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    <p>Firstly, I'd use <a href="http://joda-time.sf.net" rel="nofollow">Joda Time</a> and its <code>LocalTime</code> type to represent the times. You can then do:</p> <pre><code>static boolean isInRange(LocalTime start, LocalTime end, LocalTime value) { // Avoid equality problems - see description if (value.equals(start) || value.equals(end)) { return true; } if (start.compareTo(end) &gt; 0) { return !isInRange(end, start, value); } return start.compareTo(value) &lt; 0 &amp;&amp; value.compareTo(end) &lt; 0; } </code></pre> <p>In other words, you'd generally only end up doing the final comparison with <code>value</code> when you'd got a start/end range where <code>start &lt;= end</code> - but if you start off with the other way round, then you just invert the result of doing the reverse comparison.</p> <p>In other words, something is in the range (10pm, 2am) if it's <em>not</em> in the range (2am, 10pm).</p> <p>Alternatively, you could avoid the recursive call:</p> <pre><code>static boolean isInRange(LocalTime start, LocalTime end, LocalTime value) { return start.compareTo(end) &gt; 0 // Reversed: if it's earlier than the end point, or later than the start ? value.compareTo(end) &lt;= 0 || start.compareTo(value) &lt;= 0 : start.compareTo(value) &lt;= 0 &amp;&amp; value.compareTo(end) &lt;= 0; } </code></pre> <p>The tricky bit is equality. The code above assumes that the start/end points are both inclusive, either way round. If you want anything else, you'll need to think very carefully about every situation.</p>
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    1. POIs hour in time range that spans 2 days?
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    1. USJon Skeet
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    1. COBut, how does this solve time which spans over day? When input is just 2, 6 etc.,
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    2. CO@Nambari: Read the expanded answer.
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    3. COI would also like to thank you for the reply. It is quite extensive that even i understand the idea and logical behind the code.
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      1. UST. Akhayo
 

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