[LeetCode] 661. Image Smoother_Easy

Posted 2021-01-01

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].

 思路就是正常的, 两个for loop, 然后将8个邻居和自己相加取平均数, 最后代替原来的数即可.

 

T: O(m,n)   S; O(1)

 

Code

class Solution:
    def imageSmoother(self, M):
        dirs, lrc = [(x, y) for x in range(-1,2) for y in range(-1,2)], [len(M), len(M[0])]
        ans = [[0]* lrc[1] for _ in range(lrc[0])]
        def sum2(i, j):
            ans, count = 0, 0
            for c1, c2 in dirs:
                nr, nc = i + c1, j + c2
                if 0 <= nr < lrc[0]  and 0 <= nc < lrc[1]:
                    count += 1
                    ans += M[nr][nc]
            return ans//count
        for i in range(lrc[0]):
            for j in range(lrc[1]):
                ans[i][j] = sum2(i,j)
        return ans