leetcode 661. Image Smoother

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].
class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        int dir[8][2] = {1,0,1,1,1,-1,0,1,0,-1,-1,0,-1,1,-1,-1};
        int n = M.size();
        int m = M[n - 1].size();
        vector<vector<int> > w;
        for (int i = 0; i < n; ++i) {
            vector<int> v;
            for (int j = 0; j < m; ++j) {
                int ans = 0;
                ans += M[i][j];
                int num = 1;
                for (int k = 0; k < 8; ++k) {
                    int nx = i + dir[k][0];
                    int ny = j + dir[k][1];
                    if (nx < n && nx >= 0 && ny < m && ny >= 0) {
                        ans += M[nx][ny];
                        num++;
                    }
                }
                //cout << ans << endl;
                v.push_back(ans/num);
            }
            w.push_back(v);
            //cout << endl;
        }
        return w;
    }
};

 

posted on 2017-08-20 12:26  Beserious  阅读(235)  评论(0编辑  收藏  举报