Java中ContainsAll的成本是多少？

Question

I discovered containsAll() (a List interface method) during some coding today, and it looks pretty slick. Does anyone know how much this costs in terms of performance/iterations?

The documentation didn't offer much in terms of that.

Answer 1

• first, it iterates each element of the supplied collection
• then it iterates all elements of the list and compares the current element with them using .equals(..) (Note: this is about lists, as you specified in the question. Other collections behave differently)

So it's O(n*m), where n and m are the sizes of both collections.

public boolean containsAll(Collection<?> c) {
    Iterator<?> e = c.iterator();
    while (e.hasNext())
        if (!contains(e.next()))
        return false;
    return true;
}

Answer 2

Use the source, Luke :)

Edit: As Bozho pointed out, you're asking about List.containsAll() which overrides Collection.containsAll() . The following ramblings are mainly concerned with the latter:

Most Collection classes will use the implementation of containsAll from AbstractCollection , which does it like this:

public boolean containsAll(Collection<?> c) {
    for (Object e : c)
        if (!contains(e))
            return false;
    return true;
}

There's no guarantee that some implementation does it completely differently, though -- which could result in either better or worse runtime behavior.

The above implementation of containsAll will be at least O(n) where n is the number of items in the Collection parameter you pass in, plus whatever time contains takes:

• For a HashSet / HashMap this might be O(1) (best case, no collisions), so the overall runtime of containsAll would still be O(n)
• For an ArrayList , contains will take O(m) where m is the number items in the list (not the parameter), so the overall time for containsAll would be O(n*m)

Answer 3

if you are calling A.containsAll(B)

openjdk iterates all elements of B calling A.contains(b).

A.contains(b) iterates all elements of A calling a.equals(b)

This is taken from source for open jdk 7

Answer 4

consider

n.ContainsAll(m)

the best possible case scenario is O(m), and that's if n is a perfect hash set.

considering unsorted lists, i can come up with an O(n*log(n) + m*log(m)) algorithm