leetCode-Image Smoother

Description:
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) = 0For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0For the point (1,1): floor(8/9) = floor(0.88888889) = 0

Note:

    The value in the given matrix is in the range of [0, 255].    The length and width of the given matrix are in the range of [1, 150].

Solution:

class Solution {    public int[][] imageSmoother(int[][] M) {        int[][] N = new int[M.length][M[0].length];        for(int j = 0; j < M.length; j++){            for(int i =0; i < M[0].length; i++){                help(M, N, i,j);            }        }        return N;    }    public void help(int[][] M, int[][] N, int x, int y){        int count = 0;        int sum = 0;        for(int i = -1; i <=1; i++){            for(int j = -1; j <= 1; j++){                if(x + i >= 0 && x + i < M[0].length && y + j >=0 && y + j < M.length){                    sum += M[j + y][x + i];                    count++;                }            }        }        N[y][x] = sum / count;    }}

总结:
注意新建一个二维数组存放平滑图像绘制后的数据，另外对M行、列分别迭代，对当前元素的8+1（自己）进行判断是否越界，不越界则count+1，越界不计数。