## 题目描述

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].

## 解法一

### Python

``````class Solution:
def imageSmoother(self, M):
"""
:type M: List[List[int]]
:rtype: List[List[int]]
"""
m = len(M)
n = len(M[0])
result = [[0] * n for _ in range(m)]
for i in range(m):
for j in range(n):
current = 0
count = 0
for a in [-1, 0, 1]:
for b in [-1, 0, 1]:
row = i + a
col = j + b
if 0 <= row < m and 0 <= col < n:
current += M[row][col]
count += 1
result[i][j] = math.floor(current / count)
return result``````

### Java

``````class Solution {
public int[][] imageSmoother(int[][] M) {
int m = M.length;
int n = M[0].length;
int[][] result = new int[m][n];
int[] range = new int[] {-1, 0, 1};
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int current = 0;
int count = 0;
for (int a : range) {
for (int b : range) {
int row = i + a;
int col = j + b;
if (row >= 0 && row < m && col >= 0 && col < n) {
current += M[row][col];
count++;
}
}
}
result[i][j] = (int) Math.floor((double) current / count);
}
}
return result;
}
}``````

### C++

``````class Solution {
public:
vector<vector<int>> imageSmoother(vector<vector<int>> &M) {
int m = M.size();
int n = M[0].size();
vector<vector<int>> result(m, vector<int>(n));
int range[] = {-1, 0, 1};
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int current = 0;
int count = 0;
for (int a : range) {
for (int b : range) {
int row = i + a;
int col = j + b;
if (row >= 0 && row < m && col >= 0 && col < n) {
current += M[row][col];
count++;
}
}
}
result[i][j] = (int) floor((double) current / count);
}
}
return result;
}
};``````