题目地址：https://leetcode.com/problems/magic-squares-in-grid/description/
题目描述
A3x 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
题目大意
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。
解题方法
利用河图规律
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open in new window
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
is_repeat_right = len(set(matrix[i][j] for i in range(3) for j in range(3))) == 9
return is_number_right and is_row_right and is_col_right and is_diagonal_right and is_repeat_right
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暴力解法
二刷的时候完全忘了什么性质了……直接暴力解法，这样的话，需要按照题目做各种判断，所以函数更为复杂了。
class Solution(object):
def numMagicSquaresInside(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
M, N = len(grid), len(grid[0])
res = 0
for r in range(M - 2):
for c in range(N - 2):
curgrid = [[grid[r + i][c + j] for j in range(3)] for i in range(3)]
if self.isMagic(curgrid):
res += 1
return res
def isMagic(self, grid):
count = list(range(9))
for i in range(3):
for j in range(3):
if not (1 <= grid[i][j] <= 9):
return False
count[grid[i][j] - 1] += 1
if 0 in count: return False
row, col = [0, 0, 0], [0, 0, 0]
for i in range(3):
row[i] += sum(grid[i][j] for j in range(3))
for j in range(3):
col[j] += sum(grid[i][j] for i in range(3))
if row[0] != row[1] != row[2] or col[0] != col[1] != col[2]:
return False
if grid[0][0] + grid[2][2] != grid[0][2] + grid[2][0]:
return False
return True
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