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:

1. The value in the given matrix is in the range of [0, 255].
2. The length and width of the given matrix are in the range of [1, 150].

这道题让我们给一个图片进行平滑处理，博主其实还是有一些图像处理的背景的，一般来说都是用算子来跟图片进行卷积，但是由于这道题只是个Easy的题目，我们直接用土办法就能解了，就直接对于每一个点统计其周围点的个数，然后累加像素值，做个除法就行了，注意边界情况的处理，参见代码如下：

class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        if (M.empty() || M[0].empty()) return {};
        int m = M.size(), n = M[0].size();
        vector<vector<int>> res = M, dirs{{0,-1},{-1,-1},{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1}};
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int cnt = M[i][j], all = 1;
                for (auto dir : dirs) {
                    int x = i + dir[0], y = j + dir[1];
                    if (x < 0 || x >= m || y < 0 || y >= n) continue;
                    ++all;
                    cnt += M[x][y];
                }
                res[i][j] = cnt / all;
            }
        }
        return res;
    }
};