GPS糾偏演算法，適用於google,高德體系的地圖,精確度還比較高。我試了一下比高德本身的糾偏還精確點。

 

/**
 * gps糾偏演算法，適用於google,高德體系的地圖
 * @author Administrator
 */
public class GpsCorrect {
	final static double pi = 3.14159265358979324;
	final static double a = 6378245.0;
	final static double ee = 0.00669342162296594323;

	public static void transform(double wgLat, double wgLon, double[] latlng) {
		if (outOfChina(wgLat, wgLon)) {
			latlng[0] = wgLat;
			latlng[1] = wgLon;
			return;
		}
		double dLat = transformLat(wgLon - 105.0, wgLat - 35.0);
		double dLon = transformLon(wgLon - 105.0, wgLat - 35.0);
		double radLat = wgLat / 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);
		dLon = (dLon * 180.0) / (a / sqrtMagic * Math.cos(radLat) * pi);
		latlng[0] = wgLat + dLat;
		latlng[1] = wgLon + dLon;
	}

	private static boolean outOfChina(double lat, double lon) {
		if (lon < 72.004 || lon > 137.8347)
			return true;
		if (lat < 0.8293 || lat > 55.8271)
			return true;
		return false;
	}

	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 * 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;
	}

	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 * 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;
	}
}