Leetcode-1109. Corporate Flight Bookings

最新推荐文章于 2022-01-11 23:15:00 发布

Liuu Zn

最新推荐文章于 2022-01-11 23:15:00 发布

阅读量450

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/qq_24656853/article/details/95041169

Leetcode 专栏收录该内容

10 篇文章

订阅专栏

本文探讨了LeetCode上的航班预订问题，通过直观方法和累积求和策略对比，展现了算法优化的重要性。初始方法虽直接但效率低下，受时间限制影响。优化后的累积求和策略大幅提升了算法效率，实现O(booking+N)的时间复杂度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Original link: https://leetcode.com/problems/corporate-flight-bookings/

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

Intuitive method:
Initialize an array whose length is equal to n.
For each record, do increments to each elements.

Code:

class Solution(object):
    def corpFlightBookings(self, bookings, n):
        """
        :type bookings: List[List[int]]
        :type n: int
        :rtype: List[int]
        """
        op = [0 for i in range(n)]
        
        d = dict()
        
        for i in bookings:
            tag = str(i[0]) + "-" + str(i[1])
            
            if tag in d:
                d[tag] += i[2]
            else:
                d[tag] = i[2]
                
        
        for i in d:
            
            intervals = i.split("-")
            for t in range(int(intervals[0])-1, int(intervals[1])):
                op[t] += d[i]
        
        return op

This code runs in O(ML) where M is the number of records and L is the average length of each record.
This code failed when submitted because of Time Limit Exceeded.

https://leetcode.com/problems/corporate-flight-bookings/discuss/328856/JavaC%2B%2BPython-Straight-Forward-Solution

Inspired by the above link, the cumulative sum seems to help.
When the code contains lots of “same” array operations, cumulative sum might help.

Code:

class Solution(object):
    def corpFlightBookings(self, bookings, n):
        """
        :type bookings: List[List[int]]
        :type n: int
        :rtype: List[int]
        """

        op = [0 for i in range(n+1)]
        
        for lo, hi, num in bookings:
            
            op[lo-1] += num
            op[hi] -= num
            
        for i in range(n-1):
            op[i+1] += op[i]
            
        return op[:-1]

The new version runs in O(booking + N) time and O(N) space.