# leetcode\_661

Given a 2D integer matrix M representing the gray scale of an image, you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.

Example 1: Input: \[\[1,1,1], \[1,0,1], \[1,1,1]] Output: \[\[0, 0, 0], \[0, 0, 0], \[0, 0, 0]] Explanation: For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0 For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0 For the point (1,1): floor(8/9) = floor(0.88888889) = 0 Note: The value in the given matrix is in the range of \[0, 255]. The length and width of the given matrix are in the range of \[1, 150].

## Solutions

1. **straight forward**

```cpp
class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        int dirs[8][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};

        int m = M.size(), n = M[0].size();
        vector<vector<int>> res(m, vector<int>(n));
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++) {
                int sum = M[i][j], num = 1;
                for (auto & d : dirs) {
                    int x = i + d[0], y = j + d[1];
                    if (x < 0 || y < 0 || x >= m || y >= n)
                        continue;
                    num++;
                    sum += M[x][y];
                }
                res[i][j] = sum / num;
            }

        return res;
    }
};
```