2.contains(item)).collect(Collectors.toList());|code-block|syntax|javascript|905217|检索公共元素|905218|List<String>+result+=+list1.stream()
++++++++++++++++.distinct()
++++++++++++++++.filter(list::contains)
++++++++++++++++.collect(Collectors.toList());|905219|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Below code
Remove common elements in the list
<pre><code>List&lt;String&gt; result = list1.stream().filter(item-&gt; !list2.contains(item)).collect(Collectors.toList());
</code></pre>
Retrieve common elements
<pre><code>List&lt;String&gt; result = list1.stream()
 .distinct()
 .filter(list::contains)
 .collect(Collectors.toList());
</code></pre>

blocks|key|688994|text|这个问题涉及三个方面，许多建议都建议使用retainAll。我认为必须指出的是，随着列表大小的增长，retainAll似乎变得越来越低效。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|688995|在我的测试中，我发现转换为集合和循环大约比使用retainAll处理包含1000个项目的列表快60倍|688996|++List<Integer>+common(List<Integer>+biggerList,+List<Integer>+smallerList)+{
++++Set<Integer>+set1+=+new+HashSet<>(biggerList);
++++List<Integer>+result+=+new+ArrayList<>(smallerList.size());
++++for+(Integer+i+:+smallerList)+{
++++++if+(set1.contains(i))+{
++++++++result.add(i);
++++++}
++++}
++++return+result;
++}|code-block|syntax|javascript|688997|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The question talks about three items and many of the suggestions suggest using retainAll. I think it must be stated that as the size of the lists grown retainAll seems to become more inefficient.
In my tests I found that converting to Sets and looping is around 60 times faster than using retainAll for Lists with 1000s of items
<pre><code> List&lt;Integer&gt; common(List&lt;Integer&gt; biggerList, List&lt;Integer&gt; smallerList) {
 Set&lt;Integer&gt; set1 = new HashSet&lt;&gt;(biggerList);
 List&lt;Integer&gt; result = new ArrayList&lt;&gt;(smallerList.size());
 for (Integer i : smallerList) {
 if (set1.contains(i)) {
 result.add(i);
 }
 }
 return result;
 }
</code></pre>

I have two <code>ArrayList</code> objects with three integers each. I want to find a way to return the common elements of the two lists. Has anybody an idea how I can achieve this?

Common elements in two lists

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有两个ArrayList对象，每个对象都有三个整数。我想找到一种方法来返回这两个列表的公共元素。有谁知道我怎样才能做到这一点吗？

问两个列表中的公共元素
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问两个列表中的公共元素
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