Intersection of 2 segments

Question

public Segment(Point start, Point end) {
if (start == null || end == null)
throw new IllegalArgumentException("Arguments can't be null");
if (start.equals(end))
throw new IllegalArgumentException("The points must differ");
this.start = start;
this.end = end;
}
public Point intersection(Segment another) {

div = dy2 * dx1 - dx2 * dy1;
if (Math.abs(div) < 1.0e-10)
return null;

}

问题是当我想打印:

Segment first = new Segment(new Point(0, 0), new Point(0, 0));
Segment second = new Segment(new Point(0, 0), new Point(0, 0));
Point intersection = first.intersection(second);
System.out.println(intersection.getX());
System.out.println(intersection.getY());

当我引入所有坐标0

时出现在NullPointerException。如何解决这个问题?如果有人能帮忙，请帮忙。我不知道如何捕捉这个异常或如何解决。

Answer 1

public static Point SegIntersection(double x11, double y11, double x12, double y12,
double x21, double y21, double x22, double y22)
{
double dx1 = x12-x11;
double dy1 = y12-y11;
double dx2 = x22-x21;
double dy2 = y22-y21;
double dxx = x11-x21;
double dyy = y11-y21;
double div, t, s;

div = dy2*dx1-dx2*dy1;
if (Math.abs(div) < 1.0e-10)  //better to compare abs(div) with small Eps
return null;  //collinear case

t = (dx1*dyy-dy1*dxx) / div;
if (t < 0 || t > 1.0)
return null; //intersection outside the first segment
s = (dx2*dyy-dy2*dxx) / div;
if (s < 0 || s > 1.0)  
return null;  //intersection outside the second segment
return new Point(x11 + s * dx1, y11 + s * dy1);
}

快速测试

注意函数不检查共线段的重叠，因为问题语句忽略了这种情况。