leetcode 661. Image Smoother

1.题目

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.
翻译： 一个二维矩阵M表示一副灰度度，对每个点求以它为中心的9个灰度值的平均值。输出结果。

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

2.分析

基础的图像处理，遍历每个像素点。求mask的x,y取值范围，求和，取平均值。

3.代码

vector<vector<int>> imageSmoother(vector<vector<int>>& M) {    int X = M.size();    int Y = M[0].size();    vector<vector<int>> ans(X, vector<int>(Y, 0));    for (int i = 0; i < X; i++) {        for (int j = 0; j < Y; j++) {            int l = i - 1 > 0 ? i - 1 : 0;            int r = i + 1 < X ? i + 1 : X - 1;            int s = j - 1 > 0 ? j - 1 : 0;            int t = j + 1 < Y ? j + 1 : Y - 1;            int sum = 0;            for (int x = l; x <= r; x++)                for (int y = s; y <= t; y++)                    sum += M[x][y];            ans[i][j] = sum / ((r - l + 1)*(t - s + 1));            cout << ans[i][j] << " ";        }        cout << endl;    }    return ans;}