How do I convert a Java 8 IntStream to a List?

You can use the collect method:

IntStream.of(1, 2, 3).collect(ArrayList::new, List::add, List::addAll);

In fact, this is almost exactly what Java is doing when you call .collect(Collectors.toList()) on an object stream:

public static <T> Collector<T, ?, List<T>> toList() {
    return new Collectors.CollectorImpl(ArrayList::new, List::add, (var0, var1) -> {
        var0.addAll(var1);
        return var0;
    }, CH_ID);
}

Note: The third parameter is only required if you want to run parallel collection; for sequential collection providing just the first two will suffice.

You could also use mapToObj() on a Stream, which takes an IntFunction and returns an object-valued Stream consisting of the results of applying the given function to the elements of this stream.

List<Integer> intList = myIntStream.mapToObj(i->i).collect(Collectors.toList());

IntStream::boxed

IntStream::boxed turns an IntStream into a Stream<Integer>, which you can then collect into a List:

theIntStream.boxed().collect(Collectors.toList())

The boxed method converts the int primitive values of an IntStream into a stream of Integer objects. The word "boxing" names the int ⬌ Integer conversion process. See Oracle Tutorial.

Java 16 and later

Java 16 brought the shorter toList method. Produces an unmodifiable list. Discussed here.

theIntStream.boxed().toList() 

You can use primitive collections available in Eclipse Collections and avoid boxing.

MutableIntList list = 
    IntStream.range(1, 5)
    .collect(IntArrayList::new, MutableIntList::add, MutableIntList::addAll);

Note: I am a contributor to Eclipse Collections.