java.util.Arrays.equals
( arraylist1.toArray( new java.lang.String[ 0 ]),
 arraylist2.toArray( new java.lang.String[ 0 ]))

The == operator returns true only if both variables refer to the same
list--it does not attempt to determine if two distinct list have equivalent
contents.  .equals () won't do you any good here either.  If the lists  have
the same length and you can remove all the items of the second list from the
first, they're equal. If there will never be identical elements in any list,
you can do something like:

Set s1 = new HashSet (arraylist1);
for (Object o : arraylist2) {
 if (! s1.contains (o)) // Sets are definately not equal
 s1.remove(o);
}

If there can be identical elements, you'll have to count them or otherwise
account for multiples
Map <Object,Integer> m = new HashMap <Object,Integer> ();
for (Object o : arraylist1) {
 Integer countInt = m.get(o);
 if (countInt == null) {
   m.put (o, new Integer (1));
 } else {
   m.put (o, new Integer (1 + countInt.intValue()));
}
And then subtract the arraylist2 items int he same way.  If any item can't
be subtracted, they're not equal.

Or, sort both lists (if the values can be ordered)
Collections.sort(arraylist1);
Collections.sort(arraylist2);
And compare each position.  Any unequal item means the lists are not equal.

Cheers,
Matt Humphrey matth@ivizNOSPAM.com http://www.iviz.com/

No, that asks if the reference variables arraylist1 and arraylist2 are
equal. Two reference variables are equal if they are either both null or
refer to the same object.

Using the List equals method does not help because it takes order into
account.

How about set equality between sets containing their contents?

if( new HashSet(arraylist1).equals(new HashSet(arraylist2)) )

Patricia

There are two basic approaches:

1.If  there are no duplicate values, you can do:

   if (new HashSet(arrlist1).equals(new HashSet(arrlist2))

2A.  If you don't mind changing the order of the lists, sort them (with
Collections.sort()), and use equals() to compare the results..

2B. If you don't want to change the order, create new lists from your
existing lists and apply 2A.