# [661. Image Smoother](https://leetcode.com/problems/image-smoother/) ## 题目 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**: 1. The value in the given matrix is in the range of [0, 255]. 2. The length and width of the given matrix are in the range of [1, 150]. ## 题目大意 包含整数的二维矩阵 M 表示一个图片的灰度。你需要设计一个平滑器来让每一个单元的灰度成为平均灰度 (向下舍入) ,平均灰度的计算是周围的8个单元和它本身的值求平均,如果周围的单元格不足八个,则尽可能多的利用它们。 注意: - 给定矩阵中的整数范围为 [0, 255]。 - 矩阵的长和宽的范围均为 [1, 150]。 ## 解题思路 - 将二维数组中的每个元素变为周围 9 个元素的平均值。 - 简单题,按照题意计算平均值即可。需要注意的是边界问题,四个角和边上的元素,这些点计算平均值的时候,计算平均值都不足 9 个元素。 ## 代码 ```go func imageSmoother(M [][]int) [][]int { res := make([][]int, len(M)) for i := range M { res[i] = make([]int, len(M[0])) } for y := 0; y < len(M); y++ { for x := 0; x < len(M[0]); x++ { res[y][x] = smooth(x, y, M) } } return res } func smooth(x, y int, M [][]int) int { count, sum := 1, M[y][x] // Check bottom if y+1 < len(M) { sum += M[y+1][x] count++ } // Check Top if y-1 >= 0 { sum += M[y-1][x] count++ } // Check left if x-1 >= 0 { sum += M[y][x-1] count++ } // Check Right if x+1 < len(M[y]) { sum += M[y][x+1] count++ } // Check Coners // Top Left if y-1 >= 0 && x-1 >= 0 { sum += M[y-1][x-1] count++ } // Top Right if y-1 >= 0 && x+1 < len(M[0]) { sum += M[y-1][x+1] count++ } // Bottom Left if y+1 < len(M) && x-1 >= 0 { sum += M[y+1][x-1] count++ } //Bottom Right if y+1 < len(M) && x+1 < len(M[0]) { sum += M[y+1][x+1] count++ } return sum / count } ```