For the sake of the example lets assume you have a class Person with just a name.

private class Person {
    public String name;

    public Person(String name) {
        this.name = name;
    }
}

Jackson (Maven)

My personal favourite and probably the most widely used.

ObjectMapper mapper = new ObjectMapper();

// De-serialize to an object
Person user = mapper.readValue("{\"name\": \"John\"}", Person.class);
System.out.println(user.name); //John

// Read a single attribute
JsonNode nameNode = mapper.readTree("{\"name\": \"John\"}");
System.out.println(nameNode.get("name").asText());

Google GSON (Maven)

Gson g = new Gson();

// De-serialize to an object
Person person = g.fromJson("{\"name\": \"John\"}", Person.class);
System.out.println(person.name); //John

// Read a single attribute
JsonObject jsonObject = new JsonParser().parseString("{\"name\": \"John\"}").getAsJsonObject();
System.out.println(jsonObject.get("name").getAsString()); //John

Org.JSON (Maven)

This suggestion is listed here simply because it appears to be quite popular due to stackoverflow reference to it. I would not recommend using it as it is more a proof-of-concept project than an actual library.

JSONObject obj = new JSONObject("{\"name\": \"John\"}");

System.out.println(obj.getString("name")); //John

I had a situation where I was using a custom deserializer, but I wanted the default deserializer to do most of the work, and then using the SAME json do some additional custom work. However, after the default deserializer does its work, the JsonParser object current location was beyond the json text I needed. So I had the same problem as you: how to get access to the underlying json string.

You can use JsonParser.getCurrentLocation.getSourceRef() to get access to the underlying json source. Use JsonParser.getCurrentLocation().getCharOffset() to find the current location in the json source.

Here's the solution I used:

public class WalkStepDeserializer extends StdDeserializer<WalkStep> implements
    ResolvableDeserializer {

    // constructor, logger, and ResolvableDeserializer methods not shown

    @Override
    public MyObj deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException,
            JsonProcessingException {
        MyObj myObj = null;

        JsonLocation startLocation = jp.getCurrentLocation();
        long charOffsetStart = startLocation.getCharOffset();

        try {
            myObj = (MyObj) defaultDeserializer.deserialize(jp, ctxt);
        } catch (UnrecognizedPropertyException e) {
            logger.info(e.getMessage());
        }

        JsonLocation endLocation = jp.getCurrentLocation();
        long charOffsetEnd = endLocation.getCharOffset();
        String jsonSubString = endLocation.getSourceRef().toString().substring((int)charOffsetStart - 1, (int)charOffsetEnd);
        logger.info(strWalkStep);

        // Special logic - use JsonLocation.getSourceRef() to get and use the entire Json
        // string for further processing

        return myObj;
    }
}

And info about using a default deserializer in a custom deserializer is at How do I call the default deserializer from a custom deserializer in Jackson

If you are using maven, you can add the following dependency

<dependency>
   <groupId>javax</groupId>
   <artifactId>javaee-api</artifactId>
   <version>7.0</version>
</dependency>

you can also download the artefact manually here

http://central.maven.org/maven2/javax/javaee-api/7.0/javaee-api-7.0.jar

For Gradle, use

compile group: 'javax', name: 'javaee-api', version: '7.0'

The site https://mvnrepository.com shows you the different dependencies for different build systems.

Please note that this is only one possible dependency. You can certainly find smaller dependencies, which only contain the classes you want. Simply search for it on google or mavencentral.

The object in your file has exactly one property, named numbers.
There is no natural property.

You probably want to examine the objects inside that array.

The gson api is better and its extremely simple to use.