标签（空格分隔）： LeetCode
题目地址：https://leetcode.com/problems/magic-squares-in-grid/description/
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).
Example 1: 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
判断一个大矩阵中有多少河图。河图这个词很古典文化，其实就是1到9填在9个格子中，让横竖斜的3个数相加都相等。
直接按照河图的规定去做就OK了。用到了一个结论：河图的中心数字是5.
注意一个易忽略的点，就是所有的数字应该在1~9之间。测试用例里面出现了不在这个范围内的数字也能组成河图。
另外，关于河图，其实有很多有用的结论，我并没有使用。
河图记忆方法：偶角奇边坐心五.一线双角相对画.
这个帖子挺有意思的：1到9填在9个格子中，让横竖斜的3个数相加都相等
class Solution: def numMagicSquaresInside(self, grid): """ :type grid: List[List[int]] :rtype: int """ if len(grid) < 3 or len(grid[0]) < 3: return 0 counter = 0 for row in range(len(grid) - 2): for col in range(len(grid[0]) - 2): sub_matrix = [[grid[row + i][col + j] for j in range(3)] for i in range(3)] if self.magic_square(sub_matrix): counter += 1 return counter def magic_square(self, matrix): is_number_right = all(1 <= matrix[i][j] <= 9 for i in range(3) for j in range(3)) is_row_right = all(sum(row) == 15 for row in matrix) is_col_right = all(sum(col) == 15 for col in [[matrix[i][j] for i in range(3)] for j in range(3)]) is_diagonal_right = matrix[1][1] == 5 and matrix[0][0] + matrix[-1][-1] == 10 and matrix[0][-1] + matrix[-1][0] == 10 return is_number_right and is_row_right and is_col_right and is_diagonal_right2018 年 5 月 27 日 ―――― 周末的天气很好～