nimo23 :
I have a set to sort (with Comparators) and I dont know which version to choose:
version 1:
public static void sort(Set<User> users) {
users = users.stream()
.sorted(sort_gender.thenComparing(sort_age))
.collect(Collectors.toCollection(LinkedHashSet::new));
}
version 2:
public static Set<User> sort(Set<User> users) {
return users.stream()
.sorted(sort_gender.thenComparing(sort_age))
.collect(Collectors.toCollection(LinkedHashSet::new));
}
version 3:
public static void sort(Set<User> users) {
users.stream()
.sorted(sort_gender.thenComparing(sort_age))
.collect(Collectors.toSet());
}
version 4
public static List<User> sort(Set<User> users){
List<User> list = new ArrayList<>(users);
list.sort(sort_gender.thenComparing(sort_age));
return list;
}
All versions sorts a set and returns the sorted set. I know, only linkedHashSet can preserve ordering.
Which one should I choose, I only want to sort the input properties users and return it, so is version 1 the best for that case? (For all cases, I want the references of input users be the same as for output users.)
EDIT: I think, I will choose version 4.
Eugene :
I would add a 4-th method (if you are OK to change that method to return the sorted Set)
users.stream()
.collect(Collectors.toCollection(() -> new TreeSet<>(Comparator.comparing...)))
I would return a SortedSet to make it explicit for the caller that this is actually sorted.
If not you could do:
SortedSet<User> sorted = new TreeSet<>(Comparator.comparing...)
sorted.addAll(users);
Guess you like
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