661. Image Smoother

最新推荐文章于 2022-05-20 16:19:17 发布

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最新推荐文章于 2022-05-20 16:19:17 发布

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分类专栏： leetcode 算法 Leetcode

本文链接：https://blog.youkuaiyun.com/qq_36718317/article/details/79353237

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本文介绍了一种图像平滑算法，该算法将每个像素的颜色值替换为周围8个像素及自身的平均值（向下取整）。通过这种方法可以实现图像的模糊效果。文章提供了具体的实现代码，并解释了如何处理边界上的像素。

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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：

给一个二维数组，求它以及八个方位这九个数的平均值，如果有的方位没有的话就算那些有的。

思路：

注意判断边界以及初始化的问题。

代码:

class Solution {
    public int[][] imageSmoother(int[][] M) {
        int r=M.length;
        int c=M[0].length;
        int [][]arr=new int[r][c];
        
        for(int i=0;i<r;i++)
        {
            for(int j=0;j<c;j++)
            {
                int []a=new int [8];
                int ans=1;
                 a[0]=(i-1>=0&&j-1>=0)?M[i-1][j-1]:0;
                ans=(i-1>=0&&j-1>=0)?ans+1:ans;
                 a[1]=(i-1>=0)?M[i-1][j]:0;
                ans=(i-1>=0)?ans+1:ans;
                 a[2]=(j-1>=0)?M[i][j-1]:0;
                ans=(j-1>=0)?ans+1:ans;
                 a[3]=(i+1<r&&j-1>=0)?M[i+1][j-1]:0;
                ans=(i+1<r&&j-1>=0)?ans+1:ans;
                 a[4]=(i+1<r)?M[i+1][j]:0;
                ans=(i+1<r)?ans+1:ans;
                 a[5]=(i+1<r&&j+1<c)?M[i+1][j+1]:0;
                ans=(i+1<r&&j+1<c)?ans+1:ans;
                 a[6]=(j+1<c)?M[i][j+1]:0;
                ans=(j+1<c)?ans+1:ans;
                 a[7]=(i-1>=0&&j+1<c)?M[i-1][j+1]:0;
                ans=(i-1>=0&&j+1<c)?ans+1:ans;
                int sum=M[i][j]+a[0]+a[1]+a[2]+a[3]+a[4]+a[5]+a[6]+a[7];
                arr[i][j]=(int)Math.floor(sum/ans);
            }
        }
        return arr;
    }
}