将二维数组索引转换为一维索引

Question

I have two arrays for a chess variant I am coding in java...I have a console version so far which represents the board as a 1D array (size is 32) but I am working on making a GUI for it and I want it to appear as a 4x8 grid, so I have a 2-dimensional array of JPanels...

Question is, is there any formula that can convert the array[i][j] index into array[i] given the fact its a 4x8 array?

Answer 1

Think of it this way:

You have one array that happens to be a 1 dimensional array, which really, is just a long concatenation of items of a two dimensional array.

So, say you have a two dimensional array of size 5 x 3 (5 rows, 3 columns). And we want to make a one dimensional array. You need to decide whether you want to concatenate by rows, or by columns, for this example we'll say the concatenation is by rows. Therefore, each row is 3 columns long, so you need to think of your one-dimensional array as being defined in "steps" of 3. So, the lengthy of your one dimensional array will be 5 x 3 = 15, and now you need to find the access points.

So, say you are accessing the 2nd row and the 2nd column of your two dimensional array, then that would wind up being 3 steps (first row) + the number of steps in the second row, or 3 + 2 = 5. Since we are zero-based indexing that is -1, so that would be at index 4.

Now for the specific formulation:

int oneDindex = (row * length_of_row) + column; // Indexes

So, as an example of the above you would wind up having

oneDindex = (1 * 3) + 1

And that should be it

Answer 2

Given 4 columns by 8 rows then:

i = row * 4 + col

EDIT: My bad, nobody caught me on this mistake apparently. But it should actually be row * 4 + col .

row * 8 + col would leave unnecessary gaps in the possible indexes.

Answer 3

Every row in your 2D array is placed end to end in your 1D array. i gives which row you are in, and j gives the column (how far into that row). so if you are in the ith row, you need to place i complete rows end to end, then append j more onto that to get your single array index.

So it will be something like
singleDimIndex = array[0].length * i + j

Answer 4

i*8+j （假设8是水平宽度）

Answer 5

You could use this ArrayConvertor class to convert 2D arrays in 1D arrays and back.

Beware: Converting a 2D array to a normal one does only work with a matrix.

public class ArrayConvertor {
    static public int[] d2Tod1(int[][] array){

        int[] newArray = new int[array.length*array[0].length];

        for (int i = 0; i < array.length; ++i) 
        for (int j = 0; j < array[i].length; ++j) {
            newArray[i*array[0].length+j] = array[i][j];
        }

        return newArray;
    }

    static public int[][] d1Tod2(int[] array, int width){

        int[][] newArray = new int[array.length/width][width];

        for (int i = 0; i < array.length; ++i) {
           newArray[i/width][i%width] = array[i];
        }

        return newArray;
    }
}

And Some testing code:

public class JavaMain{
    public static void main(String[] args) {
        int[][] arr2D_1 = new int[4][8];

        byte counter=0;
        for (int i = 0; i < 4; i++) 
        for (int j = 0; j < 8; j++) {
            arr2D_1[i][j] = counter++;
        }

        int[]arr1D = ArrayConvertor.d2Tod1(arr2D_1);
        int[][] arr2D_2 = ArrayConvertor.d1Tod2(arr1D, 8);

        boolean equal = true;
        for (int i = 0; i < arr2D_1.length; i++) 
        for (int j = 0; j < arr2D_1[0].length; j++){ 
            if(arr2D_1[i][j]!=arr2D_2[i][j]) equal=false;
        }

        System.out.println("Equal: "+equal);
    }
}

Output: Equal: true

Answer 6

If 2d array is 4 width . We have pos x and pos y of an element .

We need an index of it . How to do it ?

Look at array it has 12 elements .

if you want to go to element nr 6 ,you go to element 2 in row 2 .

1 2 3 4 5 6 7 8 9 10 11 12

column is 1,2,3,4 row is: 1, 5, 9,

element 1 is (0,0) in 2d .

so use this = pos.y * width + pos.x ;