|
| 1 | +--- |
| 2 | +title: 661.图片平滑器 |
| 3 | +date: 2024-11-18 21:36:49 |
| 4 | +tags: [题解, LeetCode, 简单, 数组, 矩阵, 模拟] |
| 5 | +--- |
| 6 | + |
| 7 | +# 【LetMeFly】661.图片平滑器：模拟 |
| 8 | + |
| 9 | +力扣题目链接：[https://leetcode.cn/problems/image-smoother/](https://leetcode.cn/problems/image-smoother/) |
| 10 | + |
| 11 | +<p><strong>图像平滑器</strong> 是大小为 <code>3 x 3</code> 的过滤器，用于对图像的每个单元格平滑处理，平滑处理后单元格的值为该单元格的平均灰度。</p> |
| 12 | + |
| 13 | +<p>每个单元格的<strong> 平均灰度</strong> 定义为：该单元格自身及其周围的 8 个单元格的平均值，结果需向下取整。（即，需要计算蓝色平滑器中 9 个单元格的平均值）。</p> |
| 14 | + |
| 15 | +<p>如果一个单元格周围存在单元格缺失的情况，则计算平均灰度时不考虑缺失的单元格（即，需要计算红色平滑器中 4 个单元格的平均值）。</p> |
| 16 | + |
| 17 | +<p><img src="https://assets.leetcode.com/uploads/2021/05/03/smoother-grid.jpg" style="height: 493px; width: 493px;" /></p> |
| 18 | + |
| 19 | +<p>给你一个表示图像灰度的 <code>m x n</code> 整数矩阵 <code>img</code> ，返回对图像的每个单元格平滑处理后的图像 。</p> |
| 20 | + |
| 21 | +<p> </p> |
| 22 | + |
| 23 | +<p><strong>示例 1:</strong></p> |
| 24 | + |
| 25 | +<p><img src="https://assets.leetcode.com/uploads/2021/05/03/smooth-grid.jpg" /></p> |
| 26 | + |
| 27 | +<pre> |
| 28 | +<strong>输入:</strong>img = [[1,1,1],[1,0,1],[1,1,1]] |
| 29 | +<strong>输出:</strong>[[0, 0, 0],[0, 0, 0], [0, 0, 0]] |
| 30 | +<strong>解释:</strong> |
| 31 | +对于点 (0,0), (0,2), (2,0), (2,2): 平均(3/4) = 平均(0.75) = 0 |
| 32 | +对于点 (0,1), (1,0), (1,2), (2,1): 平均(5/6) = 平均(0.83333333) = 0 |
| 33 | +对于点 (1,1): 平均(8/9) = 平均(0.88888889) = 0 |
| 34 | +</pre> |
| 35 | + |
| 36 | +<p><strong>示例 2:</strong></p> |
| 37 | +<img alt="" src="https://assets.leetcode.com/uploads/2021/05/03/smooth2-grid.jpg" /> |
| 38 | +<pre> |
| 39 | +<strong>输入:</strong> img = [[100,200,100],[200,50,200],[100,200,100]] |
| 40 | +<strong>输出:</strong> [[137,141,137],[141,138,141],[137,141,137]] |
| 41 | +<strong>解释:</strong> |
| 42 | +对于点 (0,0), (0,2), (2,0), (2,2): floor((100+200+200+50)/4) = floor(137.5) = 137 |
| 43 | +对于点 (0,1), (1,0), (1,2), (2,1): floor((200+200+50+200+100+100)/6) = floor(141.666667) = 141 |
| 44 | +对于点 (1,1): floor((50+200+200+200+200+100+100+100+100)/9) = floor(138.888889) = 138 |
| 45 | +</pre> |
| 46 | + |
| 47 | +<p> </p> |
| 48 | + |
| 49 | +<p><strong>提示:</strong></p> |
| 50 | + |
| 51 | +<ul> |
| 52 | + <li><code>m == img.length</code></li> |
| 53 | + <li><code>n == img[i].length</code></li> |
| 54 | + <li><code>1 <= m, n <= 200</code></li> |
| 55 | + <li><code>0 <= img[i][j] <= 255</code></li> |
| 56 | +</ul> |
| 57 | + |
| 58 | + |
| 59 | + |
| 60 | +## 解题方法：模拟 |
| 61 | + |
| 62 | +依次枚举每个元素，对于某个元素枚举其3x3矩阵（若超出边界则跳过）。按题目要求，对边界问题细心一点并计算即可。 |
| 63 | + |
| 64 | ++ 时间复杂度$O(mnC^2)$ |
| 65 | ++ 空间复杂度$O(1)$：力扣返回值不计入算法空间复杂度 |
| 66 | + |
| 67 | +### AC代码 |
| 68 | + |
| 69 | +#### C++ |
| 70 | + |
| 71 | +```cpp |
| 72 | +class Solution { |
| 73 | +public: |
| 74 | + vector<vector<int>> imageSmoother(vector<vector<int>>& img) { |
| 75 | + vector<vector<int>> ans(img.size(), vector<int>(img[0].size())); |
| 76 | + for (int i = 0; i < img.size(); i++) { |
| 77 | + for (int j = 0; j < img[0].size(); j++) { |
| 78 | + int cnt = 0, s = 0; |
| 79 | + for (int x = -1; x <= 1; x++) { |
| 80 | + for (int y = -1; y <= 1; y++) { |
| 81 | + int nx = i + x, ny = j + y; |
| 82 | + if (0 <= nx && nx < img.size() && 0 <= ny && ny < img[0].size()) { |
| 83 | + s += img[nx][ny]; |
| 84 | + cnt++; |
| 85 | + } |
| 86 | + } |
| 87 | + } |
| 88 | + ans[i][j] = s / cnt; |
| 89 | + } |
| 90 | + } |
| 91 | + return ans; |
| 92 | + } |
| 93 | +}; |
| 94 | +``` |
| 95 | +
|
| 96 | +#### Python |
| 97 | +
|
| 98 | +```python |
| 99 | +from typing import List |
| 100 | +
|
| 101 | +class Solution: |
| 102 | + def imageSmoother(self, img: List[List[int]]) -> List[List[int]]: |
| 103 | + ans = [[0 for _ in range(len(img[0]))] for _ in range(len(img))] |
| 104 | + for i in range(len(img)): |
| 105 | + for j in range(len(img[0])): |
| 106 | + cnt, s = 0, 0 |
| 107 | + for dx in range(-1, 2): |
| 108 | + for dy in range(-1, 2): |
| 109 | + x, y = i + dx, j + dy |
| 110 | + if 0 <= x < len(img) and 0 <= y < len(img[0]): |
| 111 | + cnt += 1 |
| 112 | + s += img[x][y] |
| 113 | + ans[i][j] = s // cnt |
| 114 | + return ans |
| 115 | +``` |
| 116 | + |
| 117 | +#### Java |
| 118 | + |
| 119 | +```java |
| 120 | +class Solution { |
| 121 | + public int[][] imageSmoother(int[][] img) { |
| 122 | + int[][] ans = new int[img.length][img[0].length]; |
| 123 | + for (int i = 0; i < img.length; i++) { |
| 124 | + for (int j = 0; j < img[0].length; j++) { |
| 125 | + int cnt = 0, s = 0; |
| 126 | + for (int dx = -1; dx <= 1; dx++) { |
| 127 | + for (int dy = -1; dy <= 1; dy++) { |
| 128 | + int x = i + dx, y = j + dy; |
| 129 | + if (0 <= x && x < img.length && 0 <= y && y < img[0].length) { |
| 130 | + cnt++; |
| 131 | + s += img[x][y]; |
| 132 | + } |
| 133 | + } |
| 134 | + } |
| 135 | + ans[i][j] = s / cnt; |
| 136 | + } |
| 137 | + } |
| 138 | + return ans; |
| 139 | + } |
| 140 | +} |
| 141 | +``` |
| 142 | + |
| 143 | +#### Go |
| 144 | + |
| 145 | +```go |
| 146 | +package main |
| 147 | + |
| 148 | +func imageSmoother(img [][]int) (ans [][]int) { |
| 149 | + ans = make([][]int, len(img)) |
| 150 | + for i := range ans { |
| 151 | + ans[i] = make([]int, len(img[0])) |
| 152 | + for j := range ans[i] { |
| 153 | + cnt, s := 0, 0 |
| 154 | + for _, row := range img[max(0, i - 1):min(len(img), i + 2)] { |
| 155 | + for _, val := range row[max(0, j - 1):min(len(img[0]), j + 2)] { |
| 156 | + cnt++ |
| 157 | + s += val |
| 158 | + } |
| 159 | + } |
| 160 | + ans[i][j] = s / cnt |
| 161 | + } |
| 162 | + } |
| 163 | + return ans |
| 164 | +} |
| 165 | +``` |
| 166 | + |
| 167 | +> 同步发文于CSDN和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2024/11/18/LeetCode%200661.%E5%9B%BE%E7%89%87%E5%B9%B3%E6%BB%91%E5%99%A8/)哦~ |
| 168 | +> |
| 169 | +> Tisfy：[https://letmefly.blog.csdn.net/article/details/143867615](https://letmefly.blog.csdn.net/article/details/143867615) |
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