Проверка точки внутри многоугольника в Android

Вы можете использовать мою простую библиотеку именно для этого: https://github.com/snatik/polygon-contains-point.

Подготовить полигон:

Polygon polygon = Polygon.Builder()
    .addVertex(new Point(1, 3))
    .addVertex(new Point(2, 8))
    .addVertex(new Point(5, 4))
    .addVertex(new Point(5, 9))
    .addVertex(new Point(7, 5))
    .addVertex(new Point(6, 1))
    .addVertex(new Point(3, 1))
    .build();

И проверьте, находится ли точка внутри многоугольника:

Point point = new Point(4.5f, 7);
boolean contains = polygon.contains(point);

Он работает с типами float и с полигонами, содержащими дырки :)

Вы можете использовать Google Maps PolyUtil:

import com.google.maps.android.PolyUtil;

boolean inside = PolyUtil.containsLocation(new LatLng(...), poly, true);

Вот как я это сделал на Android. Он основан на этой программе Java (алгоритм литья лучей): https://gis.stackexchange.com/questions/42879/check-if-lat-long-point-is-in-a-set-of-polygons-using-google-maps/46720#46720

    public boolean pointInPolygon(LatLng point, Polygon polygon) {
        // ray casting alogrithm http://rosettacode.org/wiki/Ray-casting_algorithm
        int crossings = 0;
        List<LatLng> path = polygon.getPoints();
        path.remove(path.size()-1); //remove the last point that is added automatically by getPoints()

        // for each edge
        for (int i=0; i < path.size(); i++) {
            LatLng a = path.get(i);
            int j = i + 1;
            //to close the last edge, you have to take the first point of your polygon
            if (j >= path.size()) {
                j = 0;
            }
            LatLng b = path.get(j);
            if (rayCrossesSegment(point, a, b)) {
                crossings++;
            }
        }

        // odd number of crossings?
        return (crossings % 2 == 1);
     }

    public boolean rayCrossesSegment(LatLng point, LatLng a,LatLng b) {
                // Ray Casting algorithm checks, for each segment, if the point is 1) to the left of the segment and 2) not above nor below the segment. If these two conditions are met, it returns true
                double px = point.longitude,
                py = point.latitude,
                ax = a.longitude,
                ay = a.latitude,
                bx = b.longitude,
                by = b.latitude;
            if (ay > by) {
                ax = b.longitude;
                ay = b.latitude;
                bx = a.longitude;
                by = a.latitude;
            }
            // alter longitude to cater for 180 degree crossings
            if (px < 0 || ax <0 || bx <0) { px += 360; ax+=360; bx+=360; }
            // if the point has the same latitude as a or b, increase slightly py
            if (py == ay || py == by) py += 0.00000001;


            // if the point is above, below or to the right of the segment, it returns false
            if ((py > by || py < ay) || (px > Math.max(ax, bx))){ 
                return false;
            }
            // if the point is not above, below or to the right and is to the left, return true
            else if (px < Math.min(ax, bx)){ 
                return true;
            }
            // if the two above conditions are not met, you have to compare the slope of segment [a,b] (the red one here) and segment [a,p] (the blue one here) to see if your point is to the left of segment [a,b] or not
            else {
                double red = (ax != bx) ? ((by - ay) / (bx - ax)) : Double.POSITIVE_INFINITY;
                double blue = (ax != px) ? ((py - ay) / (px - ax)) : Double.POSITIVE_INFINITY;
                return (blue >= red);
            }

     }

Извините @sromku, я спросил себя (я никогда не пользовался такими вещами)

Вот как я решил, если у кого-то есть такой же вопрос

Builder poly2 = new Polygon.Builder();
    for(int i = 0; i< xpoints.length;i++){
        poly2.addVertex(new Point(xpoints[i],ypoints[i]));
    }
    Polygon polygon2 = poly2.build();