问题
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 m * 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)。
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
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).
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
276while this one is not:
384
519
762In 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
示例
public class Program {
public static void Main(string[] args) {
int[][] nums = null;
nums = new int[3][] {new int[]{4,3,8,4},
new int[] {9,5,1,9},
new int[] {2,7,6,2}
};
var res = NumMagicSquaresInside(nums);
Console.WriteLine(res);
nums = new int[3][] {new int[]{1,0,3,8},
new int[] {3,7,2,4},
new int[] {5,8,1,0}
};
res = NumMagicSquaresInside2(nums);
Console.WriteLine(res);
Console.ReadKey();
}
private static int NumMagicSquaresInside(int[][] grid) {
var num = 0;
for(var i = 1; i < grid.Length - 1; i++) {
for(var j = 1; j < grid[i].Length - 1; j++) {
if(grid[i][j] == 5) {
num = IsMagicSquare(grid, i, j) ? ++num : num;
}
}
}
return num;
}
private static bool IsMagicSquare(int[][] grid, int i, int j) {
//原始解法,不推荐
//记录每个值,看是不是9位,因为从1-9都必须出现
var dic = new Dictionary<int, int>();
dic[grid[i - 1][j - 1]] = grid[i - 1][j - 1];
dic[grid[i - 1][j]] = grid[i - 1][j];
dic[grid[i - 1][j + 1]] = grid[i - 1][j + 1];
dic[grid[i][j - 1]] = grid[i][j - 1];
dic[grid[i][j]] = grid[i][j];
dic[grid[i][j + 1]] = grid[i][j + 1];
dic[grid[i + 1][j - 1]] = grid[i + 1][j - 1];
dic[grid[i + 1][j]] = grid[i + 1][j];
dic[grid[i + 1][j + 1]] = grid[i + 1][j + 1];
//不为9或不在1-9范围之内,则不是幻方
if(dic.Count != 9) return false;
foreach(var item in dic) {
if(item.Value > 9 || item.Value < 0) return false;
}
//记录3行、3列、2对角线
var sum_row1 = grid[i - 1][j - 1] + grid[i - 1][j] + grid[i - 1][j + 1];
var sum_row2 = grid[i][j - 1] + grid[i][j] + grid[i][j + 1];
var sum_row3 = grid[i + 1][j - 1] + grid[i + 1][j] + grid[i + 1][j + 1];
var sum_col1 = grid[i - 1][j - 1] + grid[i][j - 1] + grid[i + 1][j - 1];
var sum_col2 = grid[i - 1][j] + grid[i][j] + grid[i + 1][j];
var sum_col3 = grid[i - 1][j + 1] + grid[i][j + 1] + grid[i + 1][j + 1];
var sum_cross1 = grid[i - 1][j - 1] + grid[i][j] + grid[i + 1][j + 1];
var sum_cross2 = grid[i - 1][j + 1] + grid[i][j] + grid[i + 1][j - 1];
var dic2 = new Dictionary<int, int>();
dic2[sum_row1] = 0;
dic2[sum_row2] = 0;
dic2[sum_row3] = 0;
dic2[sum_col1] = 0;
dic2[sum_col2] = 0;
dic2[sum_col3] = 0;
dic2[sum_cross1] = 0;
dic2[sum_cross2] = 0;
//看值是不是相同并且值之和为15
return dic2.Count == 1 && sum_row1 == 15;
}
private static int NumMagicSquaresInside2(int[][] grid) {
var num = 0;
for(var i = 1; i < grid.Length - 1; i++) {
for(var j = 1; j < grid[i].Length - 1; j++) {
if(grid[i][j] == 5) {
num = IsMagicSquare(grid, i, j) ? ++num : num;
}
}
}
return num;
}
private static bool IsMagicSquare2(int[][] grid, int row, int col) {
//值必须是1-9
for(var i = row - 1; i <= row + 1; i++)
for(var j = col - 1; j <= col + 1; j++)
if(grid[i][j] < 1 || grid[i][j] > 9) return false;
//不考虑中间的5,只需要考虑4个位置的值的和为10即可
return !(grid[row - 1][col - 1] + grid[row + 1][col + 1] != 10 ||
grid[row - 1][col] + grid[row + 1][col] != 10 ||
grid[row - 1][col + 1] + grid[row + 1][col - 1] != 10 ||
grid[row][col - 1] + grid[row][col + 1] != 10);
}
}
以上给出2种算法实现,以下是这个案例的输出结果:
1
0
分析:
显而易见,以上2种算法的时间复杂度均为: 。