超时跳出方法代码的返回值问题

Question

public static Object callMethod(final Object target , final String methodName ,final Class<?>[] parameterTypes,final Object[]params,int times){
    ExecutorService executorService = Executors.newSingleThreadExecutor();
    FutureTask<String> future = new FutureTask<String>(new Callable<String>() {
        public String call() throws Exception {
            String value = null  ;
            try {
                Method method = null ;
                method = target.getClass().getDeclaredMethod(methodName , parameterTypes ) ;
                Object returnValue = method.invoke(target, params) ;
                value = returnValue != null ? returnValue.toString() : null ;
            } catch (Exception e) {
                throw e ;
            }
            return value ;
        }
    });
    executorService.execute(future);
    String result = null;
    try{
        result = future.get(times , TimeUnit.SECONDS );
    }catch (Exception e){
        System.out.println("当前终端无法连接");
        e.printStackTrace();
    }
    executorService.shutdownNow();
    return result ;
}

这段代码 是我在网上找的可以控制等待时间运行方法的代码 可如果要运行的方法有返回值的话 我该怎么接收呢