将2D数组拆分为较小的相等大小的2D子数组

Question

Basically what I am asking is given a square 2D array and a valid patch size (the size of the 2D subarrays) how would I go about doing this. Ultimately I don't need to store the subarrays in any way, I just need to find the median of each subarray and store them in a 1D array. The median and storing to new array are simple for me, I just can't figure out how to go about the original 2D array and splitting it properly. I've attempted this several times and keep getting out of bounds errors. I have a 4x4:

[1,2,3,4] [2,3,4,1] [3,4,1,2] [4,1,2,3]

I need to split it like so

[1,2] [3,4] [2,3] [4,1]

[3,4] [1,2] [4,1] [2,3]

And then take the median of each and store them into a new 1D array.

EDIT: Solved, thanks for the help!

Answer 1

You can use Arrays.copyOfRange(Object[] src, int from, int to) for this where:

src is the source 1D array

from is the initial index of the range to be copied, inclusive.

to is the final index of the range to be copied, exclusive.

I don't prefer your code because it seems that it's time complexity is too high.

Try below code:

public class Temp {
    public static void main(String[] args) {
        int[][] arr = { { 1,2,3,4 },
                        { 2,3,4,1 },
                        { 3,4,1,2 },
                        { 4,1,2,3 } };

        int patch = 2;

        splitToSubArrays(arr, patch);

    }

    static void splitToSubArrays(int arr[][], int patch) {
        for (int i = 0; i < arr[0].length; i++) {
            int to = patch;
            for (int from = 0; to <= arr.length;) {
                int a[] = Arrays.copyOfRange(arr[i], from, to);
                // instead of printing you can store in a separate array for later usage
                System.out.println(Arrays.toString(a));
                to += patch;
                from += patch;
            }
        }
    }

}

EDIT: NB: For n*n array if n%patch is not 0 ie dimension is not divisible by patch value then you would need to use proper if condition here int a[] = Arrays.copyOfRange(arr[i], from, to); to control the index bound. Hope you are aware of this.

Output

[1, 2]
[3, 4]
[2, 3]
[4, 1]
[3, 4]
[1, 2]
[4, 1]
[2, 3]

Answer 2

I think something like this could do, altough i didn't test it, but It should give you a good idea of how to do this scan

public int[] patchArray(int[][] img, int patch)
{
    int size = img.length * (img[0].length / patch) ;
    int[] pArray = new int[size];
    int[] tmp = new int[patch];

    for (int row_i = 0; row_i < img.length; row_i++)
    {
        for (int patch_start = 0; patch_start < img[i].length; patch_start += patch)
        {
            int x = 0;
            for (int patch_i = patch_start; patch_i < (patch_start + patch); patch_i++)
            {
                tmp[patch_i - patch_start] = img[row_i][patch_i];
            }

            calculateMedian(tmp);
        }
    }

    return pArray;
}