The Image Smoother Algorithm in C++/Java

  • 时间：2020-10-05 13:15:44
  • 分类：网络文摘
  • 阅读：131 次

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].

How to Smooth Image in C++?

The following C++ code implements O(N) algorithm (where N is the number of pixels in the image) that iterates each pixel. We can’t modify the existing image, rather, it has to be done on a separate copy of image.

We can deep copy the std::vector, or just assign new pixel to it.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        int row = M.size();
        if (row == 0) return M;
        int width = M[0].size();
        if (width == 0) return M;
        vector<vector<int>> N(row, vector<int>(width)); // or N = M deep copy of vector
        for (int i = 0; i < row; ++ i) {
            for (int j = 0; j < width; ++ j) {
                int sum = 0, c = 0;
                for (int k = max(0, i - 1); k <= min(i + 1, row - 1); k ++) {
                    for (int u = max(0, j - 1); u <= min(j + 1, width - 1); u ++) {
                        sum += M[k][u];
                        c ++;
                    }
                }
                N[i][j] = sum / c;
            }
        }
        return N;
    }
};
class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
        int row = M.size();
        if (row == 0) return M;
        int width = M[0].size();
        if (width == 0) return M;
        vector<vector<int>> N(row, vector<int>(width)); // or N = M deep copy of vector
        for (int i = 0; i < row; ++ i) {
            for (int j = 0; j < width; ++ j) {
                int sum = 0, c = 0;
                for (int k = max(0, i - 1); k <= min(i + 1, row - 1); k ++) {
                    for (int u = max(0, j - 1); u <= min(j + 1, width - 1); u ++) {
                        sum += M[k][u];
                        c ++;
                    }
                }
                N[i][j] = sum / c;
            }
        }
        return N;
    }
};

We can check if the neighbour index of pixels are valid. Alternatively, we can use min/max to make sure the indices are always valid. We don’t need to use floor function as the integer division is floor anyway.

How to Smooth Image in Java?

The same image smooth algorithm can be implemented in Java as follows.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

class Solution {
    public int[][] imageSmoother(int[][] M) {
        int row = M.length;
        if (row == 0) return M;
        int width = M[0].length;
        if (width == 0) return M;
        int[][] N = new int[row][width];
        for (int i = 0; i < row; ++ i) {
            for (int j = 0; j < width; ++ j) {
                int sum = 0, c = 0;
                for (int k = Math.max(0, i - 1); k <= Math.min(i + 1, row - 1); k ++) {
                    for (int u = Math.max(0, j - 1); u <= Math.min(j + 1, width - 1); u ++) {
                        sum += M[k][u];
                        c ++;
                    }
                }
                N[i][j] = sum / c;
            }
        }
        return N;        
    }
}
class Solution {
    public int[][] imageSmoother(int[][] M) {
        int row = M.length;
        if (row == 0) return M;
        int width = M[0].length;
        if (width == 0) return M;
        int[][] N = new int[row][width];
        for (int i = 0; i < row; ++ i) {
            for (int j = 0; j < width; ++ j) {
                int sum = 0, c = 0;
                for (int k = Math.max(0, i - 1); k <= Math.min(i + 1, row - 1); k ++) {
                    for (int u = Math.max(0, j - 1); u <= Math.min(j + 1, width - 1); u ++) {
                        sum += M[k][u];
                        c ++;
                    }
                }
                N[i][j] = sum / c;
            }
        }
        return N;        
    }
}

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读：
 7 Questions You Must Answer to Get the Right Kind of Faceb　 How to Check if a Binary Tree is Balanced (Top-down and Bottom-u　 How to Check If A String Is a Number (Numeric) using Regex (C++/　 The Bash Programming Tutorial: Compute the GCD (Greatest Common 　 How to Re-Number the Files Sequentially on Windows using Batch P　 How to Split a String in Balanced Strings?　 Finding Out the Longest Arithmetic Subsequence of Given Differen　 How to Count the Prime Number of Set Bits in Binary Representati　 How to Create a Mock Echo Soap/REST API using SoapUI?　 How to Partition Array into Disjoint Intervals?　
评论列表
添加评论