一、問題描述

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.

Example1

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].

二、題意分析

將二維矩陣中每一個元素的值，設定為以當前元素為中心的3*3區域的平均值。若周圍沒有9個元素（落在四條邊上），則有多少計算多少個。

三、解題思路一

最開始的思路是，在矩陣周圍圍上一圈值為0的牆，計算每一個元素3*3區域的累加和，然後統計出的每個元素周圍所含元素的個數（不包含牆值），最後計算個平均值就可以了。 但是，這樣做的時候我在統計每個元素周圍的元素個數時，有些麻煩。

程式碼實現

class Solution {
    public int[][] imageSmoother(int[][] M) {
        int row = M.length, col = M[
 
0].length;
        System.out.println("row : " + row);
        System.out.println("col : " + col);
        int[][] result = new int[row+2][col + 2];
        // 給矩陣周圍加一圈0，方便計算
        for(int i = 0; i < row+2; i++){
            for(int j = 0; j < col+2; j++){
                if(i == 0 || i == row + 1){
                    result[i][j] = 0;
                    continue;
                }
                if(j == 0 || j == col + 1){
                    result[i][j] = 0;
                    continue;
                }
                result[i][j] = M[i-1][j-1];   
            }
        }   
        int numbers = 9;
        for(int i = 0 ; i < M.length; i++){
            for(int j = 0; j < M[i].length; j++){
                numbers = 9;
                if((i == 0 || i == row - 1) && (j == 0 || j == col -1)){
                     
                     numbers = 4;
                    if(row == 1 || col == 1)
                        numbers = 2;
                    if(col == 1 && row == 1)
                        numbers = 1;
                }else if((i == 0 || i == row -1) && (0 < j && j < col - 1)){
                    
                    numbers = 6;
                    if(row == 1 || col == 1)
                        numbers = 3;
                }else if((j == 0 || j == col - 1) && (0 < i && i < row -1)){
                    numbers = 6;
                    if(row == 1 || col == 1)
                        numbers = 3;
          
                }
                M[i][j] = calculateAvg(result, i + 1, j + 1, numbers);
    
            }
        }
        return M;
    
    }
    // 計算3*3區域的累加和
    public static int calculateAvg(int[][] result, int i, int j, int numbers){
        int sum = 0;
        for(int row = i - 1; row < i+2; row++){
            for(int col = j - 1; col < j+2; col++){
                sum +=result[row][col];            
            }
        }
        return (int) (sum/ numbers);
    }
}

時間、空間複雜度分析

1. 時間複雜度

O (n²)

2. 空間複雜度

O (n²)

四、解題思路二

直接暴力法解決

程式碼實現

class Solution {
    public int[][] imageSmoother(int[][] M) {
        int row = M.length, col = M[0].length;
        int[][] result = new int[row][col];
        int numbers = 0, sum = 0;
        for(int i = 0; i < row; i++){
            for(int j = 0; j < col; j++){
                 numbers = 0;
                 sum = 0;
                for(int k = i - 1; k < i + 2; k++){
                    // 邊界判斷
                    if(k < 0 || k >= row)
                        continue;
                    for(int c = j - 1; c < j+2; c++){
                        // 邊界判斷
                        if(c < 0 || c >= col)
                            continue;
                        sum += M[k][c];
                        numbers++;
                    }
                }
                result[i][j] = sum / numbers;
                
            }
        }
        return result;
    }
}

時間、空間複雜度分析

1. 時間複雜度

O (n²)

2. 空間複雜度

O (1)