How do I get all the items from Java list from specified position ignoring all its previous without using IndexOf/sublist() method?

Question

KeyValueAccess kva = KeyValueAccessFactoryForTestConfig.getConnection(); List> keySets = kva.getDistinctPartitionKeys(SavepointDto.class, Integer.valueOf(50));

    List<Pair<String, String>> pairs =
            keySets.stream()
                    .sorted((k1, k2) -> SavepointUtils.compareMinitues(k1.get(1), k2.get(1)))
                    .sorted((k1, k2) -> k1.get(0).compareTo(k2.get(1)))
                    .map(keySet -> Pair.of(keySet.get(0), keySet.get(1)))
                    .collect(Collectors.toList());

    System.out.println("\n");

    int index = pairs.indexOf("2016:6:29:13:27");

    for (int i = index; i < pairs.size(); i++) {
        System.out.println(pairs.get(i));
    }

}

I'm Having a List<Pair<String, String>> when I'm trying to get index of any specific element from list I'm getting ArrayIndexOutofBoundException (See above code) in my case the 1st element is date (ex: 2016:6:29:13:27 ) of String type. so can't use indexOf or substring() method.

Is there any other way to get all the elements from list from specific position by ignoring previous elements???

Answer 1

You can use stream for that ( skip first 2 elements):

Arrays.asList("a","b","c","d","e").stream()
    .skip(2)
    .collect(Collectors.toList())

Or use subList :

    List<String> list = Arrays.asList("a","b","c","d","e");
    list.subList(2, list.size());