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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