Java：AffineTransform 旋转多边形，然后得到它的点

Question

我想画一个多边形并像这样用 AffineTransform 旋转它。

float theta = 90;
Polygon p = new Polygon(new int[]{0, 4, 4, 0}, new int[]{0, 0, 4, 4}, 4);
AffineTransform transform = new AffineTransform();
transform.rotate(Math.toRadians(theta), p.xpoints[0], p.ypoints[0]);
Shape transformed = transform.createTransformedShape(p);
g2.fill(transformed);

但是，我希望能够以与多边形相同的方式访问点 (transformed.xpoints[0])。一种看待这种情况的方法是将 Shape 转换为 Polygon——但据我所知这是不可能的。

最好的选择是什么？

附带说明：这是一个创建由 4 边多边形（矩形）组成的分形树的练习。我选择使用多边形来分别将分支锚定到左上角和右上角。如果这不必要地复杂，请告诉我。

Answer 1

您也可以使用AffineTransform 来转换各个点，如下所示：

Point2D[] srcPoints = new Point2D[] { new Point(0, 0), new Point(4, 0), new Point(4, 4), new Point(4, 0) };
Point2D[] destPoints = new Point2D[4];
transform.transform(srcPoints, 0, destPoints, 0, 4);

生成的destPoints 数组如下所示：

[Point2D.Float[-0.0, 0.0], Point2D.Float[-0.0, 4.0], Point2D.Float[-4.0, 4.0], Point2D.Float[-0.0, 4.0]]

Answer 2

您可以从createTransformedShape(...)返回的路径中获取坐标。

Path2D.Double transformed = (Path2D.Double) transform.createTransformedShape(p);

List<Double> xpointsList = new ArrayList<>();
List<Double> ypointsList = new ArrayList<>();
PathIterator pi = transformed.getPathIterator(null);

while(!pi.isDone()){
    double[] coords = new double[6];
    int type = pi.currentSegment(coords);
    if(type == PathIterator.SEG_MOVETO || type == PathIterator.SEG_LINETO){ // The only types we're interested in given the original shape
        xpointsList.add(coords[0]);
        ypointsList.add(coords[1]);
    }
    pi.next();
}

我有一段时间没有这样做了，可能有一种更简单的方法可以实现您想要的。此外，转换为Path2D.Double 并不理想。

Answer 3

已经有一段时间了，但我忍不住把它放在这里，以供以后遇到此问题的人使用。

那么我们只使用一点点数学而不是 AffineTransform 怎么样？

public class ROTOR {

/**
 * Method for rotating a Point object in the XY plane or in any plane
 * parallel to the XY plane. The Point object to be rotated and the one
 * about which it is rotating must have the same Z coordinates.
 *
 * @param p a Point object in the xy plane The z coordinates of both Point
 * objects must be either the same or must both be zero.
 * @param cen a Point object about which the rotation of the first Point
 * object will occur.
 * @param angle the angle of rotation
 * @return
 */
     public static Point planarXYRotate(Point p,Point cen,double angle){

    double sin = Math.sin(angle);
    double cos = Math.cos(angle);
    double X = p.x*cos-p.y*sin+cen.x*(1-cos)+cen.y*sin;
    double Y = p.x*sin+p.y*cos+cen.y*(1-cos)-cen.x*sin;
    return new Point( (int) X, (int) Y );


    }//end method



/**
 * 
 * @param polygon The polygon to rotate
 * @param origin The point about which to rotate it.
 * @param angle The angle of rotation.
 * @return a new Polygon rotated through the specified angle and about the said point.
 */
public static Polygon rotate(Polygon polygon , Point origin , double angle){

    Point cen = new Point(origin.x , origin.y);
    Polygon newPolygon = new Polygon();

    for(int i=0;i<polygon.npoints;i++){
        Point p = new Point( polygon.xpoints[i] , polygon.ypoints[i] );
        Point point = ROTOR.planarXYRotate(p, cen, angle);
        newPolygon.addPoint((int) point.x, (int) point.y);

    }

    return newPolygon;
}

}

然后，您可以轻松获得旋转的多边形并根据需要获取其点，使用：

Polygon rotatedPolygon = ROTOR.rotate(polygon, origin, angle);