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:
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:
- The value in the given matrix is in the range of [0, 255].
- The length and width of the given matrix are in the range of [1, 150].
这道题让我们给一个图片进行平滑处理,博主其实还是有一些图像处理的背景的,一般来说都是用算子来跟图片进行卷积,但是由于这道题只是个Easy的题目,我们直接用土办法就能解了,就直接对于每一个点统计其周围点的个数,然后累加像素值,做个除法就行了,注意边界情况的处理,参见代码如下:
public: vector<vector<int>> imageSmoother(vector<vector<int>>& M) { if (M.empty() || M[0].empty()) return {}; int m = M.size(), n = M[0].size(); vector<vector<int>> res = M, dirs{{0,-1},{-1,-1},{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1}}; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { int cnt = M[i][j], all = 1; for (auto dir : dirs) { int x = i + dir[0], y = j + dir[1]; if (x < 0 || x >= m || y < 0 || y >= n) continue; ++all; cnt += M[x][y]; } res[i][j] = cnt / all; } } return res; } };
本文转自博客园Grandyang的博客,原文链接:[LeetCode] Image Smoother 图片平滑器
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