leetcode1386

先给出一个TLE的方案，思路很简单：

class Solution:
    def maxNumberOfFamilies(self, n: int, reservedSeats: 'List[List[int]]') -> int:
        dic_row = dict()
        for seats in reservedSeats:
            row,position = seats[0],seats[1]
            if row not in dic_row:
                dic_row[row] = [position]
            else:
                dic_row[row].append(position)
        count = 0
        for i in range(1,n+1):
            if i in dic_row:
                row = dic_row[i]
                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:
                    count += 2
                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):
                    count += 1
                else:
                    count += 0
            else:
                count += 2
        return count

再给一个AC的方案，简化了判断条件：

class Solution:
    def maxNumberOfFamilies(self, n: int, reservedSeats: List[List[int]]) -> int:
        seats = collections.defaultdict(int) 
        res = 0
        
        for row, col in reservedSeats:
            seats[row] = seats[row] | (1 << (col-1))

        for reserved in seats.values():
            curr = 0
            curr += (reserved & int('0111100000', 2)) == 0
            curr += (reserved & int('0000011110', 2)) == 0
            curr += (reserved & int('0001111000', 2)) == 0 and curr == 0

            res += curr    

        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。