将二维数组从行转换为块 [英] convert an 2d array from rows to blocks

问题描述

我数周内都在尝试使用此代码.我需要将二维数组中的行和列转换为块.它应该适用于大小为 n*n 的任何矩阵.(我已经获得了数组的大小)例如:这个:

I am trying to work in this code for weeks. I need to convert rows and columns in a 2d array to blocks. it should work on any matrix in size n*n. (that I have been given the size of the array) for example: this:

int[][] input =        {{1,2,3,4,5,6,7,8,9},
                       {1,2,3,4,5,6,7,8,9},
                       {1,2,3,4,5,6,7,8,9},
                       {1,2,3,4,5,6,7,8,9},
                       {1,2,3,4,5,6,7,8,9},
                       {1,2,3,4,5,6,7,8,9},
                       {1,2,3,4,5,6,7,8,9},
                       {1,2,3,4,5,6,7,8,9},
                       {1,2,3,4,5,6,7,8,9}} ;

this is need to be the output:


{{1,2,3,1,2,3,1,2,3}
{4,5,6,4,5,6,4,5,6}
{7,8,9,7,8,9,7,8,9}
{1,2,3,1,2,3,1,2,3}
{4,5,6,4,5,6,4,5,6}
{7,8,9,7,8,9,7,8,9}
{1,2,3,1,2,3,1,2,3}
{4,5,6,4,5,6,4,5,6}
{7,8,9,7,8,9,7,8,9}}

我总是卡在某个点上.

这是我写的代码:

        sqrtN is the size.

在上面提到的情况下，这里的sqrtN是3.

in the case that mentioned above, the sqrtN here is 3.

`   public static int[][] blocks(int[][] matrix, int sqrtN) {   
    int[][] blocks = matrix;
    int i = 0;
    int counter = 1;
    while(counter+1<sqrtN){ 
        int n = sqrtN;//"n" will the size of the matrix.
        while(n<blocks.length){ 
            int t = counter;
            int j = 0;
            while(t<blocks.length){
                int k = n;
                if(t==counter & i%n==0 & i>0)
                    j= t-1;
                if(j%sqrtN==0)
                    i = 0;
                while(k==n || k%sqrtN!=0){
                    int temp = blocks[t][i];
                    blocks[t][i] = blocks[j][k];
                    blocks[j][k] = temp;
                    i= i+1;
                    k= k+1; 
                }
                j=j+sqrtN;
                t=t+sqrtN;          
            }
            n= n+sqrtN;
            counter= counter+1;

        }
        i=counter;
    }
        return blocks;`

我真的很高兴能找到问题的答案.

I will really be glad to find the answer the problem.

谢谢各位.

推荐答案

无需执行复杂的逻辑.尝试使用:

No Need to perform a complex logic. Try to use :

public static int[][] blocks(int[][] matrix, int sqrtN) {   
    int[][] blocks = matrix;
    for (int i = 0; i < matrix.length; i++) {
        for (int j = 0; j < matrix.length; j++){ 
            blocks[i][j] = (i % sqrtN * sqrtN) + (j % sqrtN + 1) ;
    }
}
        return blocks;
}

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