【Leetcode】 661. 图片平滑器

An image smoother is a filter of the size 3 x 3 that can be applied to each cell of an image by rounding down the average of the cell and the eight surrounding cells (i.e., the average of the nine cells in the blue smoother). If one or more of the surrounding cells of a cell is not present, we do not consider it in the average (i.e., the average of the four cells in the red smoother).
p1
Given an m x n integer matrix img representing the grayscale of an image, return the image after applying the smoother on each cell of it.

Example 1:

Input: img = [[1,1,1],[1,0,1],[1,1,1]]
Output: [[0,0,0],[0,0,0],[0,0,0]]

Explanation:

  • For the points (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0
  • For the points (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

Example 2:
p2

Input: img = [[100,200,100],[200,50,200],[100,200,100]]
Output: [[137,141,137],[141,138,141],[137,141,137]]

Explanation:

  • For the points (0,0), (0,2), (2,0), (2,2): floor((100+200+200+50)/4) = floor(137.5) = 137
  • For the points (0,1), (1,0), (1,2), (2,1): floor((200+200+50+200+100+100)/6) = floor(141.666667) = 141
  • For the point (1,1): floor((50+200+200+200+200+100+100+100+100)/9) = floor(138.888889) = 138

Constraints:

m == img.length
n == img[i].length
1 <= m, n <= 200
0 <= img[i][j] <= 255

AC

/*
 * @lc app=leetcode.cn id=661 lang=cpp
 *
 * [661] 图片平滑器
 */

// @lc code=start
class Solution {
    
    
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& img) {
    
    
        int m = img.size(), n = img[0].size();
        vector<vector<int>> ret(m, vector<int>(n));
        for(int i = 0; i < m; i++)
        {
    
    
            for(int j = 0; j < n; j++)
            {
    
    
                int num = 0, sum = 0;
                for(int x = i - 1; x <= i + 1; x++)
                {
    
    
                    for(int y = j - 1; y <= j + 1; y++)
                    {
    
    
                        if(x >= 0 && x < m && y >= 0 && y < n)
                        {
    
    
                            num++;
                            sum += img[x][y];
                        }
                    }
                }
                ret[i][j] = sum / num;
            }
        }
        return ret;
    }
};
// @lc code=end

AC


Thought:

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转载自blog.csdn.net/qq_54053990/article/details/130821985
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