Question

我正在尝试编写一个采用二维数组“Sudoku”的代码 (matrix.length * matrix.length)我应该检查矩阵是(n^2 * n^2)还是数字(n)，我应该将行排列成块(n * n)。有什么建议？如果有任何身体可以做到4“fors”我会非常高兴

public static int[][] blocks(int[][] matrix, int sqrtN) {
    int [][] returnedMatrix = new int [matrix.length][matrix.length];
    if (matrix.length !=  Math.pow(sqrtN, 2))
            throw new RuntimeException("Matrix length is not" + Math.pow(sqrtN, 2 ));
    for(int i=0 ; i <= matrix.length ;i=i+sqrtN) {
        for(int j=sqrtN; j <= matrix.length;j=j+1) {
            int temporarily = 

        }
    }
    return returnedMatrix;
}

例如

int[][] matrix1 = {{11,12,13,14},
{15,16,17,18},
{19,20,21,22},

{23,24,25,26}} ;

int[][] matBlocks1 = blocks (matrix1, 2) ;
/*
* matBlocks1 = {{11, 12, 15, 16},
* {13, 14, 17, 18},
* {19, 20, 23, 24},
* {21, 22, 25, 26}}
*/

Answer 1

如果您定义一个类Grid来表示一个Sudoku网格，一个类Cell来引用该网格的一个单元格，您可以使用一组单元格来访问行，列和块{ {1}}：

Cell[]

要列出所有块，您可以：

public class Grid {
    public final int N;
    public final int n;
    private final int[][] matrix;

    public class Cell {
        private final int row;
        private final int col;

        private Cell(int row, int col) {
            this.row = row;
            this.col = col;
        }

        public int get() {
            return matrix[row][col];
        }

        public void set(int value) {
            matrix[row][col] = value;
        }
    }

    public Grid(int[][] matrix) throws Exception {
        N = matrix.length;
        n = (int)Math.sqrt(N);
        if (n*n != N) {
            throw new IllegalArgumentException(
                    "Matrix size must be a square number");
        }
        this.matrix = new int[N][];
        for (int i = 0; i < N; ++i) {
            int[] row = matrix[i];
            if (row.length != N) {
                throw new IllegalArgumentException("Matrix must be a square");
            }
            int[] newRow = new int[N];
            for (int j = 0; j < N; ++j) {
                newRow[j] = row[j];
            }
            this.matrix[i] = newRow;
        }
    }

    public Cell cell(int row, int col) {
        return new Cell(row, col);
    }

    public Cell[] row(int row) {
        if (row < 0 || row >= N) {
            throw new IndexOutOfBoundsException("Invalid row number: " + row);
        }
        Cell[] result = new Cell[N];
        for (int col = 0; col < N; ++col) {
            result[col] = cell(row, col);
        }
        return result;
    }

    public Cell[] colum(int col) {
        if (col < 0 || col >= N) {
            throw new IndexOutOfBoundsException("Invalid row number: " + col);
        }
        Cell[] result = new Cell[N];
        for (int row = 0; row < N; ++row) {
            result[row] = cell(row, col);
        }
        return result;
    }

    public Cell[] block(int br, int bc) {
        if (br < 0 || br >= n) {
            throw new IndexOutOfBoundsException("Invalid block row: " + br);
        }
        if (bc < 0 || bc >= n) {
            throw new IndexOutOfBoundsException("Invalid block column: " + bc);
        }
        int startRow = br*n;
        int startCol = bc*n;
        Cell[] result = new Cell[N];
        int k = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                result[k++] = cell(startRow+i, startCol+j);
            }
        }
        return result;
    }
}

使用int[][] matrix1 = { {11, 12, 13, 14}, {15, 16, 17, 18}, {19, 20, 21, 22}, {23, 24, 25, 26}}; Grid grid = new Grid(matrix1); for (int i = 0; i < grid.n; ++i) { for (int j = 0; j < grid.n; ++j) { Cell[] block = grid.block(i, j); System.out.print("{"); for (int k = 0; k < block.length; ++k) { if (k > 0) { System.out.print(", "); } System.out.print(block[k].get()); } System.out.println("}"); } }代替Cell的优势的好处是可以让您更改单元格中的值。

Answer 2

您正在寻找的公式是：

int r = (i/sqrtN)*sqrtN+j/sqrtN;
int c = (i%sqrtN)*sqrtN+j%sqrtN;
returnedMatrix[i][j] = matrix[r][c];

包含测试的完整代码：

public static int[][] blocks(int[][] matrix, int sqrtN) {
    int n = matrix.length;
    if (n != Math.pow(sqrtN, 2))
        throw new RuntimeException("Matrix length is not" + Math.pow(sqrtN, 2));
    int[][] returnedMatrix = new int[n][n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            int r = (i/sqrtN)*sqrtN+j/sqrtN;
            int c = (i%sqrtN)*sqrtN+j%sqrtN;
            returnedMatrix[i][j] = matrix[r][c];
        }
    }
    return returnedMatrix;
}

public static void print(int[][] matrix) {
    for (int i = 0; i < matrix.length; i++) {
        for (int j = 0; j < matrix[i].length; j++) {
            System.out.print(matrix[i][j] + " ");
        }
        System.out.println();
    }
    System.out.println();
}

public static void main(String args[]) {
    int[][] matrix4 = {
        {1,1,2,2},
        {1,1,2,2},
        {3,3,4,4},
        {3,3,4,4}
    };
    print(blocks(matrix4, 2));

    int[][] matrix9 = {
        {1,1,1,2,2,2,3,3,3},
        {1,1,1,2,2,2,3,3,3},
        {1,1,1,2,2,2,3,3,3},
        {4,4,4,5,5,5,6,6,6},
        {4,4,4,5,5,5,6,6,6},
        {4,4,4,5,5,5,6,6,6},
        {7,7,7,8,8,8,9,9,9},
        {7,7,7,8,8,8,9,9,9},
        {7,7,7,8,8,8,9,9,9}
    };
    print(blocks(matrix9, 3));
}

输出：

1 1 1 1 
2 2 2 2 
3 3 3 3 
4 4 4 4 

1 1 1 1 1 1 1 1 1 
2 2 2 2 2 2 2 2 2 
3 3 3 3 3 3 3 3 3 
4 4 4 4 4 4 4 4 4 
5 5 5 5 5 5 5 5 5 
6 6 6 6 6 6 6 6 6 
7 7 7 7 7 7 7 7 7 
8 8 8 8 8 8 8 8 8 
9 9 9 9 9 9 9 9 9

我正在尝试编写一个采用二维数组“Sudoku”的代码并将它们排列成块

2 个答案: