Necesita regex patrón de conseguir todo en el medio

Aleksandar L:

De esta necesidad de cuerdas que para obtener valor de autorización pero sin [\ "y \"]

[\"bla bla bla\"],\"x-user-id\":[\"\"],
\"accept\": [\"application/json, text/javascript, */*; q=0.01\"],
\"original-path\": [\"/test/test\"],
\"authorization\":[\"Bearer dadadadadadadadadadadadadadadadadadadadadadadadadadad\"],
\"requestUrl\":\"/test/test/test\",\"host\":[\"test.com\"]

He tratado con \[(.*?)\\]y devolver todos los valores entre [].

¿Cualquier sugerencia?

Azro:

Puede añadir la palabra authorizationen el patrón de obtener sólo el unoauthorization\\":\[\\"(.*?)\\"]

String value = "[\\\"bla bla bla\\\"],\\\"x-user-id\\\":[\\\"\\\"],\\\"accept\\\":[\\\"application/json, text/javascript, /; q=0.01\\\"],\\\"original-path\\\":[\\\"/test/test\\\"],\\\"authorization\\\":[\\\"Bearer dadadadadadadadadadadadadadadadadadadadadadadadadadad\\\"],\\\"requestUrl\\\":\\\"/test/test/test\\\",\\\"host\\\":[\\\"test.com\\\"]";
Pattern p = Pattern.compile("authorization\\\\\":\\[\\\\\"(.*?)\\\\\"]");
Matcher m = p.matcher(value);
if (m.find()) {
    System.out.println(m.group(1)); // Bearer dadadadadadadadadadadadadadadadadadadadadadadadadadad
}