泛型创建实例usinc jackson时，如何在Java中传递方法参数中类的类型信息？

Question

我有一个http端点，它返回不同类型的json，比如汽车、自行车、卡车。 JSON的结构如下所示，每个类都非常相似

{
    "Rows": [
        {
            "key1": "val1",
            "key2": "val2",
        }
    ]
}

我想编写一个基于类型参数创建列表对象（Car、Bike、Truck）的通用方法（我已经使用适当的jackson绑定创建了所有pojo，它们扩展了Vehicle类）

我提出了以下内容，但它在运行时抛出异常

static <T extends Vehicle> List<T> getTiles(String vehicleName, Class<T> concreteClass) throws IOException {
    Map<String, String> queryParamMap = new HashMap<>();
    queryParamMap.put("apiKey", "<KEY>");
    queryParamMap.put("type", vehicleName);
    

    ObjectMapper mapper = new ObjectMapper().configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

    OkHttpClient client = new OkHttpClient();
    HttpUrl.Builder builder = new HttpUrl.Builder()
            .scheme("https")
            .host("<URL>")
            .addPathSegment("1");
            

    queryParamMap.entrySet().stream().forEach(k -> builder.addQueryParameter(k.getKey(), k.getValue()));
    HttpUrl url = builder.build();

    Request request = new Request.Builder()
            .addHeader("accept", "application/json")
            .url(url).build();

    Response response = client.newCall(request).execute();
    Map<String, List<T>> m = mapper.readValue(response.body().string(),
            new TypeReference<Map<String, List<T>>>() {
            });

    return m.get("Rows");
}

我把它叫做

List<Car> cars = Utils.getTiles("CAR", Car.class);

我得到以下例外

Exception in thread "main" com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of `com.example.models.Vehicle` (no Creators, like default construct, exist): abstract types either need to be mapped to concrete types, have custom deserializer, or contain additional type information

我知道这与类型擦除有关，因为在运行时T绑定到车辆而不是汽车

我如何解决这个问题