leetcode 661. Image Smoother 图像细化 + 计算平均值

迷南。 2022-06-03 07:20 146阅读 0赞

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 \[1, 150\].

本题题意很简单，直接暴力遍历即可

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    class Solution 
    {
    public:
        vector<vector<int>> imageSmoother(vector<vector<int>>& m) 
        {
            int row = m.size(), col = m[0].size();
            vector<vector<int>> res(row, vector<int>(col, 0));
            for (int i = 0; i < row; i++)
            {
                for (int j = 0; j < col; j++)
                {
                    res[i][j] = getAverage(i, j, m);
                }
            }
            return res;
        }
    
        int getAverage(int x, int y, vector<vector<int>>& m)
        {
            int row = m.size(), col = m[0].size();
            int sum = 0, count = 0;
            for (int i = max(0, x - 1); i <= min(row - 1, x + 1); i++)
            {
                for (int j = max(0, y - 1); j <= min(col - 1, y + 1); j++)
                {
                    count++;
                    sum += m[i][j];
                }
            }
            return floor((double)sum / count);
        }
    };