661. Image Smoother【easy】

661. Image Smoother【easy】

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].

 

错误解法：

class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        int row = M.size();
        int col = M[0].size();
        
        vector<vector<int>> temp(row + 1, vector<int>(col + 1));
        for (int j = 0; j < col + 1; ++j) {
            temp[0][j] = 0;
        }
        for (int i = 0; i < row + 1; ++i) {
            temp[i][0] = 0;
        }
        for (int j = 0; j < col + 1; ++j) {
            temp[row][j] = 0;
        }
        for (int i = 0; i < row + 1; ++i) {
            temp[i][col] = 0;
        }
        
        for (int i = 1; i < row; ++i) {
            for (int j = 1; j < col; ++j) {
                temp[i][j] = M[i - 1][j - 1];
            }
        }
        
        for (int i = 1; i < row; ++i) {
            for (int j = 1; j < col; ++j) {
                int sum = 0;
                for (int x = -1; x <= 1; ++x) {
                    for (int y = -1; y <= 1; ++y) {
                            sum += temp[i + x][j + y];        
                    }
                }
                
                temp[i][j] = floor(sum / 9);
            }
        }
        
        vector<vector<int>> result(row, vector<int>(col));
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                result[i][j] = temp[i + 1][j + 1];
            }
        }
        
        return result;
    }
};

一开始我还想取巧，把边界扩充，想着可以一致处理，但是发现没有审清题意，坑了啊！

 

解法一：

class Solution {
private: 
    bool valid(int i,int j,vector<vector<int>>& M)
    {
        if (i >=0 && i<M.size() && j>=0 && j<M[0].size())
            return true;
        return false;
    }
    
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        vector<vector<int>> res;
        if (M.size()==0 || M[0].size()==0)
            return res;

        for (int i = 0; i< M.size(); i++)
        {            
            vector<int> cur;
            for(int j = 0; j< M[0].size(); j++)
            {
                int total = 0;
                int count = 0;
                for (int x = -1; x<2;x++)
                {
                    for (int y = -1; y<2; y++)
                    {
                        if(valid(i+x,j+y,M))
                        {
                            count++;
                            total +=M[i+x][j+y];
                        }
                    }
                }
                cur.push_back(total/count);
            }
            res.push_back(cur);
        }
        return res; 
    }
};

中规中矩的解法，完全按照题目意思搞

 

解法三：

class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        int m = M.size(), n = M[0].size();
        if (m == 0 || n == 0) return {{}};
        vector<vector<int>> dirs = {{0,1},{0,-1},{1,0},{-1,0},{-1,-1},{1,1},{-1,1},{1,-1}};
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int sum = M[i][j], cnt = 1;
                for (int k = 0; k < dirs.size(); k++) {
                    int x = i + dirs[k][0], y = j + dirs[k][1];
                    if (x < 0 || x > m - 1 || y < 0 || y > n - 1) continue;
                    sum += (M[x][y] & 0xFF);
                    cnt++;
                }
                M[i][j] |= ((sum / cnt) << 8);
            }
        }
         for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                M[i][j] >>= 8;
            }
         }
        return M;
    }

};

真正的大神解法！大神解释如下：Derived from StefanPochmann's idea in "game of life": the board has ints in [0, 255], hence only 8-bit is used, we can use the middle 8-bit to store the new state (average value), replace the old state with the new state by shifting all values 8 bits to the right.