leetcode-661 Image Smoother

661. Image Smoother

Description

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:

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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].

Analyse

使图片模糊的卷积操作，将图片(以矩阵表示)的每一个像素的灰度值变为周围8个(可以不足8个)及自身灰度值的平均值

需要注意的是矩阵的边界

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vector<vector<int>> imageSmoother(vector<vector<int>>& M)
{
    vector<vector<int>> result(M);
    for (int i = 0; i < M.size(); i++)
    {
        for (int j = 0; j < M[i].size(); j++)
        {
            int tmp = M[i][j];
            int count = 1;

            if (i != 0)
            {
                tmp += M[i-1][j];
                ++count;

                if (j != 0)
                {
                    //tmp += M[i][j-1];
                    tmp += M[i-1][j-1];
                    count += 1;
                }
                if (j != M[i].size() - 1)
                {
                    //tmp += M[i][j+1];
                    tmp += M[i-1][j+1];
                    count += 1;
                }
            }
            else if (i != M.size() - 1)
            {
                tmp += M[i+1][j];
                ++count;

                if (j != 0)
                {
                    //tmp += M[i][j-1];
                    tmp += M[i+1][j-1];
                    count += 1;
                }
                if (j != M[i].size() - 1)
                {
                    //tmp += M[i][j+1];
                    tmp += M[i+1][j+1];
                    count += 1;
                }
            }
            else
            {
                if (j != 0)
                {
                    tmp += M[i][j-1];
                    count += 1;
                }

                if (j != M[i].size() - 1)
                {
                    tmp += M[i][j+1];
                    count += 1;
                }
            }

            result[i][j] = tmp / count;
        }
    }

    return result;
}