How to check if exists any duplicate in Java 8 Streams?

Question

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?

Answer 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); } 

Answer 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.