ArrayList 中可能匹配项的 Java 迭代

Question

我的问题是关于迭代和性能的。让我们考虑以下情况:

public class Car {

    private String name;
    private int type;
    private int horsePower;

    String getKey() {
        return type + "_" + horsePower;
    }

    private final int NUM_OF_CARS = 50000;

    public void test() {
        List<Car> cars = new ArrayList<Car>(NUM_OF_CARS);

        for (int i = 0; i < NUM_OF_CARS; i++) {
            Car c = new Car();

            if (i == 0 || i == 176 || i == 895 || i == 1500 || i == 4600) {
                c.name = "Audi A4 " + i;
                c.type = 1;
                c.horsePower = 200;
            } else {
                c.name = "Default";
                c.type = 2 + i;
                c.horsePower = 201;
            }

            cars.add(c);
        }

        // Matches should contain all Audi's since they have same type and horse
        // power

        long time = SystemClock.currentThreadTimeMillis();

        HashMap<String, List<Car>> map = new HashMap<String, List<Car>>();

        for (Car c : cars) {
            if (map.get(c.getKey()) != null) {
                map.get(c.getKey()).add(c);
            } else {
                List<Car> list = new ArrayList<Car>();
                list.add(c);
                map.put(c.getKey(), list);
            }
        }

        Iterator<Entry<String, List<Car>>> iterator = map.entrySet().iterator();

        while (iterator.hasNext()) {
            if (iterator.next().getValue().size() == 1) {
                iterator.remove();
            }
        }

        Log.d("test", String.valueOf((SystemClock.currentThreadTimeMillis() - time)));
    }
}

这是在这里找到所有 Audi 的最有效方法吗？

这花了我 1700 毫秒

谢谢。

Answer 1

这取决于您进行迭代的原因。如果您确实需要访问每个底层 Car，那么您真的别无选择。但是，如果您要查找与字符串的特定匹配项，则可以考虑使用 Map。