【leetcode】1094. Car Pooling

题目如下：

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

解题思路：因为 0 <= trips[i][1] < trips[i][2] <= 1000，所以算法为O(n^2)应该是可以接受的。我的方法就是把每个location的人数都算出来，如果任意一个location的人数大于capacity即为false，否则是true。

代码如下：

class Solution(object):
    def carPooling(self, trips, capacity):
        """
        :type trips: List[List[int]]
        :type capacity: int
        :rtype: bool
        """
        max_des = 0
        for i,j,k in trips:
            max_des = max(max_des,k)
        val = [0] * (max_des + 1)
        for i, j, k in trips:
            for d in range(j,k):
                val[d] += i
                if val[d] > capacity:
                    return False
        return True