题目描述

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.

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].

解法一

思路

这题就是遍历数组元素,对于每个元素,将其值更新为其周围元素值的平均值。这里要考虑的问题是,可不可以当场更新,即不创建新的数组,遍历到某个元素时当场将其值更新?其实这样是不可以的,因为新的元素值需要是周围元素的“原始”值,而一旦周围有元素的值更新了,那么计算结果就出错了。所以这题必须要新建一个大小相等的二维数组,用来填充对应位置新的元素值。

Python

class Solution:
    def imageSmoother(self, M):
        """
        :type M: List[List[int]]
        :rtype: List[List[int]]
        """
        m = len(M)
        n = len(M[0])
        result = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                current = 0
                count = 0
                for a in [-1, 0, 1]:
                    for b in [-1, 0, 1]:
                        row = i + a
                        col = j + b
                        if 0 <= row < m and 0 <= col < n:
                            current += M[row][col]
                            count += 1
                result[i][j] = math.floor(current / count)
        return result

Java

class Solution {
    public int[][] imageSmoother(int[][] M) {
        int m = M.length;
        int n = M[0].length;
        int[][] result = new int[m][n];
        int[] range = new int[] {-1, 0, 1};
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int current = 0;
                int count = 0;
                for (int a : range) {
                    for (int b : range) {
                        int row = i + a;
                        int col = j + b;
                        if (row >= 0 && row < m && col >= 0 && col < n) {
                            current += M[row][col];
                            count++;
                        }
                    }
                }
                result[i][j] = (int) Math.floor((double) current / count);
            }
        }
        return result;
    }
}

C++

class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>> &M) {
        int m = M.size();
        int n = M[0].size();
        vector<vector<int>> result(m, vector<int>(n));
        int range[] = {-1, 0, 1};
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int current = 0;
                int count = 0;
                for (int a : range) {
                    for (int b : range) {
                        int row = i + a;
                        int col = j + b;
                        if (row >= 0 && row < m && col >= 0 && col < n) {
                            current += M[row][col];
                            count++;
                        }
                    }
                }
                result[i][j] = (int) floor((double) current / count);
            }
        }
        return result;
    }
};