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).
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Example 1:
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Input: [[4,3,8,4],
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[9,5,1,9],
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[2,7,6,2]]
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Output: 1
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Explanation:
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The following subgrid is a 3 x 3 magic square:
9
438
10
951
11
276
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while this one is not:
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384
15
519
16
762
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In total, there is only one magic square inside the given grid.
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Note:
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1 <= grid.length <= 10
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1 <= grid[0].length <= 10
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0 <= grid[i][j] <= 15
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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:
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class Solution {
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public:
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int numMagicSquaresInside(vector<vector<int>>& grid) {
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if(grid.size()<3 || grid[0].size()<3) return 0;
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int res=0;
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for(int i=0;i<grid.size()-2;i++)
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for(int j=0;j<grid[0].size()-2;j++)
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{
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bool flag=false;
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if(grid[i+1][j+1] !=5 || grid[i][j] == 5) continue;
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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;
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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;
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for(int x=0;x<3;x++)
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for(int y=0;y<3;y++)
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if(grid[i+x][j+y] >9 || grid[i+x][j+y]<0) {flag=true;break;}
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if(flag) continue;
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res++;
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}
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return res;
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}
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};
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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
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