• [Swift]LeetCode1094. 拼车 | Car Pooling


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    You are driving a vehicle that has capacity empty seats initially available for passengers.  The vehicle onlydrives east (ie. it cannot turn around and drive west.)

    Given a list of tripstrip[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

    假设你是一位顺风车司机,车上最初有 capacity 个空座位可以用来载客。由于道路的限制,车 只能 向一个方向行驶(也就是说,不允许掉头或改变方向,你可以将其想象为一个向量)。

    这儿有一份行程计划表 trips[][],其中 trips[i] = [num_passengers, start_location, end_location] 包含了你的第 i 次行程信息:

    • 必须接送的乘客数量;
    • 乘客的上车地点;
    • 以及乘客的下车地点。

    这些给出的地点位置是从你的 初始 出发位置向前行驶到这些地点所需的距离(它们一定在你的行驶方向上)。

    请你根据给出的行程计划表和车子的座位数,来判断你的车是否可以顺利完成接送所用乘客的任务(当且仅当你可以在所有给定的行程中接送所有乘客时,返回 true,否则请返回 false)。

    示例 1:

    输入:trips = [[2,1,5],[3,3,7]], capacity = 4
    输出:false
    

    示例 2:

    输入:trips = [[2,1,5],[3,3,7]], capacity = 5
    输出:true
    

    示例 3:

    输入:trips = [[2,1,5],[3,5,7]], capacity = 3
    输出:true
    

    示例 4:

    输入:trips = [[3,2,7],[3,7,9],[8,3,9]], capacity = 11
    输出:true

    提示:

    1. 你可以假设乘客会自觉遵守 “先下后上” 的良好素质
    2. trips.length <= 1000
    3. trips[i].length == 3
    4. 1 <= trips[i][0] <= 100
    5. 0 <= trips[i][1] < trips[i][2] <= 1000
    6. 1 <= capacity <= 100000

    Runtime: 36 ms
    Memory Usage: 20.9 MB
     1 class Solution {
     2     func carPooling(_ trips: [[Int]], _ capacity: Int) -> Bool {
     3         var capacity = capacity
     4         var stops:[Int] = [Int](repeating:0,count:1001)
     5         for t in trips
     6         {
     7             stops[t[1]] += t[0]
     8             stops[t[2]] -= t[0]
     9         }
    10         var i:Int = 0
    11         while(capacity >= 0 && i < 1001)
    12         {
    13             capacity -= stops[i]
    14             i += 1
    15         }
    16         return capacity >= 0
    17     }
    18 }

    44ms
     1 class Solution {
     2   func carPooling(_ trips: [[Int]], _ capacity: Int) -> Bool {
     3     var path = [Int].init(repeating: 0, count: 1001)
     4     for t in trips {
     5       path[t[1]] -= t[0]
     6       path[t[2]] += t[0]
     7     }
     8     var c = capacity
     9     for j in path {
    10       c += j
    11       if c < 0 {
    12         return false
    13       }
    14     }
    15     return true
    16   }
    17 }

    60ms

     1 class Solution {
     2     func carPooling(_ trips: [[Int]], _ capacity: Int) -> Bool {
     3         var road = Array(repeating: 0, count: 1000)
     4         for trip in trips {
     5             let count = trip[0]
     6             let from = trip[1], to = trip[2]
     7             for pos in from..<to {
     8                 road[pos] += count
     9                 if road[pos] > capacity {
    10                     return false
    11                 }
    12             }
    13         }
    14         return true
    15     }
    16 }

    68ms

     1 class Solution {
     2     struct Node: Comparable {
     3         var addP: Int
     4         var location: Int
     5         init(_ addP:Int, _ location:Int) {
     6             self.addP = addP
     7             self.location = location
     8         }
     9         
    10         static func == (lhs:Node, rhs:Node) -> Bool {
    11             return lhs.location == rhs.location
    12         }
    13         
    14         static func < (lhs:Node, rhs:Node) -> Bool {
    15             if lhs.location == rhs.location {
    16                 return lhs.addP < rhs.addP
    17             } else {
    18                 return lhs.location < rhs.location
    19             }
    20         }
    21     }
    22     func carPooling(_ trips: [[Int]], _ capacity: Int) -> Bool {
    23         var nodes = [Node]()
    24         for trip in trips {
    25             nodes.append(Node(trip[0], trip[1]))
    26             nodes.append(Node(-trip[0], trip[2]))
    27         }
    28         nodes.sort()
    29         // print(nodes)
    30         var counter = 0
    31         for node in nodes {
    32             counter += node.addP
    33             if counter > capacity {
    34                 return false
    35             }
    36         }
    37         return true
    38     }
    39 }

    80ms

     1 class Solution {
     2     func carPooling(_ trips: [[Int]], _ capacity: Int) -> Bool {
     3     let finalDestination = trips.sorted{ $0[2] < $1[2]}.last![2]
     4     // let finalDestination = trips.last![2]
     5     var capacityArr = Array(repeating: 0, count: finalDestination + 1)
     6     
     7     for trip in trips {
     8         for journey in trip[1]..<trip[2] {
     9             capacityArr[journey] += trip[0]
    10             if capacityArr[journey] > capacity {
    11                 return false
    12             }
    13         }
    14     }
    15     return true
    16   }
    17 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/11032164.html
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