LeetCode每日一题(1386. Cinema Seat Allocation)

在设有n排座位的电影院中,每个座位编号从1到10。已知部分座位已被预定,目标是最大化安排四人组坐在一起的数目。四人组必须坐在同一排的相邻座位上,但不能跨过过道。例如,[3,3]和[3,4]不视为相邻。文章介绍了如何计算在给定预定座位的情况下,最多可以安排多少个四人组的方法,并给出了一些示例。" 114138364,10539019,Java8使用Stream合并两个List并去重,"['Java编程', '数据结构', '集合框架', 'Stream']

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A cinema has n rows of seats, numbered from 1 to n and there are ten seats in each row, labelled from 1 to 10 as shown in the figure above.

Given the array reservedSeats containing the numbers of seats already reserved, for example, reservedSeats[i] = [3,8] means the seat located in row 3 and labelled with 8 is already reserved.

Return the maximum number of four-person groups you can assign on the cinema seats. A four-person group occupies four adjacent seats in one single row. Seats across an aisle (such as [3,3] and [3,4]) are not considered to be adjacent, but there is an exceptional case on which an aisle split a four-person group, in that case, the aisle split a four-person group in the middle, which means to have two people on each side.

Example 1:

Input: n = 3, reservedSeats = [[1,2],[1,3],[1,8],[2,6],[3,1],[3,10]]
Output: 4

Explanation: The figure above shows the optimal allocation for four groups, where seats mark with blue are already reserved and contiguous seats mark with orange are for one group.

Example 2:

Input: n = 2, reservedSeats = [[2,1],[1,8],[2,6]]
Output: 2

Example 3:

Input: n = 4, reservedSeats = [[4,3],[1,4],[4,6],[1,7]]
Output: 4

Constraints:

  • 1 <= n <= 10^9
  • 1 <= reservedSeats.length <= min(10*n, 10^4)
  • reservedSeats[i].length == 2
  • 1 <= reservedSeats[i][0] <= n
  • 1 <= reservedSeats[i][1] <= 10
  • All reservedSeats[i] are distinct.

座位检查比较容易, 因为一共就 4 种模式:

  1. [2, 3, 4, 5] + [6, 7, 8, 9]
  2. [2, 3, 4, 5]
  3. [4, 5, 6, 7]
  4. [6, 7, 8, 9]
    按顺序检查, 它们之间是互斥的, 有一种成立,后面的就不必检查了。

但是要注意的是,reserved_seats 中的行不是全部的行, 也就是一些行不会有保留座位, 这些行每行都可以放 2 组



impl Solution {
    pub fn max_number_of_families(n: i32, mut reserved_seats: Vec<Vec<i32>>) -> i32 {
        reserved_seats.sort_by_key(|l| (l[0], l[1]));
        let mut seats = 0;
        let mut row = 1;
        let mut ans = 0;
        for reserved in reserved_seats {
            let (r, c) = (reserved[0], reserved[1]);
            if r != row {
                if seats | 0b1000000001 == 0b1000000001 {
                    ans += 2;
                } else if seats | 0b1000011111 == 0b1000011111 {
                    ans += 1
                } else if seats | 0b1110000111 == 0b1110000111 {
                    ans += 1;
                } else if seats | 0b1111100001 == 0b1111100001 {
                    ans += 1;
                }
                ans += (r - row - 1) * 2;
                seats = 0;
                row = r;
            }
            seats |= 1 << (c - 1);
        }
        if seats | 0b1000000001 == 0b1000000001 {
            ans += 2;
        } else if seats | 0b1000011111 == 0b1000011111 {
            ans += 1
        } else if seats | 0b1110000111 == 0b1110000111 {
            ans += 1;
        } else if seats | 0b1111100001 == 0b1111100001 {
            ans += 1;
        }
        ans += (n - row) * 2;
        ans
    }
}
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