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    <p>Are these really <a href="http://java.sun.com/javase/6/docs/api/java/util/List.html" rel="noreferrer">lists</a> (ordered, with duplicates), or are they <a href="http://java.sun.com/javase/6/docs/api/java/util/Set.html" rel="noreferrer">sets</a> (unordered, no duplicates)?</p> <p>Because if it's the latter, then you can use, say, a <a href="http://java.sun.com/javase/6/docs/api/java/util/HashSet.html" rel="noreferrer"><code>java.util.HashSet&lt;E&gt;</code></a> and do this in expected linear time using the convenient <a href="http://java.sun.com/javase/6/docs/api/java/util/Set.html#retainAll(java.util.Collection)" rel="noreferrer"><code>retainAll</code></a>.</p> <pre><code> List&lt;String&gt; list1 = Arrays.asList( "milan", "milan", "iga", "dingo", "milan" ); List&lt;String&gt; list2 = Arrays.asList( "hafil", "milan", "dingo", "meat" ); // intersection as set Set&lt;String&gt; intersect = new HashSet&lt;String&gt;(list1); intersect.retainAll(list2); System.out.println(intersect.size()); // prints "2" System.out.println(intersect); // prints "[milan, dingo]" // intersection/union as list List&lt;String&gt; intersectList = new ArrayList&lt;String&gt;(); intersectList.addAll(list1); intersectList.addAll(list2); intersectList.retainAll(intersect); System.out.println(intersectList); // prints "[milan, milan, dingo, milan, milan, dingo]" // original lists are structurally unmodified System.out.println(list1); // prints "[milan, milan, iga, dingo, milan]" System.out.println(list2); // prints "[hafil, milan, dingo, meat]" </code></pre>
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    1. COwell I really don't know which data structure it should be. It has duplicates. Now you can see updated question
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    2. COWill it remove the repeated values from data set? coz I don't want to loss any value :(
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    3. CO@agazerboy: I've tried to address both questions. Feel free to ask for more clarifications.
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