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 N × 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）。

示例 1:

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


直接根据题意暴力解法：

class Solution:
    def numMagicSquaresInside(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        if len(grid) < 3 or len(grid[0]) < 3:
            return 0
        
        res = 0
        for i in range(1,len(grid)-1):
            for j in range(1,len(grid[0])-1):
                temp_set = {grid[i-1][j-1],grid[i-1][j],grid[i-1][j+1],grid[i][j-1],grid[i][j],grid[i][j+1],grid[i+1][j-1],grid[i+1][j],grid[i+1][j+1]}
                
                my_set = {1,2,3,4,5,6,7,8,9}
                
                if grid[i][j] == 5:
                    s1 = grid[i-1][j] + grid[i+1][j] + grid[i][j]
                    s2 = grid[i-1][j-1] + grid[i+1][j+1] + grid[i][j]
                    s3 = grid[i-1][j+1] + grid[i+1][j-1] + grid[i][j]
                    s4 = grid[i][j-1] + grid[i][j+1] + grid[i][j]
                    s5 = grid[i-1][j-1] + grid[i-1][j] + grid[i-1][j+1]
                    s6 = grid[i+1][j-1] + grid[i+1][j] + grid[i+1][j+1]
                    s7 = grid[i-1][j-1] + grid[i][j-1] + grid[i+1][j-1]
                    s8 = grid[i-1][j+1] + grid[i][j+1] + grid[i+1][j+1]
                                  
                    if (temp_set == my_set) and (s1 == s2 == s3 == s4 == s5 == s6 == s7 == s8 == 15):
                        res += 1
                        
        return res