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  • 【leetcode】1109. Corporate Flight Bookings

    题目如下:

    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 jinclusive.

    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

    解题思路:这种区间问题,我首先想到的是线段树,当然本题线段树似乎不是最优答案,因为我的解法耗时在2S左右。遍历bookings,把每个item的bookings[i][2]累加到相应的数的节点中,最后统一计算总数即可。

    代码如下:

    class Solution(object):
        def corpFlightBookings(self, bookings, n):
            """
            :type bookings: List[List[int]]
            :type n: int
            :rtype: List[int]
            """
            segment = [0] * (4 * n + 1)
            def recursive(start,end,low,high,inx,val):
                #print start,end,low,high,inx,val
                if low > high:
                    return
                if start == low and end == high:
                    segment[inx] += val
                    return
                mid = (low + high)/2
                #if start == low and end == high:
                #    recursive(start, mid, low, mid, inx * 2, val)
                #    recursive(mid + 1, end, mid + 1, high, inx * 2 + 1, val)
                if end <= mid:
                    recursive(start,end,low,mid,inx*2,val)
                elif start > mid:
                    recursive(start, end, mid + 1, high, inx * 2+1, val)
                else:
                    recursive(start, mid, low, mid, inx * 2, val)
                    recursive(mid+1, end, mid+1, high, inx * 2 + 1, val)
    
            res = [0] * n
            def query(inx,low,high,segment_inx,node_inx):
                if segment_inx >= len(segment):
                    return
                mid = (low + high)/2
                res[node_inx] += segment[segment_inx]
                if inx <= mid:
                    query(inx,low,mid,segment_inx*2,node_inx)
                else:
                    query(inx, mid+1, high, segment_inx * 2 + 1, node_inx)
    
            for (start,end,val) in bookings:
                recursive(start,end,1,n,1,val)
    
            for i in range(1,n+1):
                query(i,1,n,1,i-1)
            return res
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  • 原文地址:https://www.cnblogs.com/seyjs/p/11169973.html
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