获取 (int,int) 参数以接受 (int, array) 输入

Question

有什么方法可以将我的 ArrayList myZips 中的 ZipCode 转换为 Integer，以便它可以作为我的距离方法的参数？

邮编类：

public class ZipCode
{
    private int zipCode;
    private String city;
    private String state;
    private double longitude;
    private double latitude;

    public ZipCode(int pZip)
    {
        zipCode   = pZip;
        city      = "UNKOWN";
        state     = "ST";
        latitude  = 0.0;
        longitude = 0.0;
    }

    public ZipCode(int pZip, String pCity, String pState, double pLat, double pLon)
    {
        zipCode   = pZip;
        city      = pCity;
        state     = pState;
        latitude  = pLat;
        longitude = pLon;

    }

    public void setZipCode(int zipCode) {
        this.zipCode = zipCode;
    }

    public void setCity(String city) {
        this.city = city;
    }

    public void setState(String state) {
        this.state = state;
    }

    public void setLatitude(double latitude) {
        this.latitude = latitude;
    }

    public void setLongitude(double longitude) {
        this.longitude = longitude;
    }

    public int getZipCode() {
        return zipCode;
    }

    public String getCity() {
        return city;
    }

    public String getState() {
    return state;
    }

    public double getLatitude() {
        return latitude;
    }

    public double getLongitude() {
        return longitude;
    }
    public String toString() {
        return city + ", " + state + zipCode;
    }
}


public class ZipCodeDatabase {

    private ArrayList<ZipCode> myZips;

public ZipCodeDatabase()
{
    myZips = new ArrayList<ZipCode>();
}

public ZipCode findZip(int zip){
    for(ZipCode zipCode : myZips){

        if(zipCode.getZipCode() == zip){
            return zipCode;
        }
    }

    return null;

}

public double distance(int zip1, int zip2) {
    int EARTH_RADIUS = 3959;
    if (findZip(zip1) != null && findZip(zip2) != null) {
        ZipCode z1 = new ZipCode(zip1);
        ZipCode z2 = new ZipCode(zip2);
        double lat1  = Math.toRadians(z1.getLatitude());
        double long1 = Math.toRadians(z1.getLongitude());
        double lat2  = Math.toRadians(z2.getLatitude());
        double long2 = Math.toRadians(z2.getLongitude());

        double p1 = Math.cos(lat1) * Math.cos(long1) * Math.cos(lat2) * Math.cos(long2);       
        double p2 = Math.cos(lat1) * Math.sin(long1) * Math.cos(lat2) * Math.sin(long2);         
        double p3 = Math.sin(lat1) * Math.sin(lat2);         
        double distance = Math.acos(p1+p2+p3) * EARTH_RADIUS;
        int distance1 = (int)Math.round(distance);
        return distance1;
    }
    else {
        return -1;
    }
}
public ArrayList <ZipCode> withinRadius(int pZip, int pRadius){
    ArrayList<ZipCode> zips = new ArrayList<ZipCode> ();
    if (findZip(pZip) != null){
        for (int i = 0; i < myZips.size(); i++){
            if (distance(pZip, myZips.get(i)) <= pRadius) //the trouble is here, how can I convert the myZips.get(I) into the relevant 5 digit integer .
            zips.add(pZip);
        }
    }
    return zips;
}

}

我知道这是一个非常专业和基本的问题，但我是 Java 新手，似乎无法弄清楚。

Answer 1

我写了findZip method！ :)

如果你只有两个拉链可以看的话，我猜你只需要这个。

ZipCode zip1 = myZips.get(0);
ZipCode zip2 = myZips.get(1);
double d = distance(zip1.getZipCode(), zip2.getZipCode());

如果您想转换整个 ZipCode 列表，您可以这样做

List<Integer> zipCodeInts = new ArrayList<>(myZips.size());
for(ZipCode zipCode : myZips){
    zipCodeInts.add(zipCode.getZipCode());
}
// now use zipCodeInts somehow

旁注，您可以保存这两个new。而不是这样做：

if (findZip(zip1) != null && findZip(zip2) != null) {
    ZipCode z1 = new ZipCode(zip1);
    ZipCode z2 = new ZipCode(zip2);
    // do stuff
}

你可以这样做

ZipCode z1 = findZip(zip1);
ZipCode z2 = findZip(zip2);

if(z1 != null && z2 != null) {
    // do stuff
}

Answer 2

我不明白拥有findZip 背后的原因，因为我认为您进行的查找比您需要的要多，但无论如何，您已经接近了，只需要保存对您找到的ZipCode 的引用findZip 并在 distance 方法中调用 getZipCode()。

public static final int EARTH_RADIUS = 3959; // Move constants out of methods

public ZipCode findZip(int zip){
    for(ZipCode zipCode : myZips){
        if(zipCode.getZipCode() == zip){
            return zipCode;
        }
    }
    return null;
}

public int distance(int zip1, int zip2){
    // save these references so you can use them later
    ZipCode z1 = findZip(zip1);
    ZipCode z2 = findZip(zip2);

    if(z1 != null && z2 != null) {
        double lat1  = Math.toRadians(z1.getLatitude());
        double long1 = Math.toRadians(z1.getLongitude());
        double lat2  = Math.toRadians(z2.getLatitude());
        double long2 = Math.toRadians(z2.getLongitude());

        double p1 = Math.cos(lat1) * Math.cos(long1) * Math.cos(lat2) * Math.cos(long2);       
        double p2 = Math.cos(lat1) * Math.sin(long1) * Math.cos(lat2) * Math.sin(long2);         
        double p3 = Math.sin(lat1) * Math.sin(lat2);         

        double distance = Math.acos(p1+p2+p3) * EARTH_RADIUS;
        return (int)Math.round(distance);
    }
    else return -1;
}

public ArrayList <ZipCode> withinRadius(int pZip, int pRadius){
    ZipCode zip = findZip(pZip); // save this reference to calculate the distances

    ArrayList<ZipCode> zips = new ArrayList<ZipCode> ();
    if (zip != null) {
        for (int i = 0; i < myZips.size(); i++) {
            ZipCode z = myZips.get(i); // or use a for-each loop like in findZip
            if (distance(zip.getZipCode(), z.getZipCode()) <= pRadius) {
                zips.add(z);
            }
        }
    }
    return zips;
}

Answer 3

只需获取包含班级邮政编码的int：

ZipCodeDatabase.distance(myZipCode.getZipCode(), yourZipCode.getZipCode());