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:
- 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].
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; } };
原文地址:http://www.cnblogs.com/pk28/
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