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    <p>I came up against this same problem a little while ago, and after some searching, I found this to be the best solution.</p> <p>Java has a <code>Polygon</code> class with a <code>contains()</code> method that would make things really simple. Unfortunately, the <code>java.awt.Polygon</code>class is not supported in Android. However, I was able to find someone who wrote an <a href="http://www.anddev.org/other-coding-problems-f5/using-java-awt-polygon-in-android-t6521.html#wrap" rel="noreferrer">equivalent class</a>.</p> <p>I don't think you can get the individual points that make up the path from the Android <code>Path</code> class, so you will have to store the data in a different way.</p> <p>The class uses a Crossing Number algorithm to determine whether or not the point is inside of the given list of points.</p> <pre><code>/** * Minimum Polygon class for Android. */ public class Polygon { // Polygon coodinates. private int[] polyY, polyX; // Number of sides in the polygon. private int polySides; /** * Default constructor. * @param px Polygon y coods. * @param py Polygon x coods. * @param ps Polygon sides count. */ public Polygon( int[] px, int[] py, int ps ) { polyX = px; polyY = py; polySides = ps; } /** * Checks if the Polygon contains a point. * @see "http://alienryderflex.com/polygon/" * @param x Point horizontal pos. * @param y Point vertical pos. * @return Point is in Poly flag. */ public boolean contains( int x, int y ) { boolean oddTransitions = false; for( int i = 0, j = polySides -1; i &lt; polySides; j = i++ ) { if( ( polyY[ i ] &lt; y &amp;&amp; polyY[ j ] &gt;= y ) || ( polyY[ j ] &lt; y &amp;&amp; polyY[ i ] &gt;= y ) ) { if( polyX[ i ] + ( y - polyY[ i ] ) / ( polyY[ j ] - polyY[ i ] ) * ( polyX[ j ] - polyX[ i ] ) &lt; x ) { oddTransitions = !oddTransitions; } } } return oddTransitions; } } </code></pre>
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    1. COWhat would one do if they are uncertain of the number of sides the polygon has?
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    2. COThe number of sides can easily be inferred from the number of points. A rectangle has 4 points and 4 sides. A triangle has 3 points and 3 sides. Polygons drawn by continuous paths have the same number of sides and points, so if your path has "n"points, it also has "n" sides.
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    3. COThanks, it works :)
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