[leetcode] 1094. Car Pooling

[leetcode] 1094. Car Pooling

Description

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 = 4Output: false

Example 2:

Input: trips = [[2,1,5],[3,3,7]], capacity = 5Output: true

Example 3:

Input: trips = [[2,1,5],[3,5,7]], capacity = 3Output: true

Example 4:

Input: trips = [[3,2,7],[3,7,9],[8,3,9]], capacity = 11Output: true

Constraints:

trips.length <= 1000trips[i].length == 31 <= trips[i][0] <= 1000 <= trips[i][1] < trips[i][2] <= 10001 <= capacity <= 100000

分析

题目的意思是：给定trips数组和capacity，trips的格式为[[[num_passengers, start_location, end_location]], 表明开始位置和结束位置的乘客数，判断trips的乘客数不超过给定的capacity。 +我参考了一下答案的思路，记录乘客在start_location, end_location的变化的情况，然后和capacity就能够得到结果了。

用timestamp数组记录变化的乘客数，然后遍历timestamp数组就得到结果了哈。

代码

class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: timestamp=[] for trip in trips: timestamp.append([trip[1],trip[0]]) timestamp.append([trip[2],-trip[0]]) timestamp.sort() used=0 for time,change in timestamp: used+=change if(used>capacity): return False return True

参考文献

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