You are driving a vehicle that has capacity empty seats initially available for passengers. The vehicle only drives east (ie. it cannot turn around and drive west.)
Given a list of trips, trip[i] = [num_passengers, start_location, end_location] contains information about the i-th trip: the number of passengers that must be picked up, and the locations to pick them up and drop them off. The locations are given as the number of kilometers due east from your vehicle’s initial location.
Return true if and only if it is possible to pick up and drop off all passengers for all the given trips.
Example 1:
Input: trips = [[2,1,5],[3,3,7]], capacity = 4 Output: false
Example 2:
Input: trips = [[2,1,5],[3,3,7]], capacity = 5 Output: true
Example 3:
Input: trips = [[2,1,5],[3,5,7]], capacity = 3 Output: true
Example 4:
Input: trips = [[3,2,7],[3,7,9],[8,3,9]], capacity = 11 Output: true
Solution1: Min heap
Sort events by location
Time complexity: O(nlogn)
Space complexity: O(n)
C++
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 |
// Author: Huahua class Solution { public: bool carPooling(vector<vector<int>>& trips, int capacity) { priority_queue<int> q; for (const auto& trip : trips) { int pick_up = -((trip[1] << 10) | (1 << 9) | trip[0]); int drop_off = -((trip[2] << 10) | trip[0]); q.push(pick_up); q.push(drop_off); } while (q.size()) { int key = -q.top(); q.pop(); int sign = ((key >> 9) & 1) ? 1: -1; int num = key & 0xFF; if ((capacity -= sign * num) < 0) return false; } return true; } }; |
Solution 2: Preprocessing
Time complexity: O(n)
Space complexity: O(1000)
C++
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
// Author: Huahua class Solution { public: bool carPooling(vector<vector<int>>& trips, int capacity) { vector<int> d(1001); for (const auto& trip : trips) { d[trip[1]] -= trip[0]; d[trip[2]] += trip[0]; } for (const int c : d) if ((capacity += c) < 0) return false; return true; } }; |
请尊重作者的劳动成果，转载请注明出处！花花保留对文章／视频的所有权利。
如果您喜欢这篇文章／视频，欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
huahualeetcode
Be First to Comment