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
思路就是依次扫, 只是利用
Here I just want share two observatons with this 1-9 condition:
Assume a magic square:
a1,a2,a3
a4,a5,a6
a7,a8,a9
a2 + a5 + a8 = 15
a4 + a5 + a6 = 15
a1 + a5 + a9 = 15
a3 + a5 + a7 = 15
Accumulate all, then we have:sum(ai) + 3 * a5 = 60
3 * a5 = 15
a5 = 5
The center of magic square must be 5. 这个条件来去减少一些判断.
Code:
class Solution: def numMagicSquaresInside(self, grid): ans, lrc = 0, [len(grid), len(grid[0])] def checkMagic(a,b, c, d, e, f, g ,h, i): return (sorted([a,b,c,d,e,f,g,h,i]) == [i for i in range(1,10)] and (a + b+c == d + e + f == g + h + i == a + d + g == b + e + h == c +f + i == a + e + i == c + e + g == 15)) for i in range(1, lrc[0]-1): for j in range(1, lrc[1]-1): if grid[i][j] == 5 and checkMagic(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]): ans += 1 return ans
原文地址:https://www.cnblogs.com/Johnsonxiong/p/9547361.html
时间: 2024-11-06 07:17:06