661. Image Smoother

问题：

处理给定数组，（平滑处理）使得每个元素=周围+自己共9个元素的平均值。（若没有9个元素，则是周围一圈+自己元素的平均值）

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].

　　

解决方法：顺次计算 [i-1] ~ [i+1], [j-1] ~ [j+1]

tip：需要注意边界值情况

解决：if [i-1]<0 || [i+1]>row-1 || [j-1]<0 || [j+1]>col-1; then 不计算

代码参考：

class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        int rows = M.size();
        if(rows==0) return M;
        int cols = M[0].size();
        vector<vector<int>> res(rows, vector<int>(cols, 0));
        for(int i=0; i<rows; i++){
            for(int j=0; j<cols; j++){
                int sum=0;
                int count=0;
                for(int x=(i-1)>=0?i-1:0;x<=((i+1)<rows?i+1:rows-1);x++){
                    for(int y=(j-1)>=0?j-1:0;y<=((j+1)<cols?j+1:cols-1);y++){
                        sum+=M[x][y];
                        count++;
                    }
                }
                res[i][j]=sum/count;
            }
        }
        return res;
    }
};