具有长值的TreeSet自定义比较器

Question

I have a TreeSet<MyObject> where MyObject has a method getID() .

The getID() method returns a long which I want to use to sort the MyObject s within the custom comparator of a TreeSet

Comparator<MyObject> comparator = new Comparator<MyObject>() {
    public int compare(MyObject a0, MyObject a1) {
        return a0.getID()-a1.getID();
    }

};
TreeSet<MyObject> set = new TreeSet<MyObject>(comparator);

However the method compare must return an int and I'm comparing long s. Is there a way I can get around this?

Answer 1

The magnitude of the difference is not important, simply that there is a difference.

long a0long = a0.getID();
long a1long = a1.getID();
if (a0long < a1long)
{
    return -1;
}

return a0long == a1long ? 0 : 1;

Answer 2

How about just using the Long object?

Long.compare(a0.getID(),a1.getID());

Java doc:

public static int compare(long x,long y)

Compares two long values numerically. The value returned is identical to what would be returned by:

    Long.valueOf(x).compareTo(Long.valueOf(y))


Parameters:
    x - the first long to compare
    y - the second long to compare
Returns:
    the value 0 if x == y; a value less than 0 if x < y; and a value greater than 0 if x > y
Since:
    1.7

Answer 3

您可以一行完成：

return Long.valueOf(a0.getID()).compareTo(Long.valueOf(a1.getID()));