Java - Java8中的JVM垃圾收集器

Question

I have a basic SpringBoot 2.1.5.RELEASE app. Using Spring Initializer, JPA, embedded Tomcat. In terms of the Java Garbage Collection I guess that the option B of the service is better than the option A, since the object does not remain in the memory, right ?

option A:

@Override
    public List<HotelPriceSummary> overRanked7d(User user) {

        List<HotelPriceSummary> overRanked7dList =

                allNonFavoritedHotels(user)
                        .parallelStream()
                        .filter(HotelPriceSummary.overRanked7dHotelsPredicate())                            
                        .sorted(comparing((HotelPriceSummary cps) -> cps.getDailyPercentageChange()).reversed())
                        .collect(toList());

        return overRanked7dList;
    }

option B:

@Override
    public List<HotelPriceSummary> overRanked7d(User user) {

        return

                allNonFavoritedHotels(user)
                        .parallelStream()
                        .filter(HotelPriceSummary.overRanked7dHotelsPredicate())                            
                        .sorted(comparing((HotelPriceSummary cps) -> cps.getDailyPercentageChange()).reversed())
                        .collect(toList());


    }

Answer 1

From JVM implementation point of view, the only difference is declaring a new object reference to the returned object. Hence, no new object is being created in either case and no new memory is being used. From coding perspective, option B is preferred.