在国内，我们使用高德地图所获取的经纬度由于保密原因实际上经过了加偏处理
所以需要从国测局（GCJ02）坐标转换成基于WGS-84坐标系的GPS坐标
可在Android项目中直接使用，已经经过本人检验，建议建立GCJ2WGSUtils类，直接调用static方法计算
效果图：将转换前后都转为WGS-84观察，精度已经相当高了
WGSLat函数
输入：GCJ-02坐标系的经纬度输出：WGS-84坐标系下的纬度
WGSLon函数
输入：GCJ-02坐标系的经纬度输出：WGS-84坐标系下的经度
代码如下：
public class GCJ2WGSUtils {//输入GCJ经纬度 转WGS纬度public static double WGSLat (double lat,double lon) {double PI = 3.14159265358979324;//圆周率double a = 6378245.0;//克拉索夫斯基椭球参数长半轴adouble ee = 0.00669342162296594323;//克拉索夫斯基椭球参数第一偏心率平方double dLat = transformLat(lon - 105.0, lat - 35.0);double radLat = lat / 180.0 * 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) * PI);return (lat - dLat);}//输入GCJ经纬度 转WGS经度public static double WGSLon (double lat,double lon) {double PI = 3.14159265358979324;//圆周率double a = 6378245.0;//克拉索夫斯基椭球参数长半轴adouble ee = 0.00669342162296594323;//克拉索夫斯基椭球参数第一偏心率平方double dLon = transformLon(lon - 105.0, lat - 35.0);double radLat = lat / 180.0 * PI;double magic = Math.sin(radLat);magic = 1 - ee * magic * magic;double sqrtMagic = Math.sqrt(magic);dLon = (dLon * 180.0) / (a / sqrtMagic * Math.cos(radLat) * PI);return (lon - dLon);}//转换经度所需public static double transformLon(double x, double y) {double PI = 3.14159265358979324;//圆周率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 * PI) + 20.0 * Math.sin(2.0 * x * PI)) * 2.0 / 3.0;ret += (20.0 * Math.sin(x * PI) + 40.0 * Math.sin(x / 3.0 * PI)) * 2.0 / 3.0;ret += (150.0 * Math.sin(x / 12.0 * PI) + 300.0 * Math.sin(x / 30.0 * PI)) * 2.0 / 3.0;return ret;}//转换纬度所需public static double transformLat(double x, double y) {double PI = 3.14159265358979324;//圆周率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 * PI) + 20.0 * Math.sin(2.0 * x * PI)) * 2.0 / 3.0;ret += (20.0 * Math.sin(y * PI) + 40.0 * Math.sin(y / 3.0 * PI)) * 2.0 / 3.0;ret += (160.0 * Math.sin(y / 12.0 * PI) + 320 * Math.sin(y * PI / 30.0)) * 2.0 / 3.0;return ret;}}
