0853. Car Fleet

# 853. Car Fleet#

## 题目 #

`N` cars are going to the same destination along a one lane road. The destination is `target` miles away.

Each car `i` has a constant speed `speed[i]` (in miles per hour), and initial position `position[i]` miles towards the target along the road.

A car can never pass another car ahead of it, but it can catch up to it, and drive bumper to bumper at the same speed.

The distance between these two cars is ignored - they are assumed to have the same position.

car fleet is some non-empty set of cars driving at the same position and same speed. Note that a single car is also a car fleet.

If a car catches up to a car fleet right at the destination point, it will still be considered as one car fleet.

How many car fleets will arrive at the destination?

Example 1:

``````Input: target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3]
Output: 3
Explanation:
The cars starting at 10 and 8 become a fleet, meeting each other at 12.
The car starting at 0 doesn't catch up to any other car, so it is a fleet by itself.
The cars starting at 5 and 3 become a fleet, meeting each other at 6.
Note that no other cars meet these fleets before the destination, so the answer is 3.
``````

Note:

1. `0 <= N <= 10 ^ 4`
2. `0 < target <= 10 ^ 6`
3. `0 < speed[i] <= 10 ^ 6`
4. `0 <= position[i] < target`
5. All initial positions are different.

## 题目大意 #

N  辆车沿着一条车道驶向位于 target 英里之外的共同目的地。每辆车 i 以恒定的速度 speed[i] （英里/小时），从初始位置 position[i] （英里） 沿车道驶向目的地。

问最后会有多少车队到达目的地?

## 解题思路 #

• 根据每辆车距离终点和速度，计算每辆车到达终点的时间，并按照距离从大到小排序(position 越大代表距离终点越近)
• 从头往后扫描排序以后的数组，时间一旦大于前一个 car 的时间，就会生成一个新的 car fleet，最终计数加一即可。

## 代码 #

``````
package leetcode

import (
"sort"
)

type car struct {
time     float64
position int
}

// ByPosition define
type ByPosition []car

func (a ByPosition) Len() int           { return len(a) }
func (a ByPosition) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a ByPosition) Less(i, j int) bool { return a[i].position > a[j].position }

func carFleet(target int, position []int, speed []int) int {
n := len(position)
if n <= 1 {
return n
}
cars := make([]car, n)
for i := 0; i < n; i++ {
cars[i] = car{float64(target-position[i]) / float64(speed[i]), position[i]}
}
sort.Sort(ByPosition(cars))
fleet, lastTime := 0, 0.0
for i := 0; i < len(cars); i++ {
if cars[i].time > lastTime {
lastTime = cars[i].time
fleet++
}
}
return fleet
}

``````

Apr 8, 2023