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    <p><a href="http://java.sun.com/javase/6/docs/api/java/util/Collections.html" rel="noreferrer"><code>Collections.binarySearch()</code></a>.</p> <p>Assuming the cats are sorted by birthday, this will give the index of one of the cats with the correct birthday. From there, you can iterate backwards and forwards until you hit one with a different birthday.</p> <p>If the list is long and/or not many cats share a birthday, this should be a significant win over straight iteration.</p> <p>Here's the sort of code I'm thinking of. Note that I'm assuming a <a href="http://java.sun.com/javase/6/docs/api/java/util/RandomAccess.html" rel="noreferrer">random-access</a> list; for a linked list, you're pretty much stuck with iteration. (Thanks to fred-o for pointing this out in the comments.)</p> <pre><code>List&lt;Cat&gt; cats = ...; // sorted by birthday List&lt;Cat&gt; catsWithSameBirthday = new ArrayList&lt;Cat&gt;(); Cat key = new Cat(); key.setBirthday(...); final int index = Collections.binarySearch(cats, key); if (index &lt; 0) return catsWithSameBirthday; catsWithSameBirthday.add(cats.get(index)); // go backwards for (int i = index-1; i &gt; 0; i--) { if (cats.get(tmpIndex).getBirthday().equals(key.getBirthday())) catsWithSameBirthday.add(cats.get(tmpIndex)); else break; } // go forwards for (int i = index+1; i &lt; cats.size(); i++) { if (cats.get(tmpIndex).getBirthday().equals(key.getBirthday())) catsWithSameBirthday.add(cats.get(tmpIndex)); else break; } return catsWithSameBirthday; </code></pre>
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    1. USMichael Myers
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    1. COCollections.binarySearch() return a single element and makes no guarantees about elements which are considered identical.
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      1. USJake
    2. COMaybe that will teach me to read the question before answering. :)
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    3. COAlso, Collections.binarySearch() is only efficient for random access lists.
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      1. USfred-o
 

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