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
这道题定义了一种神奇正方形,是一个3x3大小,且由1到9中到数字组成,各行各列即对角线和都必须相等。那么其实这个神奇正方形的各行各列及对角线之和就已经被限定了,必须是15才行,而且最中间的位置必须是5,否则根本无法组成满足要求的正方形。博主也没想出啥特别巧妙的方法,就老老实实的遍历所有的3x3大小的正方形呗,我们写一个子函数来检测各行各列及对角线的和是否为15,在调用子函数之前,先检测一下中间的数字是否为5,是的话再进入子函数。在子函数中,先验证下该正方形中的数字是否只有1到9中的数字,且不能由重复出现,使用一个一维数组来标记出现过的数字,若当前数字已经出现了,直接返回true。之后便是一次计算各行各列及对角线之和是否为15了,若全部为15,则返回true,参见代码如下:
class Solution {
public:
int numMagicSquaresInside(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[].size(), res = ;
for (int i = ; i < m - ; ++i) {
for (int j = ; j < n - ; ++j) {
if (grid[i + ][j + ] == && isValid(grid, i, j)) ++res;
}
}
return res;
}
bool isValid(vector<vector<int>>& grid, int i, int j) {
vector<int> cnt();
for (int x = i; x < i + ; ++x) {
for (int y = j; y < j + ; ++y) {
int k = grid[x][y];
if (k < || k > || cnt[k] == ) return false;
cnt[k] = ;
}
}
if ( != grid[i][j] + grid[i][j + ] + grid[i][j + ]) return false;
if ( != grid[i + ][j] + grid[i + ][j + ] + grid[i + ][j + ]) return false;
if ( != grid[i + ][j] + grid[i + ][j + ] + grid[i + ][j + ]) return false;
if ( != grid[i][j] + grid[i + ][j] + grid[i + ][j]) return false;
if ( != grid[i][j + ] + grid[i + ][j + ] + grid[i + ][j + ]) return false;
if ( != grid[i][j + ] + grid[i + ][j + ] + grid[i + ][j + ]) return false;
if ( != grid[i][j] + grid[i + ][j + ] + grid[i + ][j + ]) return false;
if ( != grid[i + ][j] + grid[i + ][j + ] + grid[i][j + ]) return false;
return true;
}
};
参考资料: