- Corporate Flight Bookings
Medium
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
解法1：差分数组。注意booking数组里面每一项是按index从1开始。
class Solution {
public:
vector<int> corpFlightBookings(vector<vector<int>>& bookings, int n) {
int bookingcount = bookings.size();
if (bookingcount == 0) return {
};
vector<int> res(n, 0);
vector<int> diffs(n, 0);
for (int i = 0; i < bookingcount; i++) {
diffs[bookings[i][0] - 1] += bookings[i][2];
if (bookings[i][1] < n) diffs[bookings[i][1]] -= bookings[i][2];
}
res[0] = diffs[0];
for (int i = 1; i < n; i++) {
res[i] += res[i - 1] + diffs[i];
}
return res;
}
};
解法2：线段树
解法3：树状数组