Image Smoother问题及解法

问题描述：

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.

示例：

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

问题分析：
对每一个点及其周围的值求平均值作为新的值赋予该点。

过程详见代码：

class Solution {
public:
vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
vector<vector<int>> res;
int m = M.size();
if (!m) return res;
res = M;
int n = M[0].size();
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
int sum = M[i][j];
int count = 1;
if (i > 0)
{
sum += M[i - 1][j];
count++;
}
if (i + 1 < m)
{
sum += M[i + 1][j];
count++;
}
if (j + 1 < n)
{
sum += M[i][j + 1];
count++;
}
if (j > 0)
{
sum += M[i][j - 1];
count++;
}
if (i > 0 && j > 0)
{
sum += M[i - 1][j - 1];
count++;
}
if (i > 0 && j + 1 < n)
{
sum += M[i -1][j + 1];
count++;
}
if (i + 1 < m && j > 0)
{
sum += M[i + 1][j - 1];
count++;
}
if (i + 1 < m && j + 1 < n)
{
sum += M[i + 1][j + 1];
count++;
}
res[i][j] = (int)floor(sum / (double)count);
}
}
return res;
}
};