Fastest way to determine the lowest available key in Java HashMap?

问题

Imagine a situation like this: I have a HashMap<Integer, String>, in which I store the connected clients. It is HashMap, because the order does not matter and I need speed. It looks like this:

{
    3: "John",
    528: "Bob",
    712: "Sue"
}

Most of the clients disconnected, so this is why I have the large gap. If I want to add a new client, I need a key and obviously the usage of _map.size() to get a key is incorrect.

So, currently I use this function to get he lowest available key:

private int lowestAvailableKey(HashMap<?, ?> _map) {
    if (_map.isEmpty() == false) {
        for (int i = 0; i <= _map.size(); i++) {
            if (_map.containsKey(i) == false) {
                return i;
            }
        }
    }

    return 0;
}

In some cases, this is really slow. Is there any faster or more professional way to get the lowest free key of a HashMap?

回答1:

Any reason to use a HashMap? If you used TreeMap instead, the map would be ordered by key automatically. Yes, you end up with O(log n) access instead of O(1), but it's the most obvious approach.

Of course you could always maintain both a HashMap and a TreeSet, making sure you add entries and remove entries from both together, if you really needed to. The TreeSet would just act as an ordered set of keys for the map.

来源：https://stackoverflow.com/questions/18345131/fastest-way-to-determine-the-lowest-available-key-in-java-hashmap