661. Image Smoother 图片平滑器
2022年3月7日
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/
@TOC
题目地址:https://leetcode.com/problems/image-smoother/description/
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:
- 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].
题目大意
对一张图片取模糊,每个位置的数值等于9联通区域内的所有数字的平均值取整。
解题方法
方法一:暴力解决
非常粗暴的解法。判断数组中每个元素的周边的所有元素的和,并记录元素的个数。用判断边界的方式,防止过界。
from copy import deepcopy as copy
class Solution(object):
def imageSmoother(self, M):
"""
:type M: List[List[int]]
:rtype: List[List[int]]
"""
if not M or not M[0]:
return M
rows = len(M)
cols = len(M[0])
isValid = lambda i,j: i >=0 and i < rows and j >= 0 and j < cols
row, col = 0, 0
answer = copy(M)
for row in xrange(rows):
for col in xrange(cols):
_sum, count = 0, 0
for i in xrange(-1, 2):
for j in xrange(-1, 2):
if isValid(row + i, col + j):
_sum += M[row + i][col + j]
count += 1
answer[row][col] = _sum / count
return answer
日期
2018 年 1 月 24 日 2018 年 11 月 16 日 —— 又到周五了!