Fixed SubList Count but Dynamic Members in Java

Question

I have the following Integer list

List<Integer> arrayList = new ArrayList<Integer>();
   for (int i = 0; i < 7; i++) {
    arrayList.add(i);
}

So the list is like this [0,1,2,3,4,5,6]. My scenario is

If I give the value = 5 as parameter then I would like to split 5 sub list like this

[0,5], [1,6] , [2], [3], [4]

If I give the value = 4 as parameter then I would like to split 4 sub list like this

[0,4], [1,5], [2,6] , [3]

If I give the value = 3 as parameter then I would like to split 3 sub list like this

[0,3,6], [1,4], [2,5]

I already tested with below function but it is not my need.

public List<List<Integer>> chopped(List<Integer> list, final int splitCount) {
        List<List<Integer>> parts = new ArrayList<List<Integer>>();
        final int N = list.size();
        for (int i = 0; i < N; i += splitCount) {
            parts.add(new ArrayList<Notification>(list.subList(i, Math.min(N, i + splitCount))));
        }
        return parts;
    }

At the above function, I give splitCount to 5 then the function returns

[0,1,2,3,4], [5,6]

The result I expect is [0,5], [1,6] , [2], [3], [4]

Could you please specify the rule that you want to put into place? Instead of just giving examples?I thought I figured your requirement - but then I saw the N=3 example. — GhostCat
– GhostCat, Commented Aug 21, 2017 at 8:18
@GhostCat Sorry i do not understand what you said? Which parts don't you understand in my question. Sorry my English is not good. — Sai Ye Yan Naing Aye
– Sai Ye Yan Naing Aye, Commented Aug 21, 2017 at 8:24
I am saying: do not describe your requirement via examples only. You should also describe the requirement directly. I do not understand how to use "list-length" and "split factor" to determine the expected result. — GhostCat
– GhostCat, Commented Aug 21, 2017 at 8:27

Maurice Perry · Accepted Answer · 2017-08-21 08:23:20Z

6

How about:

public List<List<Integer>> chopped(List<Integer> list, final int splitCount) {
    List<List<Integer>> parts = new ArrayList<>(splitCount);
    for (int i = 0; i < splitCount; ++i) {
        parts.add(new ArrayList<>());
    }
    final int N = list.size();
    for (int i = 0; i < N; ++i) {
        parts.get(i % splitCount).add(list.get(i));
    }
    return parts;
}

answered Aug 21, 2017 at 8:23

Maurice Perry

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1 Comment

GhostCat Over a year ago

You probably want to add an explanation that the whole topic is about doing some sort of "module" distribution of the elements. And: I like your answer - it is simple and easy to understand!

WSM RWL · Accepted Answer · 2017-08-21 10:10:40Z

0

Simple Solution Hope This Helps

    ArrayList<Integer> al1 = new ArrayList<>();
    ArrayList<ArrayList<Integer>> al2 = new ArrayList<>();
    // Add objects in  al1 ...
    int numberToSplit = 2;
    if (numberToSplit > al1.size()) {
        throw new Exception("Your message");

    } else {
        ArrayList<Integer> newArr;
        int counter = 0;
        for (int j = 0; j < numberToSplit; j++) {
            newArr = new ArrayList<>();
            for (int k = counter; k < al1.size(); k = k + numberToSplit) {
                newArr.add(ALI.get(k));

            }

            al2.add(newArr);
            counter++;

        }
    }

answered Aug 21, 2017 at 10:10

WSM RWL

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Comments

Boris van Katwijk · Accepted Answer · 2017-08-21 10:12:48Z

0

A functional implementation using Streams without variables:

public static void main(String[] args) {
    List<Integer> asList = Arrays.asList(0, 1, 2, 3, 4, 5, 6);
    System.out.println(chopped(asList, 3));
}

private static List<List<Integer>> chopped(List<Integer> source, int splitCount) {
    return IntStream.range(0, splitCount)
        .boxed()
        .map(index -> getElements(source, index, splitCount))
        .collect(Collectors.toList());
}

private static List<Integer> getElements(List<Integer> source, int index, int splitCount) {
    return IntStream.range(0, source.size())
            .filter(i -> i % splitCount  == index)
            .map(i -> source.get(i))
            .boxed()
            .collect(Collectors.toList());
}

edited Aug 21, 2017 at 10:12

answered Aug 21, 2017 at 8:28

Boris van Katwijk

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