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107 lines
2.5 KiB
Markdown
107 lines
2.5 KiB
Markdown
# [661. Image Smoother](https://leetcode.com/problems/image-smoother/)
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## 题目
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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.
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**Example 1**:
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```
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Input:
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[[1,1,1],
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[1,0,1],
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[1,1,1]]
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Output:
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[[0, 0, 0],
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[0, 0, 0],
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[0, 0, 0]]
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Explanation:
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For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0
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For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0
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For the point (1,1): floor(8/9) = floor(0.88888889) = 0
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```
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**Note**:
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1. The value in the given matrix is in the range of [0, 255].
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2. The length and width of the given matrix are in the range of [1, 150].
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## 题目大意
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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。
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注意:
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- 给定矩阵中的整数范围为 [0, 255]。
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- 矩阵的长和宽的范围均为 [1, 150]。
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## 解题思路
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- 将二维数组中的每个元素变为周围 9 个元素的平均值。
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- 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。
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## 代码
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```go
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func imageSmoother(M [][]int) [][]int {
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res := make([][]int, len(M))
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for i := range M {
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res[i] = make([]int, len(M[0]))
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}
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for y := 0; y < len(M); y++ {
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for x := 0; x < len(M[0]); x++ {
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res[y][x] = smooth(x, y, M)
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}
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}
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return res
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}
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func smooth(x, y int, M [][]int) int {
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count, sum := 1, M[y][x]
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// Check bottom
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if y+1 < len(M) {
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sum += M[y+1][x]
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count++
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}
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// Check Top
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if y-1 >= 0 {
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sum += M[y-1][x]
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count++
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}
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// Check left
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if x-1 >= 0 {
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sum += M[y][x-1]
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count++
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}
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// Check Right
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if x+1 < len(M[y]) {
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sum += M[y][x+1]
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count++
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}
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// Check Coners
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// Top Left
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if y-1 >= 0 && x-1 >= 0 {
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sum += M[y-1][x-1]
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count++
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}
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// Top Right
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if y-1 >= 0 && x+1 < len(M[0]) {
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sum += M[y-1][x+1]
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count++
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}
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// Bottom Left
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if y+1 < len(M) && x-1 >= 0 {
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sum += M[y+1][x-1]
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count++
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}
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//Bottom Right
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if y+1 < len(M) && x+1 < len(M[0]) {
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sum += M[y+1][x+1]
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count++
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}
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return sum / count
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}
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``` |