C#LeetCode刷题之#840-矩阵中的幻方（Magic Squares In Grid）

问题

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3 x 3 的幻方是一个填充有从 1 到 9 的不同数字的 3 x 3 矩阵，其中每行，每列以及两条对角线上的各数之和都相等。

给定一个由整数组成的 m * n 矩阵，其中有多少个 3 × 3 的 “幻方” 子矩阵？（每个子矩阵都是连续的）。

输入: [[4,3,8,4],

      [9,5,1,9],

      [2,7,6,2]]

输出: 1

解释: 

下面的子矩阵是一个 3 x 3 的幻方：

438

951

276

而这一个不是：

384

519

762

总的来说，在本示例所给定的矩阵中只有一个 3 x 3 的幻方子矩阵。

提示:

1 <= grid.length = grid[0].length <= 10

0 <= grid[i][j] <= 15

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A 3 x 3 magic square is a 3 x 3 grid filled with distinct numbers from 1 to 9 such that each row, column, and both diagonals all have the same sum.

Given an grid of integers, how many 3 x 3 "magic square" subgrids are there?  (Each subgrid is contiguous).

Input: [[4,3,8,4],

        [9,5,1,9],

        [2,7,6,2]]

Output: 1

Explanation: 

The following subgrid is a 3 x 3 magic square:

438

951

276

while this one is not:

384

519

762

In total, there is only one magic square inside the given grid.

Note:

1 <= grid.length <= 10

1 <= grid[0].length <= 10

0 <= grid[i][j] <= 15

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示例

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public class Program {

    public static void Main(string[] args) {
        int[][] nums = null;

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

        var res = NumMagicSquaresInside(nums);
        Console.WriteLine(res);

        nums = new int[3][] {new int[]{1,0,3,8},
            new int[] {3,7,2,4},
            new int[] {5,8,1,0}
        };

        res = NumMagicSquaresInside2(nums);
        Console.WriteLine(res);

        Console.ReadKey();
    }

    private static int NumMagicSquaresInside(int[][] grid) {
        var num = 0;
        for(var i = 1; i < grid.Length - 1; i++) {
            for(var j = 1; j < grid[i].Length - 1; j++) {
                if(grid[i][j] == 5) {
                    num = IsMagicSquare(grid, i, j) ? ++num : num;
                }
            }
        }
        return num;
    }

    private static bool IsMagicSquare(int[][] grid, int i, int j) {
        //原始解法，不推荐
        //记录每个值，看是不是9位，因为从1-9都必须出现
        var dic = new Dictionary<int, int>();
        dic[grid[i - 1][j - 1]] = grid[i - 1][j - 1];
        dic[grid[i - 1][j]] = grid[i - 1][j];
        dic[grid[i - 1][j + 1]] = grid[i - 1][j + 1];

        dic[grid[i][j - 1]] = grid[i][j - 1];
        dic[grid[i][j]] = grid[i][j];
        dic[grid[i][j + 1]] = grid[i][j + 1];

        dic[grid[i + 1][j - 1]] = grid[i + 1][j - 1];
        dic[grid[i + 1][j]] = grid[i + 1][j];
        dic[grid[i + 1][j + 1]] = grid[i + 1][j + 1];

        //不为9或不在1-9范围之内，则不是幻方
        if(dic.Count != 9) return false;

        foreach(var item in dic) {
            if(item.Value > 9 || item.Value < 0) return false;
        }

        //记录3行、3列、2对角线
        var sum_row1 = grid[i - 1][j - 1] + grid[i - 1][j] + grid[i - 1][j + 1];
        var sum_row2 = grid[i][j - 1] + grid[i][j] + grid[i][j + 1];
        var sum_row3 = grid[i + 1][j - 1] + grid[i + 1][j] + grid[i + 1][j + 1];

        var sum_col1 = grid[i - 1][j - 1] + grid[i][j - 1] + grid[i + 1][j - 1];
        var sum_col2 = grid[i - 1][j] + grid[i][j] + grid[i + 1][j];
        var sum_col3 = grid[i - 1][j + 1] + grid[i][j + 1] + grid[i + 1][j + 1];

        var sum_cross1 = grid[i - 1][j - 1] + grid[i][j] + grid[i + 1][j + 1];
        var sum_cross2 = grid[i - 1][j + 1] + grid[i][j] + grid[i + 1][j - 1];

        var dic2 = new Dictionary<int, int>();
        dic2[sum_row1] = 0;
        dic2[sum_row2] = 0;
        dic2[sum_row3] = 0;

        dic2[sum_col1] = 0;
        dic2[sum_col2] = 0;
        dic2[sum_col3] = 0;

        dic2[sum_cross1] = 0;
        dic2[sum_cross2] = 0;

        //看值是不是相同并且值之和为15
        return dic2.Count == 1 && sum_row1 == 15;
    }

    private static int NumMagicSquaresInside2(int[][] grid) {
        var num = 0;
        for(var i = 1; i < grid.Length - 1; i++) {
            for(var j = 1; j < grid[i].Length - 1; j++) {
                if(grid[i][j] == 5) {
                    num = IsMagicSquare(grid, i, j) ? ++num : num;
                }
            }
        }
        return num;
    }

    private static bool IsMagicSquare2(int[][] grid, int row, int col) {
        //值必须是1-9
        for(var i = row - 1; i <= row + 1; i++)
            for(var j = col - 1; j <= col + 1; j++)
                if(grid[i][j] < 1 || grid[i][j] > 9) return false;
        //不考虑中间的5，只需要考虑4个位置的值的和为10即可
        return !(grid[row - 1][col - 1] + grid[row + 1][col + 1] != 10 ||
          grid[row - 1][col] + grid[row + 1][col] != 10 ||
          grid[row - 1][col + 1] + grid[row + 1][col - 1] != 10 ||
          grid[row][col - 1] + grid[row][col + 1] != 10);
    }

}

以上给出2种算法实现，以下是这个案例的输出结果：

该文章的最新版本已迁移至个人博客【比特飞】，单击链接 https://www.byteflying.com/archives/3752 访问。

1
0

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分析：

显而易见，以上2种算法的时间复杂度均为：  。