Error comparing two ArrayLists in java

Question

I'm trying to write a method that takes 2 ArrayLists of doubles and returns all the values in set1 that aren't found in set2. These numbers should be returned in set3. I keep getting an out of memory error. Can anyone point me in the right direction?

ArrayList<Double> setDiff(ArrayList<Double> set1, ArrayList<Double> set2){
    ArrayList<Double> set3 = new ArrayList<Double>();
    int count = 0;
    while(count < set1.size()){
        boolean inList = false;
        while(inList == false){
            int count2 = 0;
            while(count2 < set2.size() && set1.get(count) == set2.get(count2)){
                count2++;
            }
            if(count2 != set2.size()){
                set3.add(set1.get(count));
            }
            else{
                inList = true;
                count++;
            }
        }
    }

    return set3;
}

Minor note: you can write !inList instead of inlist == false (The first one is less readable, but also more common. Your choice :p) — keyser
– keyser, Commented Jan 31, 2013 at 22:05
Unfortunately I cannot use .contains() or .remove() because of the specifications of the assignment. — user2028306
– user2028306, Commented Jan 31, 2013 at 22:05
possible duplicate of How can I calculate the difference between two ArrayLists? — Zakaria
– Zakaria, Commented Jan 31, 2013 at 22:06
Then replace contains() with a loop (see Gothmog's answer below). — keyser
– keyser, Commented Jan 31, 2013 at 22:06

Gothmog · Accepted Answer · 2013-01-31 22:30:10Z

2

Some loop is likely not stopping as you would expect.

The following code snippet would accomplish pretty much the same you are trying to do.

for (Double d : set1) {
    if (!set2.contains(d)) {
        set3.add(d);
    }
}

UPDATE: since you say you cannot use contains(), you can perform the checks by yourself:

for (Double d : set1) {
        boolean found = false;
        for (int i=0; i<set2.size() && !found; i++) {
                if (d.equals(set2.get(i))) {
                    found = true;
            }
        }
        if (!found) {
            set3.add(d);
        }
}

EDIT: Furthermore, the problem in your code lies in the line

  if(count2 != set2.size()){

You should change != with >, since in the case of being count2 less than set2, the external count variable will not increase, resulting in an infinite loop, and after a few seconds, an OutOfMemoryError.

Also, your algorithm wasn't 100% correct either, since the looping through the second list was not consistent. You can see a similar approach with while-loops below:

                int count = 0;
                while (count < set1.size()) {
                    boolean inList = false;
                    int count2 = 0;
                    while (inList == false && count2 < set2.size()) {
                        if (set1.get(count).equals(set2.get(count2))) {
                            inList = true;
                        }
                        count2++;
                    }
                    if (!inList) {
                            set3.add(set1.get(count));
                    }
                    count++;
               }

edited Jan 31, 2013 at 22:30

answered Jan 31, 2013 at 22:01

Gothmog

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3 Comments

Jeroen Vannevel Over a year ago

Don't forget to invert (!) your if statement.

user2028306 Over a year ago

I just ran this through the program and this is exactly what I'm looking for, but my professor isn't allowing us to use .contains()

user2028306 Over a year ago

Thank you so much. I really appreciate your help. Stupid mistake on my part.

Woot4Moo · Accepted Answer · 2013-01-31 22:00:44Z

2

I would recommend using Collection utils Disjunction

Returns a Collection containing the exclusive disjunction (symmetric difference) of the given Collections.

The cardinality of each element e in the returned Collection will be equal to max(cardinality(e,a),cardinality(e,b)) - min(cardinality(e,a),cardinality(e,b)).

This is equivalent to subtract(union(a,b),intersection(a,b)) or union(subtract(a,b),subtract(b,a)).

answered Jan 31, 2013 at 22:00

Woot4Moo

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Comments

Peter Elliott · Accepted Answer · 2013-01-31 22:02:13Z

2

it may be advantageous to sort your lists before doing these comaprisons, then searching for items can be performed much more efficiently.

you can also try doing this:

set1.removeAll(set2)

items remaining in set1 were the items not in set2

answered Jan 31, 2013 at 22:02

Peter Elliott

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