用矩形检测线的交点

Question

我想检测带有随机矩形的随机线的交点。我的Line类获得4个坐标，并从中创建一条线。我的交集函数获取2条线并返回交点，如果没有点，则返回null。

我有2个功能：一个找到最接近的交点并返回它（我认为这里无关紧要）：位于Line.java

public Point closestIntersectionToStartOfLine(Rectangle rect){
    Line[] rectLines = new Line[4];
    Point[] IntersectionPoints = new Point[4];
    double minDistance=0;
    //up
    rectLines[0] = new Line(rect.getUpperLeft().getX(), rect.getUpperLeft().getY(), rect.getWidth()+rect.getUpperLeft().getX(), rect.getUpperLeft().getY());
    //right
    rectLines[1] = new Line(rect.getWidth()+rect.getUpperLeft().getX(), rect.getUpperLeft().getY(), rect.getUpperLeft().getX()+rect.getWidth(), rect.getHeight()+rect.getUpperLeft().getY());
    //down
    rectLines[2] = new Line(rect.getUpperLeft().getX()+rect.getWidth(), rect.getHeight()+rect.getUpperLeft().getY(), rect.getUpperLeft().getX(), rect.getUpperLeft().getY()+rect.getHeight());
    //left
    rectLines[3] = new Line(rect.getUpperLeft().getX(), rect.getUpperLeft().getY()+rect.getHeight(), rect.getUpperLeft().getX(), rect.getUpperLeft().getY());

    for (int i=0; i<4; i++) {
        if (intersectionWith(rectLines[i]) != null)
            IntersectionPoints[i] = intersectionWith(rectLines[i]);
    }
    for (int i=0; i<IntersectionPoints.length; i++) {
        if (i == 0)
            minDistance = rect.getUpperLeft().distance(IntersectionPoints[i]);
        else {
            if (rect.getUpperLeft().distance(IntersectionPoints[i]) < minDistance)
                Point.closestIntersectionToStartOfLine = IntersectionPoints[i];
        }
    }
    return Point.closestIntersectionToStartOfLine;
}

并且此函数将给定线的所有交点保留在列表中。位于Rectangle.java

 // Return a (possibly empty) List of intersection points
// with the specified line.
public java.util.List<Point> intersectionPoints(Line line) {
    List<Point> intersectionPointsList = new ArrayList<Point>();
    Line[] rectangleSides = new Line[4];
    Point[] corners = new Point[4];
    rectangleCorners(corners);
    recatngleSides(rectangleSides, corners);
    for (int i = 0; i < 4; i++) {
        Point p = line.intersectionWith(rectangleSides[i]);
        if (p != null && !intersectionPointsList.contains(p)) {
            intersectionPointsList.add(p);
        }
    }
    return intersectionPointsList;
}
public void rectangleCorners(Point[] corners) {
    double width = this.getWidth();
    double height = this.getHeight();
    double x = this.upperLeft.getX();
    double y = this.upperLeft.getY();
    //upper left
    corners[0] = this.getUpperLeft();
    //upper right
    corners[1] = new Point(x + width, y);
    //down right
    corners[2] = new Point(x + width, y + height);
    //down left
    corners[3] = new Point(x, y + height);
}

/**
 * recatngleSides.
 * Saves the lines in the 4 rectangle edges.
 *
 * @param sides  - an empty array of lines
 * @param corners - an array of edge points.
 */
public void recatngleSides(Line[] sides, Point[] corners) {
    // up
    sides[0] = new Line(corners[0], corners[1]);
    // right
    sides[1] = new Line(corners[1], corners[2]);
    // down
    sides[2] = new Line(corners[2], corners[3]);
    // left
    sides[3] = new Line(corners[3], corners[0]);
}

}

我不知道为什么，但是它可以将随机点识别为相交点，如图所示。

Answer 1

因为，如果您查看“随机”点的位置，那么实际上，如果将矩形的边延伸到无穷大，它们将位于不相交的矩形的边。您实际上需要检查与线段的相交点是否为矩形。