840.Magic Squares In Grid

840.Magic Squares In Grid

难度:Easy

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

矩阵元素为1-9，不能相同，而且还要和都相等，最中间元素肯定是5,其他和为10的元素分别与5对称分布。

代码如下：

class Solution {
public:
    int numMagicSquaresInside(vector<vector<int>>& grid) {
        if(grid.size()<3 || grid[0].size()<3) return 0;
        int res=0;
        for(int i=0;i<grid.size()-2;i++)
            for(int j=0;j<grid[0].size()-2;j++)
            {
                bool flag=false;
                if(grid[i+1][j+1] !=5 || grid[i][j] == 5) continue;
                if(grid[i][j]+grid[i+2][j+2] !=10 || grid[i+2][j]+grid[i][j+2] !=10 || grid[i+1][j]+grid[i+1][j+2] != 10 || grid[i][j+1] +grid[i+2][j+1] != 10) continue;
                if(grid[i][j]+grid[i][j+1]+grid[i][j+2] !=15 || grid[i][j]+grid[i+1][j]+grid[i+2][j] !=15) continue;
                for(int x=0;x<3;x++)
                for(int y=0;y<3;y++)
                    if(grid[i+x][j+y] >9 || grid[i+x][j+y]<0) {flag=true;break;}
                if(flag) continue;
                res++;

            }
        return res;
        
    
    }
};

执行用时 :4 ms, 在所有 C++ 提交中击败了91.95%的用户 内存消耗 :8.3 MB, 在所有 C++ 提交中击败了100.00%的用户