【LeetCode】661. Image Smoother
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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) = 0For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0For 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].
题意:求矩阵每个位置的相邻8个位置加上本身的值的平均值。
class Solution {public: int find(vector<vector<int>>& M,int x,int y){ int dx[9]={0,0,1,1,1,0,-1,-1,-1}; int dy[9]={0,-1,-1,0,1,1,1,0,-1}; int r=M.size(),c=M[0].size(); int sum=0,n=0; for(int i=0;i<9;i++){ int xi=x+dx[i]; int yi=y+dy[i]; if(xi>=0&&xi<r&&yi>=0&&yi<c){ sum+=M[xi][yi]; n++; } } return sum/n; } vector<vector<int>> imageSmoother(vector<vector<int>>& M) { int r=M.size(),c=M[0].size(); vector<vector<int>> ans(r, vector<int>(c)); for(int i=0;i<r;i++){ for(int j=0;j<c;j++){ ans[i][j]=find(M,i,j); } } return ans; }};
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