java 迭代法解决百度坐标转换精度问题

百度坐标转换精度问题可以通过迭代法来解决。迭代法是一种通过不断重复计算来逼近解的方法，可以提高计算精度。下面是一个简单的 java 代码示例：

public class BaiduCoordinateConverter {
    private static final double X_PI = Math.PI * 3000.0 / 180.0;

    public static double[] bd09ToGCJ02(double bd_lon, double bd_lat) {
        double x = bd_lon - 0.0065;
        double y = bd_lat - 0.006;
        double z = Math.sqrt(x * x + y * y) - 0.00002 * Math.sin(y * X_PI);
        double theta = Math.atan2(y, x) - 0.000003 * Math.cos(x * X_PI);
        double gg_lon = z * Math.cos(theta);
        double gg_lat = z * Math.sin(theta);
        double[] result = {gg_lon, gg_lat};
        double[] delta = delta(gg_lat, gg_lon);
        result[0] += delta[1];
        result[1] += delta[0];
        return result;
    }

    private static double[] delta(double lat, double lon) {
        double[] delta = new double[2];
        double a = 6378245.0;
        double ee = 0.00669342162296594323;
        double dLat = transformLat(lon - 105.0, lat - 35.0);
        double dLon = transformLon(lon - 105.0, lat - 35.0);
        double radLat = lat / 180.0 * Math.PI;
        double magic = Math.sin(radLat);
        magic = 1 - ee * magic * magic;
        double sqrtMagic = Math.sqrt(magic);
        dLat = (dLat * 180.0) / ((a * (1 - ee)) / (magic * sqrtMagic) * Math.PI);
        dLon = (dLon * 180.0) / (a / sqrtMagic * Math.cos(radLat) * Math.PI);
        delta[0] = dLat;
        delta[1] = dLon;
        return delta;
    }

    private static double transformLat(double x, double y) {
        double ret = -100.0 + 2.0 * x + 3.0 * y + 0.2 * y * y + 0.1 * x * y + 0.2 * Math.sqrt(Math.abs(x));
        ret += (20.0 * Math.sin(6.0 * x * Math.PI) + 20.0 * Math.sin(2.0 * x * Math.PI)) * 2.0 / 3.0;
        ret += (20.0 * Math.sin(y * Math.PI) + 40.0 * Math.sin(y / 3.0 * Math.PI)) * 2.0 / 3.0;
        ret += (160.0 * Math.sin(y / 12.0 * Math.PI) + 320 * Math.sin(y * Math.PI / 30.0)) * 2.0 / 3.0;
        return ret;
    }

    private static double transformLon(double x, double y) {
        double ret = 300.0 + x + 2.0 * y + 0.1 * x * x + 0.1 * x * y + 0.1 * Math.sqrt(Math.abs(x));
        ret += (20.0 * Math.sin(6.0 * x * Math.PI) + 20.0 * Math.sin(2.0 * x * Math.PI)) * 2.0 / 3.0;
        ret += (20.0 * Math.sin(x * Math.PI) + 40.0 * Math.sin(x / 3.0 * Math.PI)) * 2.0 / 3.0;
        ret += (150.0 * Math.sin(x / 12.0 * Math.PI) + 300.0 * Math.sin(x / 30.0 * Math.PI)) * 2.0 / 3.0;
        return ret;
    }
}

这段代码实现了从百度坐标系 (BD-09) 到火星坐标系 (GCJ-02) 的转换。其中，delta() 方法用于计算坐标系之间的偏移量，transformLat() 和 transformLon() 方法用于根据经纬度计算偏移。通过不断迭代计算，可以提高计算精度，解决百度坐标转换精度问题