Java—实现判断一个经纬度坐标是否在一个多边形内

阅读完需：约 2 分钟

创建一个实体类：Point

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;
    }
}

判断坐标的工具类：

public class GPSjudgment {
    public static void main(String[] args) {
        //114.331951,30.64091#114.341049,30.610185#114.331436,30.588058#114.312038,30.56393#114.293498,30.558609#114.267922,30.563784#114.231185,30.57945#114.212303,30.601616#114.235649,30.626878#114.280624,30.646818#
        Map[] map=new Map[]{};
        Long s=System.currentTimeMillis();
        Point[] ps = new Point[] { new Point(111.283365,40.099701),
                new Point(111.277957,40.099659),
                new Point(111.277589,40.098224),
                new Point(111.278083,40.097183),
                new Point(111.27917,40.097038),
                new Point(111.27917,40.097038),
                new Point(111.279997,40.098086),
                new Point(111.281317,40.0981),
                new Point(111.281371,40.097755),
                new Point(111.28218,40.096989),
                new Point(111.28218,40.096989),
                new Point(111.28341,40.099715)};

        Point n1 = new Point(111.280652,40.098169);
        System.out.println( "n1:" + isPtInPoly(n1.getX() , n1.getY() , ps));
        Long e=System.currentTimeMillis();
        System.out.println(e-s);
    }
    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;
    }
}

原理：

采取的一个方法就是，从这个点『往右水平出发』判断经过了多少条『多边形的边』：

如果是奇数条边，则这个点在多边形内，如果是偶数条，则在多边形外。