LeetCode--661. Image Smoother

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最新推荐文章于 2025-04-03 07:15:00 发布

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本文介绍了一种基于周围像素平均值的图像平滑算法实现方法，通过遍历矩阵并计算每个像素与其周围像素的平均灰度值来达到平滑效果。

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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].

题意：给定一个二维矩阵，让你把每个元素求取其与周围八个元素的平均值，不足八个元素的按最大数量求取。

思路：

1.做一个周围元素的数组x[]和y[],逐个遍历矩阵求取平均值即可。

参考代码：

class Solution {
public:
    int x[9]={-1,-1,-1,0,0,0,1,1,1};
    int y[9]={-1,0,1,-1,0,1,-1,0,1};
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        int m=M.size(),n=M[0].size();
         vector<vector<int>> ans(m);
        for(int i=0;i<m;++i){
            ans[i].resize(n);
            for(int j=0;j<n;++j){
                double t=0;
                int cnt=0;
                for(int k=0;k<9;++k){
                    if(i+x[k]>=0&&i+x[k]<m&&j+y[k]>=0&&j+y[k]<n){
                        t+=M[i+x[k]][j+y[k]];
                        cnt++;
                    }
                }
                ans[i][j]=floor(t/cnt);
            }
        }
        return ans;
    }
};