根据已知经纬度和指定的范围求出最大及最小的经纬度

public class LatLonUtil {

private static final double PI = 3.14159265;
private static final double EARTH_RADIUS = 6378137;
private static final double RAD = Math.PI / 180.0;

//@see http://snipperize.todayclose.com/snippet/php/SQL-Query-to-Find-All-Retailers-Within-a-Given-Radius-of-a-Latitude-and-Longitude--65095/ //The circumference of the earth is 24,901 miles.
//24,901/360 = 69.17 miles / degree
/**
* @param raidus 单位米
* return minLat,minLng,maxLat,maxLng
*/
public static double[] getAround(double lat,double lon,int raidus){

Double latitude = lat;
Double longitude = lon;

Double degree = (24901*1609)/360.0;
double raidusMile = raidus;

Double dpmLat = 1/degree;
Double radiusLat = dpmLat*raidusMile;
Double minLat = latitude - radiusLat;
Double maxLat = latitude + radiusLat;

Double mpdLng = degree*Math.cos(latitude * (PI/180));
Double dpmLng = 1 / mpdLng;
Double radiusLng = dpmLng*raidusMile;
Double minLng = longitude - radiusLng;
Double maxLng = longitude + radiusLng;
//System.out.println("["+minLat+","+minLng+","+maxLat+","+maxLng+"]");
return new double[]{minLat,minLng,maxLat,maxLng};
}

/**
* 根据两点间经纬度坐标（double值），计算两点间距离，单位为米
* @param lng1
* @param lat1
* @param lng2
* @param lat2
* @return
*/
public static double getDistance(double lng1, double lat1, double lng2, double lat2)
{
double radLat1 = lat1*RAD;
double radLat2 = lat2*RAD;
double a = radLat1 - radLat2;
double b = (lng1 - lng2)*RAD;
double s = 2 * Math.asin(Math.sqrt(Math.pow(Math.sin(a/2),2) +
Math.cos(radLat1)*Math.cos(radLat2)*Math.pow(Math.sin(b/2),2)));
s = s * EARTH_RADIUS;
s = Math.round(s * 10000) / 10000;
return s;
}

public static void main(String[] args){
Double lat1 = 34.264648;
Double lon1 = 108.952736;

int radius = 1000;
//[34.25566276027792,108.94186385411045,34.27363323972208,108.96360814588955]
getAround(lat1,lon1,radius);

//911717.0   34.264648,108.952736,39.904549,116.407288
double dis = getDistance(108.952736,34.264648,116.407288,39.904549);
System.out.println(dis);
}
}