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:
问题解析:
给定表示图像灰度的二维整数矩阵M,平滑图像,使每个单元格的灰度值成为所有包括周围8个单元格及其本身的平均灰度(舍入)。如果周围的单元格小于8个,则使用尽可能多的单元格。
题目简单,遍历实现即可。
class Solution {
public int[][] imageSmoother(int[][] M) {
if (M == null) return null;
int rows = M.length;
if (rows == 0) return new int[0][];
int cols = M[0].length;
int result[][] = new int[rows][cols];
for (int row = 0; row < rows; row++){
for (int col = 0; col < cols; col++){
int count = 0;
int sum = 0;
for (int rInc : new int[]{-1, 0, 1}){
for (int cInc : new int[]{-1, 0, 1}){
if (isValid(row + rInc, col + cInc, rows, cols)){
count++;
sum += M[row + rInc][col + cInc];
}
}
}
result[row][col] = sum / count;
}
}
return result;
}
private boolean isValid(int x, int y, int rows, int cols){
return x >=0 && x < rows && y >=0 && y < cols;
}
}