leetcode.840矩阵中的幻方
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