LeetCode 661. Image Smoother （图像平滑器）

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].

---

题目标签：Array

　　题目给了我们一个2d M array，让我们平滑处理图片。对于每一个cell，把它更新为 以自身为中心 3x3 的平均值。

　　就用常规方法做，新设一个 res[][] array，遍历M，对于每一个cell， 遍历以它为中心的3x3的cells，得到平均值，存入res。

　　需要注意的就是，3x3的边界问题。

Java Solution:

Runtime beats 72.97% 

完成日期：10/19/2017

关键词：Array

关键点：处理3x3的边界问题

class Solution 
{
    public int[][] imageSmoother(int[][] M) 
    {
        int rows = M.length;
        int cols = M[0].length;
        int[][] res = new int[rows][cols];
        
        for(int i=0; i<rows; i++)
        {
            for(int j=0; j<cols; j++)
            {
                int sum = 0;
                int count = 0;
                // sum 3x3 area and take care of the boundary
                for(int x=Math.max(0,i-1); x<=Math.min(rows-1, i+1); x++)
                {
                    for(int y=Math.max(0, j-1); y<=Math.min(cols-1, j+1); y++)
                    {
                        sum += M[x][y]; // sum up cells value
                        count++; // count cells number
                    }
                }
                
                res[i][j] = sum / count; // get average value
            }
        }
        
        return res;
    }
}

参考资料：N/A

LeetCode 题目列表 - LeetCode Questions List