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：

 1 class Solution {
 2 public:
 3     vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
 4         int row = M.size();
 5         int col = M[0].size();
 6
 7         vector<vector<int>> temp(row + 1, vector<int>(col + 1));
 8         for (int j = 0; j < col + 1; ++j) {
 9             temp[0][j] = 0;
10         }
11         for (int i = 0; i < row + 1; ++i) {
12             temp[i][0] = 0;
13         }
14         for (int j = 0; j < col + 1; ++j) {
15             temp[row][j] = 0;
16         }
17         for (int i = 0; i < row + 1; ++i) {
18             temp[i][col] = 0;
19         }
20
21         for (int i = 1; i < row; ++i) {
22             for (int j = 1; j < col; ++j) {
23                 temp[i][j] = M[i - 1][j - 1];
24             }
25         }
26
27         for (int i = 1; i < row; ++i) {
28             for (int j = 1; j < col; ++j) {
29                 int sum = 0;
30                 for (int x = -1; x <= 1; ++x) {
31                     for (int y = -1; y <= 1; ++y) {
32                             sum += temp[i + x][j + y];
33                     }
34                 }
35
36                 temp[i][j] = floor(sum / 9);
37             }
38         }
39
40         vector<vector<int>> result(row, vector<int>(col));
41         for (int i = 0; i < row; ++i) {
42             for (int j = 0; j < col; ++j) {
43                 result[i][j] = temp[i + 1][j + 1];
44             }
45         }
46
47         return result;
48     }
49 };

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：

 1 class Solution {
 2 private:
 3     bool valid(int i,int j,vector<vector<int>>& M)
 4     {
 5         if (i >=0 && i<M.size() && j>=0 && j<M[0].size())
 6             return true;
 7         return false;
 8     }
 9
10 public:
11     vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
12         vector<vector<int>> res;
13         if (M.size()==0 || M[0].size()==0)
14             return res;
15
16         for (int i = 0; i< M.size(); i++)
17         {
18             vector<int> cur;
19             for(int j = 0; j< M[0].size(); j++)
20             {
21                 int total = 0;
22                 int count = 0;
23                 for (int x = -1; x<2;x++)
24                 {
25                     for (int y = -1; y<2; y++)
26                     {
27                         if(valid(i+x,j+y,M))
28                         {
29                             count++;
30                             total +=M[i+x][j+y];
31                         }
32                     }
33                 }
34                 cur.push_back(total/count);
35             }
36             res.push_back(cur);
37         }
38         return res;
39     }
40 };

中规中矩的解法，完全按照题目意思搞

解法三：

 1 class Solution {
 2 public:
 3     vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
 4         int m = M.size(), n = M[0].size();
 5         if (m == 0 || n == 0) return {{}};
 6         vector<vector<int>> dirs = {{0,1},{0,-1},{1,0},{-1,0},{-1,-1},{1,1},{-1,1},{1,-1}};
 7         for (int i = 0; i < m; i++) {
 8             for (int j = 0; j < n; j++) {
 9                 int sum = M[i][j], cnt = 1;
10                 for (int k = 0; k < dirs.size(); k++) {
11                     int x = i + dirs[k][0], y = j + dirs[k][1];
12                     if (x < 0 || x > m - 1 || y < 0 || y > n - 1) continue;
13                     sum += (M[x][y] & 0xFF);
14                     cnt++;
15                 }
16                 M[i][j] |= ((sum / cnt) << 8);
17             }
18         }
19          for (int i = 0; i < m; i++) {
20             for (int j = 0; j < n; j++) {
21                 M[i][j] >>= 8;
22             }
23          }
24         return M;
25     }
26
27 };

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.