How to check if exists any duplicate in Java 8 Streams?
In java 8, what's the best way to check if a List contains any duplicate?
My idea was something like:
list.size() != list.stream().distinct().count()
Is it the best way?
Solution 1:
Your code would need to iterate over all elements. If you want to make sure that there are no duplicates simple method like
public static <T> boolean areAllUnique(List<T> list){
Set<T> set = new HashSet<>();
for (T t: list){
if (!set.add(t))
return false;
}
return true;
}
would be more efficient since it can give you false
immediately when first non-unique element would be found.
This method could also be rewritten as (assuming non-parallel streams and thread-safe environment) using Stream#allMatch
which also is short-circuit (returns false immediately for first element which doesn't fulfill provided condition)
public static <T> boolean areAllUnique(List<T> list){
Set<T> set = new HashSet<>();
return list.stream().allMatch(t -> set.add(t));
}
or as @Holger mentioned in comment
public static <T> boolean areAllUnique(List<T> list){
return list.stream().allMatch(new HashSet<>()::add);
}
Solution 2:
I used the following:
1. return list.size() == new HashSet<>(list).size();
.
I'm not sure how it compares to:
2. return list.size() == list.stream().distinct().count();
and
3. return list.stream().sequential().allMatch(new HashSet<>()::add);
in terms of performance.
The last one (#3) has possibility to handle not only collections (e.g. lists), but also streams (without explicitly collecting them).
Upd.: The last one (#3) seems to be the best not only because it can handle pure streams, but also because it stops on the first duplicate (while #1 and #2 always iterate till the end) — as @Pshemo said in comment.