Storing arrays in Set and avoiding duplicates

Question

HashSet boog = new HashSet();
boog.add(new String[]{\"a\", \"b\", \"c\"});
boog.add(new String[]{\"a\", \"b\", \"c\"});
boog.add(new          


        

Answer 1

The "better way" is to use collections. Use a List instead of a String[]:

Set<List<String>> boog = //...
boog.add(Arrays.asList("a", "b", "c"));
boog.add(Arrays.asList("a", "b", "c"));
boog.add(Arrays.asList("a", "b", "d"));

System.out.println(boog.size()); // 2

Edit

If you absolutely needed to use arrays as keys, you could build a transparent wrapper around each key and put that in the map. Some libraries help you with that. For example, here's how you could do a Set<String[]> using Trove:

Set<String[]> boog = new TCustomHashSet<String[]>(new ArrayHashingStrategy());

boog.add(new String[]{"a", "b", "c"});
boog.add(new String[]{"a", "b", "c"});
boog.add(new String[]{"a", "b", "d"});

System.out.println(boog.size()); // 2

//...
public class ArrayHashingStrategy extends HashingStrategy<Object[]> {

   public int computeHashCode(Object[] array) {
      return Arrays.hashCode(array);
   }

   public boolean equals(Object[] arr1, Object[] arr2) {
      return Arrays.equals(arr1, arr2);
   }
}        

Answer 2

You can't. arrays use the default identity-based Object.hashCode() implementation and there's no way you can override that. Don't use Arrays as keys in a HashMap / HashSet!

Use a Set of Lists instead.

Answer 3

Actually, you can. You can use TreeSet with provided Comparator. In your case it will be something like:

Set<String[]> boog = new TreeSet<>((o1, o2) -> {
    for (int i = 0; i < o1.length; i++){
        int cmp = o1[i].compareTo(o2[i]);
        if (cmp != 0) {
            return cmp;
        }
    }
    return o1.length - o2.length;
});

Under the hood it will looks like alphabetic sorted tree.

Answer 4

You are actually using the default hashCode method returning different values for all your different arrays!

The best way to solve this is either to use a Collection (such as a List or a Set) or to define your own wrapper class such as:

public class StringArray {
    public String[] stringArray;

    [...] // constructors and methods

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        for(String string : stringArray){
            result = prime * result + ((string == null) ? 0 : string.hashCode());
        }
    }
}

This class actually uses pretty much the same hashCode method as the one for the List.

You now handle:

HashSet<StringArray> boog = new HashSet<StringArray>();

Answer 5

hashCode() of arrays uses the default implementation, which doesn't take into account the elements, and you can't change that.

You can use a List instead, with a hashCode() calculated based on the hashcodes of its elements. ArrayList (as most implementations) uses such function.

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Alternatively (but less preferable, unless you are forced somehow to use arrays), you can use a 'special' HashSet where instead of invoking key.hashCode() invoke Arrays.hashCode(array). To implement that extend HashMap and then use Collections.newSetFromMap(map)