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  • 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。

    版权声明:本文为博主原创文章,未经博主允许不得转载。
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  • 原文地址:https://www.cnblogs.com/kevincong/p/7887601.html
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