Comparing Java Lists

Question

How can I return true if list1 and list2 share at least 1 item?

Example: list1 = (1,2,3) ... list2 = (2,3,4)

someFunction(list1, list2); // returns true

possible duplicate of [Java Compare Two Lists](http://stackoverflow.com/questions/2762093/java-compare-two-lists) — , Apr 16 '13 at 05:06
@chris, what user `loki` below said, but it seemed non-optimal. — Kevin Meredith, Apr 16 '13 at 05:10

TofuBeer · Accepted Answer · 2013-04-16T05:40:40.837

8

Take a look at the Collections.disjoint method. If it is true there are no items in common.

edited Apr 16 '13 at 05:40

answered Apr 16 '13 at 05:09

TofuBeer

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+1 Looks like that method attempts to optimize by deciding which collection to iterate and which to check with `contains`, based on a few factors. [\[source\]](http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/7-b147/java/util/Collections.java#Collections.disjoint%28java.util.Collection%2Cjava.util.Collection%29) – Paul Bellora Apr 16 '13 at 05:23

jcreason · Answer 2 · 2013-04-16T05:39:48.030

1

Like TofuBeer said, take a look at Collections.disjoint (I'd upvote his if I had any reputation...):

public void main() {
    List<Integer> list1 = Arrays.asList(1,2,3);
    List<Integer> list2 = Arrays.asList(2,3,4); 
    someFunction(list1, list2);
}

private boolean someFunction(List<Integer> list1, List<Integer> list2) {
    return ! Collections.disjoint(list1, list2);
}

edited Apr 16 '13 at 05:39

answered Apr 16 '13 at 05:22

jcreason

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score 0 · Answer 3 · answered Apr 16 '13 at 05:09

0

Iterate over one list using "Iterator" and use contains method of other list to check the element. Isn't it?

answered Apr 16 '13 at 05:09

Lokesh

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I thought of doing that, but figured there must be a better way. – Kevin Meredith Apr 16 '13 at 05:10

shiladitya · Answer 4 · 2013-04-16T05:20:04.710

0

public boolean someFunction(List<T> l1,List<T> l2)
{
    Iterator<T> i = l1.iterator();
    while(i.hasNext())
    {
        if(l2.contains(i.next())
            return true;
    }
    return false;
}

edited Apr 16 '13 at 05:20

answered Apr 16 '13 at 05:12

shiladitya

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score 0 · Answer 5 · answered Apr 16 '13 at 05:26

If space is not an issue, why not just use a HashMap?

Iterator it1 = list1.iterator(), it2 = list2.iterator();
Map <K, Integer> listmap = new HashMap <K, Integer> ();

while (it1.hasNext() && it2.hasNext()) {
    K elem1 = it1.next(), elem2 = it2.next();
    if ((listmap.get(elem1) != null && listmap.get(elem1) == 2) || 
        (listmap.get(elem2) != null && listmap.get(elem2) == 1)) {
        return false;
    }
    else {
        listmap.put(elem1, 1);
        listmap.put(elem2, 2);
    }

}
return true

This way, you don't have to loop through the entire second array to check each element of the first one, since adding elements to a hash table happens in amortized constant time.

By the way, a faster solution would be to use IntHashMap from Apache commons (or SparseArray on Android).

Comparing Java Lists

5 Answers5