> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/661.image_smoother.md).

# 661.Image Smoother

**661.Image Smoother**

难度:Easy

> 包含整数的二维矩阵 M 表示一个图片的灰度。你需要设计一个平滑器来让每一个单元的灰度成为平均灰度 (向下舍入) ，平均灰度的计算是周围的8个单元和它本身的值求平均，如果周围的单元格不足八个，则尽可能多的利用它们。

```
示例 1:

输入:
[[1,1,1],
 [1,0,1],
 [1,1,1]]
输出:
[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]
解释:
对于点 (0,0), (0,2), (2,0), (2,2): 平均(3/4) = 平均(0.75) = 0
对于点 (0,1), (1,0), (1,2), (2,1): 平均(5/6) = 平均(0.83333333) = 0
对于点 (1,1): 平均(8/9) = 平均(0.88888889) = 0
注意:

给定矩阵中的整数范围为 [0, 255]。
矩阵的长和宽的范围均为 [1, 150]。
```

注意一下特殊情况就行，其他按题意平滑。

```
class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        int r=M.size();
        int c=M[0].size();
        if(r==1 && c==1) return M;
        vector<vector<int>> res(M);
        if(r==1)
        {
            res[0][0]=(M[0][0]+M[0][1])/2;
            res[0][c-1]=(M[0][c-1]+M[0][c-2])/2;
            for(int i=1;i<c-1;i++)
            res[0][i] = (M[0][i] + M[0][i-1] +M[0][i+1] )/3;
            return res;
        }
       if(c==1)
        {
            res[0][0]=(M[0][0]+M[1][0])/2;
            res[r-1][0]=(M[r-1][0]+M[r-2][0])/2;
            for(int i=1;i<r-1;i++)
            res[i][0] = (M[i][0] + M[i-1][0] +M[i+1][0])/3;
            return res;
        }

        for(int i=1;i<r-1;i++)
            for(int j=1;j<c-1;j++)
                res[i][j] =(M[i][j] +M[i-1][j]+ M[i][j-1] + M[i-1][j-1]+ M[i][j+1] + M[i+1][j] + M[i+1][j+1] + M[i-1][j+1] + M[i+1][j-1])/9;
        for(int i=1;i<c-1;i++)
            res[0][i] = (M[0][i] + M[0][i-1] +M[0][i+1] +M[1][i] + M[1][i-1]+M[1][i+1])/6;
        for(int i=1;i<c-1;i++)
             res[r-1][i] = (M[r-1][i] + M[r-1][i-1] +M[r-1][i+1] +M[r-2][i] + M[r-2][i-1]+M[r-2][i+1])/6;
        for(int i=1;i<r-1;i++)
            res[i][0] = (M[i][0] + M[i-1][0] +M[i+1][0] +M[i][1] + M[i-1][1]+M[i+1][1])/6;
        
        for(int i=1;i<r-1;i++)
                res[i][c-1] = (M[i][c-1] + M[i-1][c-1] +M[i+1][c-1] +M[i][c-2] + M[i-1][c-2]+M[i+1][c-2])/6;
        
        res[0][0]=(M[0][0]+M[0][1] +M[1][1]+M[1][0])/4;
        res[0][c-1]=(M[0][c-1]+M[0][c-2]+M[1][c-1]+M[1][c-2])/4;
        res[r-1][0]=(M[r-1][0]+M[r-1][1]+M[r-2][0]+M[r-2][1])/4;
        res[r-1][c-1]=(M[r-1][c-1]+M[r-2][c-1]+M[r-2][c-2]+M[r-1][c-2])/4;
        return res;
    }
        
};
```

> 执行用时 :328ms, 在所有 C++ 提交中击败了37.50%的用户\
> 内存消耗 :17.5MB, 在所有 C++ 提交中击败了86.36%的用户


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