Java-get most common element in a list

In statistics, this is called the "mode". A vanilla Java 8 solution looks like this:

Stream.of(1, 3, 4, 3, 4, 3, 2, 3, 3, 3, 3, 3)
      .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))
      .entrySet()
      .stream()
      .max(Map.Entry.comparingByValue())
      .ifPresent(System.out::println);

Which yields:

3=8

jOOλ is a library that supports mode() on streams. The following program:

System.out.println(
    Seq.of(1, 3, 4, 3, 4, 3, 2, 3, 3, 3, 3, 3)
       .mode()
);

Yields:

Optional[3]

For simplicity's sake, I omitted using BigDecimal. The solution would be the same, though.

(disclaimer: I work for the company behind jOOλ)

This is fairly easy to implement yourself:

public static <T> T mostCommon(List<T> list) {
    Map<T, Integer> map = new HashMap<>();

    for (T t : list) {
        Integer val = map.get(t);
        map.put(t, val == null ? 1 : val + 1);
    }

    Entry<T, Integer> max = null;

    for (Entry<T, Integer> e : map.entrySet()) {
        if (max == null || e.getValue() > max.getValue())
            max = e;
    }

    return max.getKey();
}

---

List<Integer> list = Arrays.asList(1,3,4,3,4,3,2,3,3,3,3,3);
System.out.println(mostCommon(list));

3

If you want to handle cases where there's more then one most frequent element, you can scan the list once to determine how many times the most frequent element(s) occur, and then scan the list again, put those elements in a set and return that.

Probably the simplest solution with Guava looks like

Multiset<BigDecimal> multiset = HashMultiset.create(listOfNumbers);
BigDecimal maxElement = null;
int maxCount = 0;
for (Multiset.Entry<BigDecimal> entry : multiset.entrySet()) {
  if (entry.getCount() > maxCount) {
    maxElement = entry.getElement();
    maxCount = entry.getCount();
  }
}

That's a complete solution, and shorter than the other alternatives I see discussed.