Java-正确的直线相交检查
我需要确定两条直线段是否相交，但使用line2D.linesIntersect方法存在问题。即使这些行只共享一个端点，该方法也会返回一个真正的结果，如下所示：Java-正确的直线相交检查,java,line,point,Java,Line,Point,我需要确定两条直线段是否相交，但使用line2D.linesIntersect方法存在问题。即使这些行只共享一个端点，该方法也会返回一个真正的结果，如下所示： Point2D.Double temp1 = new Point2D.Double(0, 0); Point2D.Double temp2 = new Point2D.Double(0, 1); Point2D.Double temp3 = new Point2D.Double(1, 0); if(Line2D.linesInterse
Point2D.Double temp1 = new Point2D.Double(0, 0);
Point2D.Double temp2 = new Point2D.Double(0, 1);
Point2D.Double temp3 = new Point2D.Double(1, 0);
if(Line2D.linesIntersect(temp1.x, temp1.y, temp2.x, temp2.y, temp1.x, temp1.y, temp3.x, temp3.y){
System.out.println("Lines share an endpoint.");
}
else{
System.out.println("Lines don't intersect.");
}
在这种情况下，我将始终收到“lines share a endpoint”消息。当然，在某些直线共享端点的情况下，它们可能会无限多次相交（（0,0）到（0,1）被（0,0）到（0,2）相交），这显然会返回真实结果。但是，在其他情况下，如果只共享端点，而不发生其他交集，程序将无法正常工作。有什么方法可以避免这个问题吗？这是我用我的基本数学知识可以得出的答案。希望对你有帮助。给定4个点，它会告诉您两条线（从这四个点开始）是否相交、是否共享一个端点或两者都不共享
//starting point of line 1
Point2D.Double temp1 = new Point2D.Double(0 , 1);
//ending point of line 1
Point2D.Double temp2 = new Point2D.Double(0, -1);
//starting point of line 2
Point2D.Double temp3 = new Point2D.Double(-1, 0);
//ending point of line 2
Point2D.Double temp4 = new Point2D.Double(1, 0);
//determine if the lines intersect
boolean intersects = Line2D.linesIntersect(temp1.x, temp1.y, temp2.x, temp2.y, temp3.x, temp3.y, temp4.x, temp4.y);
//determines if the lines share an endpoint
boolean shareAnyPoint = shareAnyPoint(temp1, temp2, temp3, temp4);
if (intersects && shareAnyPoint) {
System.out.println("Lines share an endpoint.");
} else if (intersects && !shareAnyPoint) {
System.out.println("Lines intersect.");
} else {
System.out.println("Lines neither intersect nor share a share an endpoint.");
}
public static boolean shareAnyPoint(Point2D.Double A, Point2D.Double B, Point2D.Double C, Point2D.Double D) {
if (isPointOnTheLine(A, B, C)) return true;
else if (isPointOnTheLine(A, B, D)) return true;
else if (isPointOnTheLine(C, D, A)) return true;
else if (isPointOnTheLine(C, D, B)) return true;
else return false;
}
试试看，让我知道结果 如果不局限于Point2D和Line2D对象，则可以使用（Java拓扑套件）
- 将线创建为线字符串对象
- 使用该方法
LineString lineA = new GeometryFactory().createLineString(new Coordinate[]{new Coordinate(0, 0), new Coordinate(0, 10)});
LineString lineB = new GeometryFactory().createLineString(new Coordinate[]{new Coordinate(-5,5), new Coordinate(5,5)});
boolean intersect = lineA.intersects(lineB);
我在网络上找到了一个解决方案，并对其进行了调整和手动测试。 它是自包含的，只需导入java.awt.geom.Point2D即可
/**
* Test if the first point lies in the bounding box denote by the other two
* points.
*/
public static boolean isBetween(Point2D pToTest, Point2D p1, Point2D p2) {
return isBetween(pToTest.getX(), pToTest.getY(), p1.getX(), p1.getY(), p2.getX(), p2.getY());
}
/**
* Called {@link #isBetween(Point2D, Point2D, Point2D)} passing those points
* coordinates.
*/
public static boolean isBetween(double pxToTest, double pyToTest, double px1, double py1, double px2, double py2) {
double w, h;
// taking inspiration from Rectangle's class
w = px1 - px2;
if (w < 0)
w = -w;
h = py1 - py2;
if (h < 0)
h = -h;
// use p1 as the left-top corner of the rectangle
// (the left-top corner is considered as the origin (0;0))
if (px1 > px2)
px1 = px2;//
if (py1 > py2)
py1 = py2;//
if (pxToTest < px1 || pyToTest < py1) {
return false;
}
w += px1;
h += py1;
// overflow || intersect
return ((w < px1 || w > pxToTest) && (h < py1 || h > pyToTest));
}
public static double slope(double xa, double ya, double xb, double yb) {
if (xb == xa)
return Double.POSITIVE_INFINITY;
return (yb == ya) ? 0.0 : ((yb - ya) / (xb - xa));
}
/**
* Called by {@link #areLinesIntersecting(Point2D, Point2D, Point2D, Point2D)}
* by providing each of those point's coordinates.
*/
public static Point2D areLinesIntersecting(double pxStart1, double pyStart1, double pxEnd1, double pyEnd1,
double pxStart2, double pyStart2, double pxEnd2, double pyEnd2) {
double slope_ab, slope_cd, numerator, denominator, q_ab, q_cd, x, y;
Point2D p;
slope_ab = slope(pxStart1, pyStart1, pxEnd1, pyEnd1);
slope_cd = slope(pxStart2, pyStart2, pxEnd2, pyEnd2);
q_ab = (pyStart1 - slope_ab * pxStart1);
q_cd = (pyStart2 - slope_cd * pxStart2);
if ((slope_ab == slope_cd) // parallel?
&& (
// overlapping?
((slope_ab == Double.POSITIVE_INFINITY || slope_ab == Double.NaN) && pxStart1 == pxStart2) //
|| //
// overlapping?
(slope_ab == 0.0 && pyStart1 == pyStart2) //
|| //
// if different costant parts of lines, then parallel BUT not overlapping
(q_ab == q_cd)//
)) {
if (isBetween(pxStart2, pyStart2, pxStart1, pyStart1, pxEnd1, pyEnd1))
return new Point2D.Double(pxStart2, pyStart2);
if (isBetween(pxEnd2, pyEnd2, pxStart1, pyStart1, pxEnd1, pyEnd1))
return new Point2D.Double(pxEnd2, pyEnd2);
if (isBetween(pxStart1, pyStart1, pxStart2, pyStart2, pxEnd2, pyEnd2))
return new Point2D.Double(pxStart1, pyStart1);
if (isBetween(pxEnd1, pyEnd1, pxStart2, pyStart2, pxEnd2, pyEnd2))
return new Point2D.Double(pxEnd1, pyEnd1);
}
if (slope_ab == Double.POSITIVE_INFINITY) {
// ab vertical line: all a&b's x-es are equals
x = pxStart1;
y = (q_cd + (slope_cd * x));
// it's a cross
if ((pyStart1 <= pyEnd1) ? (y < pyStart1 || y > pyEnd1)
// point are reversed
: (y > pyStart1 || y < pyEnd1))
return null;
if ((pxStart2 < pxEnd2) ? (pxStart2 <= x && x <= pxEnd2)//
: (pxEnd2 <= x && x <= pxStart2))
return new Point2D.Double(x, y);
else
return null;
}
if (slope_cd == Double.POSITIVE_INFINITY) {
// cd vertical line: all c&d's x-es are equals
x = pxStart2;
y = (q_ab + (slope_ab * x));
// it's a cross
if ((pyStart2 <= pyEnd2) ? (y < pyStart2 || y > pyEnd2)
// point are reversed
: (y > pyStart2 || y < pyEnd2))
return null;
// if the y lies inside the line a-b, then intersection
if ((pxStart1 < pxEnd1) ? (pxStart1 <= x && x <= pxEnd1)//
: (pxEnd1 <= x && x <= pxStart1))
return new Point2D.Double(x, y);
else
return null;
}
// slopes cannot be infinity
if (slope_ab == 0.0) {
y = pyStart1;
// slope_cd cannot be Infinity (second group of checks) and zero (first ones)
x = (y - q_cd) / slope_cd;
if ((pxStart1 <= pxEnd1) ? (x < pxStart1 || x > pxEnd1)
// point are reversed
: (x > pxStart1 || x < pxEnd1))
return null;
if ((pxStart2 <= pxEnd2) ? (x < pxStart2 || x > pxEnd2)
// point are reversed
: (x > pxStart2 || x < pxEnd2))
return null;
if ((pyStart2 < pyEnd2) ? (pyStart2 <= y && y <= pyEnd2)//
: (pyEnd2 <= y && y <= pyStart2))
return new Point2D.Double(x, y);
else
return null;
}
if (slope_cd == 0.0) {
y = pyStart2;
// slope_ab cannot be Infinity (second group of checks) and zero (first ones)
x = (y - q_ab) / slope_ab;
if ((pxStart2 <= pxEnd2) ? (x < pxStart2 || x > pxEnd2)
// point are reversed
: (x > pxStart2 || x < pxEnd2))
return null;
if ((pxStart1 <= pxEnd1) ? (x < pxStart1 || x > pxEnd1)
// point are reversed
: (x > pxStart1 || x < pxEnd1))
return null;
if ((pyStart1 < pyEnd1) ? (pyStart1 <= y && y <= pyEnd1)//
: (pyEnd1 <= y && y <= pyStart1))
return new Point2D.Double(x, y);
else
return null;
}
denominator = slope_cd - slope_ab;
numerator = q_ab - q_cd;
x = (numerator / denominator);
y = (q_ab + (slope_ab * x));
p = new Point2D.Double(x, y);
if (isBetween(p.getX(), p.getY(), pxStart1, pyStart1, pxEnd1, pyEnd1)
&& isBetween(p.getX(), p.getY(), pxStart2, pyStart2, pxEnd2, pyEnd2))
return p;
y = (q_cd + (slope_cd * x));
p = new Point2D.Double(x, y);
if ((isBetween(p.getX(), p.getY(), pxStart1, pyStart1, pxEnd1, pyEnd1)
&& isBetween(p.getX(), p.getY(), pxStart2, pyStart2, pxEnd2, pyEnd2)))
return p;
return null;
}
/**
*测试第一个点是否位于由其他两个点表示的边界框中
*要点。
*/
公共静态布尔值介于之间（点2D测试、点2D p1、点2D p2）{
返回isBetween（pToTest.getX（），pToTest.getY（），p1.getX（），p1.getY（），p2.getX（），p2.getY（））；
}
/**
*通过这些点调用{@link#isBetween（Point2D，Point2D，Point2D）}
*坐标。
*/
公共静态布尔值isBetween（双pxToTest、双pyToTest、双px1、双py1、双px2、双py2）{
双w，h；
//从矩形的课程中获得灵感
w=px1-px2；
if（w<0）
w=-w；
h=py1-py2；
if（h<0）
h=-h；
//使用p1作为矩形的左上角
//（左上角视为原点（0；0））
如果（px1>px2）
px1=px2//
如果（py1>py2）
py1=py2//
if（pxToTestpxToTest）和&（hpyToTest））；
}
公共静态双斜率（双xa、双ya、双xb、双yb）{
如果（xb==xa）
返回双正无穷大；
收益率（yb==ya）？0.0:（（yb-ya）/（xb-xa））；
}
/**
*由{@link#areLinesIntersecting（Point2D，Point2D，Point2D，Point2D）调用
*通过提供每个点的坐标。
*/
公共静态点2D是线性插入检测（双pxStart1、双pyStart1、双pxEnd1、双pyEnd1、，
双pxStart2，双pyStart2，双pxEnd2，双pyEnd2）{
双斜率ab，斜率cd，分子，分母，q_ab，q_cd，x，y；
点2dp；
斜率=斜率（pxStart1，pyStart1，pxEnd1，pyEnd1）；
斜率=斜率（pxStart2，pyStart2，pxEnd2，pyEnd2）；
q_ab=（pyStart1-斜率_ab*pxStart1）；
q_cd=（pyStart2-斜率_cd*pxStart2）；
如果（（斜率=slope\u ab==slope\u cd）//平行？
&& (
//重叠？
（斜率| ab==Double.POSITIVE |斜率| ab==Double.NaN）和&pxStart1==pxStart2）//
|| //
//重叠？
（斜率=0.0&&pyStart1==pyStart2）//
|| //
//如果线的不同共有部分，则平行但不重叠
（q_ab==q_cd）//
)) {
if（isBetween（pxStart2、pyStart2、pxStart1、pyStart1、pxEnd1、pyEnd1））
返回新的点2d.Double（pxStart2，pyStart2）；
if（isBetween（pxEnd2，pyEnd2，pxStart1，pyStart1，pxEnd1，pyEnd1））
返回新的点2d.Double（pxEnd2，pyEnd2）；
if（isBetween（pxStart1、pyStart1、pxStart2、pyStart2、pxEnd2、pyEnd2））
返回新的点2d.Double（pxStart1，pyStart1）；
if（isBetween（pxEnd1，pyEnd1，pxStart2，pyStart2，pxEnd2，pyEnd2））
返回新的点2d.Double（pxEnd1，pyEnd1）；
}
if（斜率=双正无穷大）{
//ab垂直线：所有a&b的x-E相等
x=pxStart1；
y=（q_cd+（斜率_cd*x））；
//这是一个十字架
if（（pyStart1 pyEnd1）
//这一点是相反的
：（y>pyStart1 | | y/**
* Test if the first point lies in the bounding box denote by the other two
* points.
*/
public static boolean isBetween(Point2D pToTest, Point2D p1, Point2D p2) {
return isBetween(pToTest.getX(), pToTest.getY(), p1.getX(), p1.getY(), p2.getX(), p2.getY());
}
/**
* Called {@link #isBetween(Point2D, Point2D, Point2D)} passing those points
* coordinates.
*/
public static boolean isBetween(double pxToTest, double pyToTest, double px1, double py1, double px2, double py2) {
double w, h;
// taking inspiration from Rectangle's class
w = px1 - px2;
if (w < 0)
w = -w;
h = py1 - py2;
if (h < 0)
h = -h;
// use p1 as the left-top corner of the rectangle
// (the left-top corner is considered as the origin (0;0))
if (px1 > px2)
px1 = px2;//
if (py1 > py2)
py1 = py2;//
if (pxToTest < px1 || pyToTest < py1) {
return false;
}
w += px1;
h += py1;
// overflow || intersect
return ((w < px1 || w > pxToTest) && (h < py1 || h > pyToTest));
}
public static double slope(double xa, double ya, double xb, double yb) {
if (xb == xa)
return Double.POSITIVE_INFINITY;
return (yb == ya) ? 0.0 : ((yb - ya) / (xb - xa));
}
/**
* Called by {@link #areLinesIntersecting(Point2D, Point2D, Point2D, Point2D)}
* by providing each of those point's coordinates.
*/
public static Point2D areLinesIntersecting(double pxStart1, double pyStart1, double pxEnd1, double pyEnd1,
double pxStart2, double pyStart2, double pxEnd2, double pyEnd2) {
double slope_ab, slope_cd, numerator, denominator, q_ab, q_cd, x, y;
Point2D p;
slope_ab = slope(pxStart1, pyStart1, pxEnd1, pyEnd1);
slope_cd = slope(pxStart2, pyStart2, pxEnd2, pyEnd2);
q_ab = (pyStart1 - slope_ab * pxStart1);
q_cd = (pyStart2 - slope_cd * pxStart2);
if ((slope_ab == slope_cd) // parallel?
&& (
// overlapping?
((slope_ab == Double.POSITIVE_INFINITY || slope_ab == Double.NaN) && pxStart1 == pxStart2) //
|| //
// overlapping?
(slope_ab == 0.0 && pyStart1 == pyStart2) //
|| //
// if different costant parts of lines, then parallel BUT not overlapping
(q_ab == q_cd)//
)) {
if (isBetween(pxStart2, pyStart2, pxStart1, pyStart1, pxEnd1, pyEnd1))
return new Point2D.Double(pxStart2, pyStart2);
if (isBetween(pxEnd2, pyEnd2, pxStart1, pyStart1, pxEnd1, pyEnd1))
return new Point2D.Double(pxEnd2, pyEnd2);
if (isBetween(pxStart1, pyStart1, pxStart2, pyStart2, pxEnd2, pyEnd2))
return new Point2D.Double(pxStart1, pyStart1);
if (isBetween(pxEnd1, pyEnd1, pxStart2, pyStart2, pxEnd2, pyEnd2))
return new Point2D.Double(pxEnd1, pyEnd1);
}
if (slope_ab == Double.POSITIVE_INFINITY) {
// ab vertical line: all a&b's x-es are equals
x = pxStart1;
y = (q_cd + (slope_cd * x));
// it's a cross
if ((pyStart1 <= pyEnd1) ? (y < pyStart1 || y > pyEnd1)
// point are reversed
: (y > pyStart1 || y < pyEnd1))
return null;
if ((pxStart2 < pxEnd2) ? (pxStart2 <= x && x <= pxEnd2)//
: (pxEnd2 <= x && x <= pxStart2))
return new Point2D.Double(x, y);
else
return null;
}
if (slope_cd == Double.POSITIVE_INFINITY) {
// cd vertical line: all c&d's x-es are equals
x = pxStart2;
y = (q_ab + (slope_ab * x));
// it's a cross
if ((pyStart2 <= pyEnd2) ? (y < pyStart2 || y > pyEnd2)
// point are reversed
: (y > pyStart2 || y < pyEnd2))
return null;
// if the y lies inside the line a-b, then intersection
if ((pxStart1 < pxEnd1) ? (pxStart1 <= x && x <= pxEnd1)//
: (pxEnd1 <= x && x <= pxStart1))
return new Point2D.Double(x, y);
else
return null;
}
// slopes cannot be infinity
if (slope_ab == 0.0) {
y = pyStart1;
// slope_cd cannot be Infinity (second group of checks) and zero (first ones)
x = (y - q_cd) / slope_cd;
if ((pxStart1 <= pxEnd1) ? (x < pxStart1 || x > pxEnd1)
// point are reversed
: (x > pxStart1 || x < pxEnd1))
return null;
if ((pxStart2 <= pxEnd2) ? (x < pxStart2 || x > pxEnd2)
// point are reversed
: (x > pxStart2 || x < pxEnd2))
return null;
if ((pyStart2 < pyEnd2) ? (pyStart2 <= y && y <= pyEnd2)//
: (pyEnd2 <= y && y <= pyStart2))
return new Point2D.Double(x, y);
else
return null;
}
if (slope_cd == 0.0) {
y = pyStart2;
// slope_ab cannot be Infinity (second group of checks) and zero (first ones)
x = (y - q_ab) / slope_ab;
if ((pxStart2 <= pxEnd2) ? (x < pxStart2 || x > pxEnd2)
// point are reversed
: (x > pxStart2 || x < pxEnd2))
return null;
if ((pxStart1 <= pxEnd1) ? (x < pxStart1 || x > pxEnd1)
// point are reversed
: (x > pxStart1 || x < pxEnd1))
return null;
if ((pyStart1 < pyEnd1) ? (pyStart1 <= y && y <= pyEnd1)//
: (pyEnd1 <= y && y <= pyStart1))
return new Point2D.Double(x, y);
else
return null;
}
denominator = slope_cd - slope_ab;
numerator = q_ab - q_cd;
x = (numerator / denominator);
y = (q_ab + (slope_ab * x));
p = new Point2D.Double(x, y);
if (isBetween(p.getX(), p.getY(), pxStart1, pyStart1, pxEnd1, pyEnd1)
&& isBetween(p.getX(), p.getY(), pxStart2, pyStart2, pxEnd2, pyEnd2))
return p;
y = (q_cd + (slope_cd * x));
p = new Point2D.Double(x, y);
if ((isBetween(p.getX(), p.getY(), pxStart1, pyStart1, pxEnd1, pyEnd1)
&& isBetween(p.getX(), p.getY(), pxStart2, pyStart2, pxEnd2, pyEnd2)))
return p;
return null;
}