Question

661. 图片平滑器

English Version

题目描述

图像平滑器 是大小为 3 x 3 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。

每个单元格的  平均灰度 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）。

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）。

给你一个表示图像灰度的 m x n 整数矩阵 img ，返回对图像的每个单元格平滑处理后的图像 。

 

示例 1:

输入:img = [[1,1,1],[1,0,1],[1,1,1]]
输出:[[0, 0, 0],[0, 0, 0], [0, 0, 0]]
解释:
对于点 (0,0), (0,2), (2,0), (2,2): 平均(3/4) = 平均(0.75) = 0
对于点 (0,1), (1,0), (1,2), (2,1): 平均(5/6) = 平均(0.83333333) = 0
对于点 (1,1): 平均(8/9) = 平均(0.88888889) = 0

示例 2:

输入: img = [[100,200,100],[200,50,200],[100,200,100]]
输出: [[137,141,137],[141,138,141],[137,141,137]]
解释:
对于点 (0,0), (0,2), (2,0), (2,2): floor((100+200+200+50)/4) = floor(137.5) = 137
对于点 (0,1), (1,0), (1,2), (2,1): floor((200+200+50+200+100+100)/6) = floor(141.666667) = 141
对于点 (1,1): floor((50+200+200+200+200+100+100+100+100)/9) = floor(138.888889) = 138

 

提示:

• m == img.length
• n == img[i].length
• 1 <= m, n <= 200
• 0 <= img[i][j] <= 255

解法

方法一

class Solution:
  def imageSmoother(self, img: List[List[int]]) -> List[List[int]]:
    m, n = len(img), len(img[0])
    ans = [[0] * n for _ in range(m)]
    for i in range(m):
      for j in range(n):
        s = cnt = 0
        for x in range(i - 1, i + 2):
          for y in range(j - 1, j + 2):
            if 0 <= x < m and 0 <= y < n:
              cnt += 1
              s += img[x][y]
        ans[i][j] = s // cnt
    return ans

class Solution {
  public int[][] imageSmoother(int[][] img) {
    int m = img.length;
    int n = img[0].length;
    int[][] ans = new int[m][n];
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        int s = 0;
        int cnt = 0;
        for (int x = i - 1; x <= i + 1; ++x) {
          for (int y = j - 1; y <= j + 1; ++y) {
            if (x >= 0 && x < m && y >= 0 && y < n) {
              ++cnt;
              s += img[x][y];
            }
          }
        }
        ans[i][j] = s / cnt;
      }
    }
    return ans;
  }
}

class Solution {
public:
  vector<vector<int>> imageSmoother(vector<vector<int>>& img) {
    int m = img.size(), n = img[0].size();
    vector<vector<int>> ans(m, vector<int>(n));
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        int s = 0, cnt = 0;
        for (int x = i - 1; x <= i + 1; ++x) {
          for (int y = j - 1; y <= j + 1; ++y) {
            if (x < 0 || x >= m || y < 0 || y >= n) continue;
            ++cnt;
            s += img[x][y];
          }
        }
        ans[i][j] = s / cnt;
      }
    }
    return ans;
  }
};

func imageSmoother(img [][]int) [][]int {
  m, n := len(img), len(img[0])
  ans := make([][]int, m)
  for i, row := range img {
    ans[i] = make([]int, n)
    for j := range row {
      s, cnt := 0, 0
      for x := i - 1; x <= i+1; x++ {
        for y := j - 1; y <= j+1; y++ {
          if x >= 0 && x < m && y >= 0 && y < n {
            cnt++
            s += img[x][y]
          }
        }
      }
      ans[i][j] = s / cnt
    }
  }
  return ans
}

function imageSmoother(img: number[][]): number[][] {
  const m = img.length;
  const n = img[0].length;
  const locations = [
    [-1, -1],
    [-1, 0],
    [-1, 1],
    [0, -1],
    [0, 0],
    [0, 1],
    [1, -1],
    [1, 0],
    [1, 1],
  ];

  const res = [];
  for (let i = 0; i < m; i++) {
    res.push([]);
    for (let j = 0; j < n; j++) {
      let sum = 0;
      let count = 0;
      for (const [y, x] of locations) {
        if ((img[i + y] || [])[j + x] != null) {
          sum += img[i + y][j + x];
          count++;
        }
      }
      res[i].push(Math.floor(sum / count));
    }
  }
  return res;
}

impl Solution {
  pub fn image_smoother(img: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
    let m = img.len();
    let n = img[0].len();
    let locations = [
      [-1, -1],
      [-1, 0],
      [-1, 1],
      [0, -1],
      [0, 0],
      [0, 1],
      [1, -1],
      [1, 0],
      [1, 1],
    ];

    let mut res = vec![];
    for i in 0..m {
      res.push(vec![]);
      for j in 0..n {
        let mut sum = 0;
        let mut count = 0;
        for [y, x] in locations.iter() {
          let i = (i as i32) + y;
          let j = (j as i32) + x;
          if i < 0 || i == (m as i32) || j < 0 || j == (n as i32) {
            continue;
          }
          count += 1;
          sum += img[i as usize][j as usize];
        }
        res[i].push(sum / count);
      }
    }
    res
  }
}