661. Image Smoother

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].

题意：给定二维矩阵，表示图像的灰度。对图像进行平滑处理，将每一个像素的值换成其周围9个单元格的平均值。

思路
1.题目要求矩阵中元素周围一圈元素和它本身累加之后求平均值。
2.因为对于每个元素，计算之后的值覆盖原值会影响其他元素的计算，所以要创建一个新的二维数组来保存。
3.比较老实（呆）的计算M[i][j]的方式：M[i-1][n-1]+M[i-1][n]+M[i-1][n+1]+...+M[i+1][n+1]，再来求平均。再来判断元素是在四个角上还是边上还是内部的，这样做太麻烦了。
4.现在需要一个比较好的方式来指代M[i][j]的周围元素和其本身。假设M[i+m][j+n]是周围元素和其本身元素中的一个，那么m和n属于[-1, 0, 1]。所以可以建两个数组：

    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[][] rets = 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 incR : new int[]{-1, 0, 1}) {
                    for (int incC : new int[]{-1, 0, 1}) {
                        int r = row + incR;
                        int c = col + incC;
                        if (0 <= r && r < rows && 0 <= c && c < cols) {
                            count++;
                            sum += M[r][c];
                        }
                    }
                }
                rets[row][col] = sum / count;
            }
        }
        return rets;
    }