Java 8 Stream API

Question

I have this simple instruction

Stream.concat(manager.getChild().stream(), 
 manager1.getChild().stream())
 .map(dev -> dev.getSalary())
 .reduce(0, Integer::max);

that concats two List and return the developer that earning more. This returns the maximum salary of the objects in the stream, but how can I retrieve the object that has the maximum salary?

• Sergii Bishyr: Why are you using reduce here?
• Timothy Truckle: What do you mean by "all the object"?
• Jesper: Remove the map and reduce and use max instead with an appropriate Comparator that compares by salary.
• Dennis A. Boanini: I try to use this Employee rich = Stream.concat(manager.getChild().stream(), manager1.getChild().stream()) .max((dev1, dev2) -> Integer.max(dev1.getSalary(), dev2.getSalary())).get(); but not work very well.
• Joshua: Please, next time, choose a more fitting question title

Answer 1

Use Stream.max(Comparator<? super T> comparator) method:

Stream.concat(manager.getChild().stream(),
 manager1.getChild().stream())
 .max(Comparator.comparingInt(dev -> dev.getSalary())

Answer 2

Unfortunately you can't reach back to a previous step to get the stream element that provided the maximum salary; you have to compare the elements by an attribute.

To find the subordinate with the highest salary:

Employee max = Stream.of(manager, manager1)
 .map(Employee::getChild)
 .flatMap(Collection::stream)
 .max(Comparators.comparing(Employee::getSalary))
 .orElse(null);

I have assumed a class of Employee so method references can be used, and also refactored the code to a IMHO more usable/standard approach of starting by streaming the managers.

• ZhekaKozlov: flatMap is overkill here
• Bohemian: @ZhekaKozlov but it's more flexible: You can plug this flow on to any stream of managers, eg a List<Employee> via .stream(), or an Employee[] via Arrays.stream() etc. Conversely, hardcoding the stream calls is not scaleable or flexible and breaks DRY