确定Java HashMap中最低可用密钥的最快方法? [英] Fastest way to determine the lowest available key in Java HashMap?

问题描述

想象一下这样的情况: 我有一个HashMap<Integer, String>，用于存储连接的客户端.它是HashMap，因为顺序无关紧要，我需要速度.看起来像这样:

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"
}

大多数客户端都断开连接，因此这就是我有很大差距的原因. 如果要添加新客户端，则需要密钥，显然使用_map.size()来获取密钥是不正确的.

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.

因此，当前，我使用此功能来获取最低可用密钥:

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;
}

在某些情况下，这确实很慢. 是否有更快或更专业的方法来获取HashMap的最低可用密钥?

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

推荐答案

是否有使用HashMap的理由?如果您使用的是 TreeMap ，地图将通过按键自动排序.是的，您最终获得的是O(log n)而不是O(1)的访问权限，但这是最明显的方法.

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.

当然，您始终可以同时维护HashMap 和 TreeSet，如果确实需要，请确保同时添加条目和从条目中删除条目. TreeSet只是充当地图的有序键集.

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.

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