Java: Split array by value

Question

I'm trying to split an array into subarrays by value. For example (with the main function delaration and class name removed):

int[] arr = {1, 5, 3, 4, 2, 2, 5, 3, 1};
for (int i : splitByVal(arr)) {
    System.out.println(i);
}

I already have one way to do this:

public int[][] splitByVal(arr) {
    int prevItem = arr[0];
    ArrayList<Integer[]> chunks = new ArrayList<Integer[]>();
    ArrayList<Integer> currentChunk = new ArrayList<Integer>();
    for (int i : arr) {
        if (i == prevItem) {
            currentChunk.add(i);
        }
        else {
            chunks.add((Integer[]) currentChunk.toArray());
            currentChunk.clear();
        }
        prevItem = i;
    }
    return chunks.toArray()
}

What I want to know is: Is there a better way to do this?

what is slope? if current element is not the same as prevItem then prevItem = i;. and you should put .add(i) outside of if-else block — mangusta
– mangusta, Commented May 4, 2020 at 4:58
@mangusta Sorry, I was reading something else while writing this question. slope is supposed to be i. — fwoosh
– fwoosh, Commented May 5, 2020 at 3:48

score 1 · Accepted Answer · 2020-05-10 10:06:11Z

1

As it was correctly mentioned in the comment you haven't provided definition of the slope in your code. So it's not completely clear what your code is doing.

However, this is just one of the possible ways to do it if I understood you correctly:

public int[][] splitByValue(final int[] arr) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int number : arr) {
        map.put(number, map.getOrDefault(number, 0) + 1);
    }
    int[][] result = new int[map.keySet().size()][];
    int i = 0;
    for (Integer key : map.keySet()) {
        result[i] = new int[map.get(key)];
        for (int j = 0; j < result[i].length; j++) {
            result[i][j] = key;
        }
        i++;
    }        
    return result;
}

The assumption here is that you are only dealing with array of integers.

edited May 10, 2020 at 10:06

answered May 4, 2020 at 5:15

user671150

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