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) = 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:

    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。

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