Check if an ArrayList contains every element from another ArrayList (or Collection)

Question

There is probably a simple one-liner that I am just not finding here, but this is my question:

How do I check if an ArrayList contains all of the objects in another

Answer 1

Per the List interface:

myList.containsAll(...);

Answer 2

You can use containsAll method of the list to do the check. However, this is a linear operation. If the list is large, you should convert it to HashSet first, and then perform containsAll:

HashSet tmp = new HashSet(one);
if (tmp.containsAll(two)) {
    ...
}

If the length of one is N and the length of two is M, this solution has time complexity of O(M+N); the "plain" containsAll has the complexity of O(M*N), which may be significantly worse.

Answer 3

Here is another example use of containsAll() that I have used for asserting that two arrays are equal in JUnit testing:

List<String> expected = new ArrayList<String>();
expected.add("this");
expected.add("that");
expected.add("another");

List<String> actual = new ArrayListString();
actual.add("another");
actual.add("that");
actual.add("this");

Assert.assertTrue("The lists do not match!", expected.containsAll(actual));

Answer 4

Take a look at containsAll(Collection<?> c) method from List interface. I think it is what you are looking for.

Answer 5

There is a method called containsAll declared in the java.util.Collection interface. In your setting one.containsAll(two) gives the desired answer.

Answer 6

Your code in the example doesn't make sense, but here's an example anyway.

ArrayList<Integer> one, two;
//initialize
boolean good = true;
for (int i = 0; i < two.size(); i ++) {
    if (!(one.contains(two.get(i))) {
        good = false;
        break;
    }
}

It simply loops through all of two's elements and checks to see if they are in one.

Then the boolean good contains the value you want.

See ArrayList#contains.

EDIT: oh wow, I totally forgot containsAll. Oh well, this is an alternate way to do it if you really want to understand it.