这个算法就是判断一个点向左的射线跟一个多边形的交叉点有几个
如果结果为奇数的话那么说明这个点落在多边形中
反之则不在。直接贴代码吧
public class Point {
private Double x;
private Double y;
public Point(Double x, Double y) {
this.x = x;
this.y = y;
}
public Double getX() {
return x;
}
public void setX(Double x) {
this.x = x;
}
public Double getY() {
return y;
}
public void setY(Double y) {
this.y = y;
}
}
public static void main(String[] args) {
//跑道经纬度[]
Point[] ps = new Point[] { new Point(118.26763298804507,29.727463756843807), new Point(118.26623051693838,29.725847852307695), new Point(118.24472520887039,29.73991959541906), new Point(120.1798 , 30.2781), new Point(118.24612753025814,29.741535726649143) };
//航班经纬度
Point n1=new Point(119.56083333333333,28.36916666666667);
System.out.println( "n1:" + isPtInPoly(n1.getX() , n1.getY() , ps));
}
public static boolean isPtInPoly(double ALon , double ALat , Point[] ps) {
int iSum, iCount, iIndex;
double dLon1 = 0, dLon2 = 0, dLat1 = 0, dLat2 = 0, dLon;
if (ps.length < 3) {
return false;
}
iSum = 0;
iCount = ps.length;
for (iIndex = 0; iIndex<iCount;iIndex++) {
if (iIndex == iCount - 1) {
dLon1 = ps[iIndex].getX();
dLat1 = ps[iIndex].getY();
dLon2 = ps[0].getX();
dLat2 = ps[0].getY();
} else {
dLon1 = ps[iIndex].getX();
dLat1 = ps[iIndex].getY();
dLon2 = ps[iIndex + 1].getX();
dLat2 = ps[iIndex + 1].getY();
}
// 以下语句判断A点是否在边的两端点的水平平行线之间,在则可能有交点,开始判断交点是否在左射线上
if (((ALat >= dLat1) && (ALat < dLat2)) || ((ALat >= dLat2) && (ALat < dLat1))) {
if (Math.abs(dLat1 - dLat2) > 0) {
//得到 A点向左射线与边的交点的x坐标:
dLon = dLon1 - ((dLon1 - dLon2) * (dLat1 - ALat) ) / (dLat1 - dLat2);
// 如果交点在A点左侧(说明是做射线与 边的交点),则射线与边的全部交点数加一:
if (dLon < ALon) {
iSum++;
}
}
}
}
if ((iSum % 2) != 0) {
return true;
}
return false;
}
}
转自:https://blog.youkuaiyun.com/qq_22929803/article/details/46818009