标签：题意 pre solution 空间 present rip str return 思路
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:
原题地址: Image Smoother
难度: Easy
题意: 平滑图片, 每一个点值为其四周(包含自身)的平均值
思路:
一个二维数组, 按照题意,点可以分为三类
(1)四个角的点, 其周围有4个点(包含自身)
(2)二维数组最外层除了四个角的点,其周围有6个点(包含自身)
(3)其他的点(内层点),其周围有9个点(包含自身)
直接暴力解决,遍历数组
代码:
class Solution(object): def imageSmoother(self, M): """ :type M: List[List[int]] :rtype: List[List[int]] """ m = len(M) n = len(M[0]) res = [[0] * n for i in range(m)] if m <= 1 and n <= 1: return M if m == 1 and n > 1: res[0][0] = (M[0][0] + M[0][1]) // 2 res[0][-1] = (M[0][-1] + M[0][-2]) // 2 for j in range(1, n-1): res[0][j] = sum(M[0][j-1: j+2]) // 3 return res if n == 1 and m > 1: res[0][0] = (M[0][0] + M[1][0]) // 2 res[-1][0] = (M[-1][0] + M[-2][0]) // 2 for i in range(1, m-1): res[i][0] = (M[i-1][0] + M[i][0] + M[i+1][0]) // 3 return res for i in range(m): for j in range(n): if i == 0 and j == 0: res[i][j] = (M[i][j] + M[i][j+1] + M[i+1][j] + M[i+1][j+1]) // 4 if i == 0 and j == n-1: res[i][j] = (M[i][j] + M[i][j-1] + M[i+1][j] + M[i+1][j-1]) // 4 if i == m-1 and j == 0: res[i][j] = (M[i][j] + M[i][j+1] + M[i-1][j] + M[i-1][j+1]) // 4 if i == m-1 and j == n-1: res[i][j] = (M[i][j] + M[i][j-1] + M[i-1][j] + M[i-1][j-1]) // 4 if i == 0 and 0 < j < n-1: res[i][j] = (sum(M[i][j-1: j+2]) + sum(M[i+1][j-1: j+2])) // 6 if i == m-1 and 0 < j < n-1: res[i][j] = (sum(M[i][j-1: j+2]) + sum(M[i-1][j-1: j+2])) // 6 if j == 0 and 0 < i < m-1: res[i][j] = (sum(M[i][j: j+2]) + sum(M[i-1][j: j+2]) + sum(M[i+1][j: j+2])) // 6 if j == n-1 and 0 < i < m-1: res[i][j] = (sum(M[i][j-1: j+1]) + sum(M[i-1][j-1: j+1]) + sum(M[i+1][j-1: j+1])) // 6 if 0 < i < m -1 and 0 < j < n-1: res[i][j] = (sum(M[i][j-1: j+2]) + sum(M[i-1][j-1: j+2]) + sum(M[i+1][j-1: j+2])) // 9 return res
时间复杂度: O(mn)
空间复杂度: O(1)
标签：题意 pre solution 空间 present rip str return 思路
原文地址：https://www.cnblogs.com/chimpan/p/9727238.html
