Groovy: the simplest way to detect duplicate, non-consecutive values in a list

This should do it:

List list = ["a", "b", "c", "a", "d", "c", "a"]

and

list.countBy{it}.grep{it.value > 1}.collect{it.key}

To determine whether a collection contains non-unique items (your first two examples), you can do something like this:

def a = [1, 2, 3, 1]   
boolean nonUnique = a.clone().unique().size() != a.size()

(Note that unique() modifies the list).

Meanwhile, Collection.unique() seems to do what you asked as far as 'grouping' items (your last three examples).

Edit: unique() works properly regardless of whether the collection is sorted.

In case you need to obtain duplicate elements:

    def nonUniqueElements = {list ->
        list.findAll{a -> list.findAll{b -> b == a}.size() > 1}.unique()
    }

    assert nonUniqueElements(['a', 'b', 'b', 'c', 'd', 'c']) == ['b', 'c']