LeetCode 1109. Corporate Flight Bookings

最新推荐文章于 2023-02-10 15:25:09 发布

taoqick

最新推荐文章于 2023-02-10 15:25:09 发布

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本文介绍了一种高效处理航班预订统计的算法，通过两次不同的实现方式，展示了如何计算每班航班的预订座位数。首次尝试使用排序和累加的方法，但存在处理结束条件的难点；第二次实现则直接利用前一次的结果进行累加，避免了排序，简化了解决方案。该算法适用于处理大量航班预订数据，确保准确计算每班航班的总预订座位。

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There are n flights, and they are labeled from 1 to n.

We have a list of flight bookings. The i-th booking bookings[i] = [i, j, k] means that we booked k seats from flights labeled i to j inclusive.

Return an array answer of length n, representing the number of seats booked on each flight in order of their label.

Example 1:

Input: bookings = [[1,2,10],[2,3,20],[2,5,25]], n = 5
Output: [10,55,45,25,25]

Constraints:

1 <= bookings.length <= 20000
1 <= bookings[i][0] <= bookings[i][1] <= n <= 20000
1 <= bookings[i][2] <= 10000

-------------------------------------------------------

受到 Merge Intervals 这题的影响，上来先排了个序，对结束条件开始没处理好，还不行bug了一下，代码如下：

class Solution:
    def corpFlightBookings(self, bookings, n):
        lst,res = [(item[0]-1,0,item[2]) for item in bookings]+[(item[1]-1,1,item[2]) for item in bookings],[0 for i in range(n)]
        lst.sort()
        pre,accu=0,0
        for pos,tag,book in lst:
            while(pre<pos):
                res[pre] = accu
                pre += 1
            if (tag == 0):
                accu += book
                res[pre] = accu
            else:
                if (pos == pre): #bug: 开始没有考虑清楚这坑
                    res[pre] = accu
                    pre += 1
                accu -= book
        return res

后来去discussion看，根本不用预先排序，累加之前的结果就好，由于输出形式不同，trick也不一样，慢慢感受一下：

class Solution:
    def corpFlightBookings(self, bookings, n):
        res = [0 for i in range(n+1)]
        for booking in bookings:
            start,end,seat = booking[0]-1,booking[1]-1,booking[2]
            res[start] += seat
            res[end+1] -= seat
        for i in range(1,n):
            res[i] += res[i-1]
        return res[:-1]