---
# System prepended metadata

title: 853. Car Fleet(Medium)

---

## 853. Car Fleet(Medium)

### 題目描述
There are n cars going to the same destination along a one-lane road. The destination is target miles away.

You are given two integer array position and speed, both of length n, where position[i] is the position of the ith car and speed[i] is the speed of the ith car (in miles per hour).

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 faster car will slow down to match the slower car's speed. The distance between these two cars is ignored (i.e., they are assumed to have the same position).

A 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.

Return the number of car fleets that 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 (speed 2) and 8 (speed 4) become a fleet, meeting each other at 12.
The car starting at 0 does not catch up to any other car, so it is a fleet by itself.
The cars starting at 5 (speed 1) and 3 (speed 3) become a fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target.
Note that no other cars meet these fleets before the destination, so the answer is 3.
Example 2:

Input: target = 10, position = [3], speed = [3]
Output: 1
Explanation: There is only one car, hence there is only one fleet.
Example 3:

Input: target = 100, position = [0,2,4], speed = [4,2,1]
Output: 1
Explanation:
The cars starting at 0 (speed 4) and 2 (speed 2) become a fleet, meeting each other at 4. The fleet moves at speed 2.
Then, the fleet (speed 2) and the car starting at 4 (speed 1) become one fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target.
 

Constraints:

n == position.length == speed.length
1 <= n <= 10^5
0 < target <= 10^6
0 <= position[i] < target
All the values of position are unique.
0 < speed[i] <= 10^6

### 解題思路

* 計算各車到達目標所需的時間
* 按位置從小到大排序
* 從最靠近目標的位置往回數，比較前後車的時間，若後車花的時間較久，代表前車自己形成一個車隊
* 回傳車隊數量
![](https://hackmd.io/_uploads/Hk8RWpiWp.png)

### 程式碼

```cpp=
#include <stdio.h>
#include <stdlib.h>

// 結構體表示汽車訊息
typedef struct {
    int position;   // 汽車位置
    float time;    // 汽車到達目標位置所需時間
}Car;

int cmp(const void* a, const void* b){
  Car *carsA = (Car*)a;
  Car *carsB = (Car*)b;
  return (carsA->position - carsB->position);
}

int carFleet(int target, int* position, int positionSize, int* speed, int speedSize){
    int n = positionSize;

    // 建立汽車結構體數組並計算時間
    Car *cars = malloc(n * sizeof(Car));
    for (int i = 0; i < n; i++) {
        cars[i].position = position[i];
        cars[i].time = (float)(target - position[i]) / speed[i];
    }

    // 按位置從小到大排序
    qsort(cars,n,sizeof(Car),cmp);

    int fleetcnt = 1;
    float cur = cars[n-1].time;
    for(int i = n-1; i >= 0; i--){
        //printf("%d %f\n",cars[i].position,cars[i].time);
        if(cars[i].time > cur){
            cur = cars[i].time;
            fleetcnt++;
        }
    }
    return fleetcnt;
}
```

### 時間/空間複雜度
* 時間複雜度: $O(nlogn)$
* 空間複雜度: $O(n)$
![](https://hackmd.io/_uploads/HkTHVA3Zp.png)

