1094. Car Pooling

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

 

Constraints:

1. trips.length <= 1000
2. trips[i].length == 3
3. 1 <= trips[i][0] <= 100
4. 0 <= trips[i][1] < trips[i][2] <= 1000
5. 1 <= capacity <= 100000

class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        Map<Integer, Integer> map = new TreeMap();
        for(int[] trip : trips) {
            map.put(trip[1], map.getOrDefault(trip[1], 0) + trip[0]);
            map.put(trip[2], map.getOrDefault(trip[2], 0) - trip[0]);
        }
        for(int i : map.values()) {
            capacity -= i;
            if(capacity < 0) return false;
        } 
        return true;
    }
}

https://leetcode.com/problems/car-pooling/discuss/317610/JavaC%2B%2BPython-Meeting-Rooms-III

 map记录每个station需要的capacity，用treemap从小到大排列，然后用capacity减去每个站的capacity，看够不够

public boolean carPooling(int[][] trips, int capacity) {    
  int stops[] = new int[1001]; 
  for (int t[] : trips) {
      stops[t[1]] += t[0];
      stops[t[2]] -= t[0];
  }
  for (int i = 0; capacity >= 0 && i < 1001; ++i) capacity -= stops[i];
  return capacity >= 0;
}

或者根据题意，一共有1001个站，然后操作。牛蛋