Sorting an ArrayList<String> with custom Comparator

Question

I am trying to sort an ArrayList<String> using custom Comparator. My requirement is that XX String should be the first String, others should follow natural ordering.

I need : [XX, XX, 1, 5, 9, A, D, G, Q, Z]
What I am getting is [1, 5, 9, A, D, G, Q, Z, XX, XX]

Following is my code:

public class Test {
    public static void main(String[] args) 
    {
        List<String> list = new ArrayList<String>();
        list.add("Z");
        list.add("5");
        list.add("D");
        list.add("G");
        list.add("XX");     
        list.add("9");
        list.add("Q");
        list.add("XX");
        list.add("1");
        list.add("A");      
        Collections.sort(list, new CustomComparator());     
        System.out.println(list);       
    }
}
class CustomComparator implements Comparator<String>
{
    @Override
    public int compare(String o1, String o2) {      
        if(o2.equals("XX")) {
            return -1;
        }       
        return o1.compareTo(o2);
    }   
}

EDIT: if i change comparator logic to:

@Override
    public int compare(String o1, String o2) {      
        if(o2.equals("XX")) {
            return 1;
        }       
        return o1.compareTo(o2);
    }

I am getting :

[1, XX, 9, A, Q, XX, 5, D, G, Z]

Please let me know how to proceed. Thanks in advance.

Meno Hochschild · Accepted Answer · 2014-04-25 15:10:38Z

2

Use this comparator implementation:

@Override
public int compare(String o1, String o2) {      
    if(o2.equals("XX")) {
        return o1.equals("XX") ? 0 : 1;
    } else if (o1.equals("XX")) {
        return -1;
    }
    return o1.compareTo(o2);
}

Reason: Beyond the question when to use -1 or 1 it is important to guarantee a clear order for all possible tuples of o1 and o2, see javadoc:

[...] The implementor must ensure that sgn(compare(x, y)) == -sgn(compare(y, x)) for all x and y. (This implies that compare(x, y) must throw an exception if and only if compare(y, x) throws an exception.)

The implementor must also ensure that the relation is transitive: ((compare(x, y)>0) && (compare(y, z)>0)) implies compare(x, z)>0. [...]

edited Apr 25, 2014 at 15:10

answered Apr 25, 2014 at 15:04

Meno Hochschild

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2 Comments

nikis Over a year ago

congratulations, your solution is only valid one

Meno Hochschild Over a year ago

@nikis Thank you, I am used to such comparators because my employer often ask for exotic and overly complex comparison strategies, this here is just very simplified ;-)

La-comadreja · Accepted Answer · 2014-04-25 15:05:05Z

1

Concise way to write the method:

public int compare(String o1, String o2) {
    if (o1.equals(o2)) return 0;
    if (o1.equals("XX")) return -1;
    return (o2.equals("XX")) ? 1 : o1.compareTo(o2);
}

answered Apr 25, 2014 at 15:05

La-comadreja

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