leetcode1386

先给出一个TLE的方案，思路很简单：

 1 class Solution:
 2     def maxNumberOfFamilies(self, n: int, reservedSeats: 'List[List[int]]') -> int:
 3         dic_row = dict()
 4         for seats in reservedSeats:
 5             row,position = seats[0],seats[1]
 6             if row not in dic_row:
 7                 dic_row[row] = [position]
 8             else:
 9                 dic_row[row].append(position)
10         count = 0
11         for i in range(1,n+1):
12             if i in dic_row:
13                 row = dic_row[i]
14                 if 2 not in row and 3 not in row and 4 not in row and 5 not in row and 6 not in row and 7 not in row and 8 not in row and 9 not in row:
15                     count += 2
16                 elif (2 not in row and 3 not in row and 4 not in row and 5 not in row) or (4 not in row and 5 not in row and 6 not in row and 7 not in row) or (6 not in row and 7 not in row and 8 not in row and 9 not in row):
17                     count += 1
18                 else:
19                     count += 0
20             else:
21                 count += 2
22         return count

再给一个AC的方案，简化了判断条件：

 1 class Solution:
 2     def maxNumberOfFamilies(self, n: int, reservedSeats: List[List[int]]) -> int:
 3         seats = collections.defaultdict(int) 
 4         res = 0
 5         
 6         for row, col in reservedSeats:
 7             seats[row] = seats[row] | (1 << (col-1))
 8 
 9         for reserved in seats.values():
10             curr = 0
11             curr += (reserved & int('0111100000', 2)) == 0
12             curr += (reserved & int('0000011110', 2)) == 0
13             curr += (reserved & int('0001111000', 2)) == 0 and curr == 0
14 
15             res += curr    
16 
17         return res + 2 * (n - len(seats))

参考地址：https://leetcode.com/problems/cinema-seat-allocation/discuss/548251/Python3-Easy-Python-using-bitmask-(Detailed-Solution)

这种解决方案很巧妙，佩服。

posted on 2020-03-28 15:39 Sempron2800+ 阅读(193) 评论(0) 编辑收藏举报