List difference in java

Question

I have two ArrayList as follows:

original: 12, 16, 17, 19, 101

selected: 16, 19, 107, 108, 109

Answer 1

List<Integer> original = Arrays.asList(12,16,17,19,101);
List<Integer> selected = Arrays.asList(16,19,107,108,109);

ArrayList<Integer> add = new ArrayList<Integer>(selected);
add.removeAll(original);
System.out.println("Add: " + add);

ArrayList<Integer> remove = new ArrayList<Integer>(original);
remove.removeAll(selected);
System.out.println("Remove: " + remove);

Output:

Add: [107, 108, 109]
Remove: [12, 17, 101]

Uses Collection's removeAll method. See javadocs.

Answer 2

As an alternative, you could use CollectionUtils from Apache commons library. It has static intersection, union and subtract methods suitable for your case.

Answer 3

Use this method if you want to get intersection of a list of lists

List<Address> resultsIntersectionSet( List<Set<Address>> inputListOfLists )
{
    Set<Address> intersection = new HashSet<>();

    if ( !inputListOfLists.isEmpty() )
        intersection = inputListOfLists.get( 0 );

    for ( Set<Address> filterResultList : inputListOfLists )
    {
        intersection.retainAll( filterResultList );
    }

    return new ArrayList<>( intersection );
}

Answer 4

 List<Integer> original;
 List<Integer> selected;

 List<Integer> add = new ArrayList<Integer>(selected);
 add.removeAll(original);

 List<Integer> remove = new ArrayList<Integer>(original);
 remove.removeAll(selected);

Be careful with border cases around duplicate elements. Should cardinality be respected? As in, if I had 5, 6 originally and 5, 5, 6 after, should add be 5? The above works better with Sets, since they don't have duplicates (plus contains() lookups are faster since they are indexed by the data they contain).

Answer 5

For intersection and union operations the natural collection type is a Set rather than a List, its also more efficient to use.