[LeetCode] 1109. Corporate Flight Bookings 公司航班预订
There are n
flights that are labeled from 1
to n
.
You are given an array of flight bookings bookings
, where bookings[i] = [firsti, lasti, seatsi]
represents a booking for flights firsti
through lasti
(inclusive) with seatsi
seats reserved for each flight in the range.
Return *an array answer
of length n
, where answer[i]
is the total number of seats reserved for flight *i
.
Example 1:
Input: bookings = [[1,2,10],[2,3,20],[2,5,25]], n = 5
Output: [10,55,45,25,25]
Explanation:
Flight labels: 1 2 3 4 5
Booking 1 reserved: 10 10
Booking 2 reserved: 20 20
Booking 3 reserved: 25 25 25 25
Total seats: 10 55 45 25 25
Hence, answer = [10,55,45,25,25]
Example 2:
Input: bookings = [[1,2,10],[2,2,15]], n = 2
Output: [10,25]
Explanation:
Flight labels: 1 2
Booking 1 reserved: 10 10
Booking 2 reserved: 15
Total seats: 10 25
Hence, answer = [10,25]
Constraints:
1 <= n <= 2 * 104
1 <= bookings.length <= 2 * 104
bookings[i].length == 3
1 <= firsti <= lasti <= n
1 <= seatsi <= 104
这道题说是有n个航班,标号从1到n,每次公司可以连续预定多个航班上的座位,用一个三元数组 [i, j, k],表示分别预定航班i到j上的k个座位,最后问每个航班上总共被预定了多少个座位。博主先试了一下暴力破解,毫无意外的超时了,想想为啥会超时,因为对于每个预定的区间,都遍历一次的话,最终可能达到n的平方级的复杂度。所以就需要想一些节省运算时间的办法,其实这道的解法很巧妙,先来想想,假如只有一个预定,是所有航班上均订k个座位,那么暴力破解的方法就是从1遍历到n,然后每个都加上k,但还有一种方法,就是只在第一天加上k,然后计算累加和数组,这样之后的每一天都会被加上k。如果是预定前一半的航班,那么暴力破解的方法就是从1遍历到 n/2,而这里的做法是在第一个天加上k,在第 n/2 + 1 天减去k,这样再求累加和数组时,后一半的航班就不会加上k了。对于所有的预定都可以采用这种做法,在起始位置加上k,在结束位置加1处减去k,最后再整体算累加和数组,这样就把平方级的时间复杂度缩小到了线性,完美通过 OJ,参见代码如下:
class Solution {
public:
vector<int> corpFlightBookings(vector<vector<int>>& bookings, int n) {
vector<int> res(n);
for (auto booking : bookings) {
res[booking[0] - 1] += booking[2];
if (booking[1] < n) res[booking[1]] -= booking[2];
}
for (int i = 1; i < n; ++i) {
res[i] += res[i - 1];
}
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1109
参考资料:
https://leetcode.com/problems/corporate-flight-bookings/
https://leetcode.com/problems/corporate-flight-bookings/discuss/328871/C%2B%2BJava-with-picture-O(n)
https://leetcode.com/problems/corporate-flight-bookings/discuss/328856/JavaC%2B%2BPython-Sweep-Line