Python习题集（1）Moore Neighbourhood

yanshicj

有意思

conn

谢谢分享

爱琴海里的星

大神们~

旅程

学习学习

skyflyingsnow

即使纯粹把所有的情况列出来也只占20多行啊。

1. def count_neighbours(grid, row, col):
   
2.     if (row < 0 or row >= len(grid)) and (col < 0 or col >= len(grid[0])):
   
3.         return 0
   
4.     elif (row > 0 and row < len(grid)-1) and (col > 0 and col < len(grid[0])-1) :#in the middle
   
5.         return grid[row-1][col-1] + grid[row-1][col] + grid[row-1][col+1] + grid[row][col-1] + grid[row][col+1] + grid[row+1][col-1] + grid[row+1][col] + grid[row+1][col+1]
   
6.     elif row == 0 and col == 0:# in four corners
   
7.         return grid[row][col+1] + grid[row+1][col+1] + grid[row+1][col]
   
8.     elif row == 0 and col == len(grid[0])-1 :
   
9.         return grid[row][col-1] + grid[row+1][col-1] + grid[row+1][col]
   
10.     elif row == len(grid)-1 and col == 0 :
    
11.         return grid[row-1][col] + grid[row-1][col+1] + grid[row][col+1]
    
12.     elif row == len(grid)-1 and col == len(grid[0])-1:
    
13.         return grid[row-1][col] + grid[row][col-1] + grid[row-1][col-1]
    
14.     elif row == 0 and (col > 0 and col < len(grid[0])-1): #on four sides
    
15.         return grid[row][col-1] + grid[row][col+1] + grid[row+1][col-1] + grid[row+1][col] + grid[row+1][col+1]
    
16.     elif row == len(grid) -1 and (col > 0 and col < len(grid[0])-1):
    
17.         return grid[row][col-1] + grid[row][col+1] + grid[row-1][col-1] + grid[row-1][col] + grid[row-1][col+1]
    
18.     elif (row > 0 or row < len(grid)-1) and col == 0:
    
19.         return grid[row-1][col] + grid[row+1][col] + grid[row-1][col+1] + grid[row][col+1] + grid[row+1][col+1]
    
20.     elif (row > 0 or row < len(grid)-1) and col == len(grid[0])-1:
    
21.         return grid[row-1][col] + grid[row+1][col] + grid[row-1][col-1] + grid[row][col-1] + grid[row+1][col-1]
    
22.     else:
    
23.         return 0

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其实在Checkio上面有人用4行就解决了这个问题，那个人真的好厉害啊：

1. def count_neighbours(grid, row, col):
   
2.     rows = range(max(0, row - 1), min(row + 2, len(grid)))
   
3.     cols = range(max(0, col - 1), min(col + 2, len(grid[0])))
   
4.     return sum(grid[r][c] for r in rows for c in cols) - grid[row][col]

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彬雪

我是来学习的