# [LeetCode] 853. Car Fleet 车队

2021年09月15日 阅读数：3

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

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.java

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.python

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

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.app

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

How many car fleets will arrive at the destination?post

Example 1:url

```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:spa

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.

N辆车沿着一条车道驶向位于target英里以外的共同目的地。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。

Java:

```public int carFleet(int target, int[] pos, int[] speed) {
TreeMap<Integer, Double> m = new TreeMap<>();
for (int i = 0; i < pos.length; ++i) m.put(-pos[i], (double)(target - pos[i]) / speed[i]);
int res = 0; double cur = 0;
for (double time : m.values()) {
if (time > cur) {
cur = time;
res++;
}
}
return res;
}
```

Java:

```class Solution {
public int carFleet(int target, int[] position, int[] speed) {
int N = position.length;
int res = 0;
//创建位置-时间的N行2列的二维数组
double[][] cars = new double[N];
for (int i = 0; i < N; i++) {
cars[i] = new double[]{position[i], (double) (target - position[i]) / speed[i]};
}
Arrays.sort(cars, (a, b) -> Double.compare(a, b));
double cur = 0;

//从后往前比较
for (int i = N - 1; i >= 0; i--) {
if (cars[i] > cur) {
cur = cars[i];
res++;
}
}
return res;
}

}　　```

Python:

``` def carFleet(self, target, pos, speed):
time = [float(target - p) / s for p, s in sorted(zip(pos, speed))]
res = cur = 0
for t in time[::-1]:
if t > cur:
res += 1
cur = t
return res
```

Python:

```class Solution:
def carFleet(self, target, position, speed):
"""
:type target: int
:type position: List[int]
:type speed: List[int]
:rtype: int
"""
cars = [(pos, spe) for pos, spe in zip(position, speed)]
sorted_cars = sorted(cars, reverse=True)
times = [(target - pos) / spe for pos, spe in sorted_cars]
stack = []
for time in times:
if not stack:
stack.append(time)
else:
if time > stack[-1]:
stack.append(time)
return len(stack)
```

Python:

```# Time:  O(nlogn)
# Space: O(n)
class Solution(object):
def carFleet(self, target, position, speed):
"""
:type target: int
:type position: List[int]
:type speed: List[int]
:rtype: int
"""
times = [float(target-p)/s for p, s in sorted(zip(position, speed))]
result, curr = 0, 0
for t in reversed(times):
if t > curr:
result += 1
curr = t
return result　```

C++:

```int carFleet(int target, vector<int>& pos, vector<int>& speed) {
map<int, double> m;
for (int i = 0; i < pos.size(); i++) m[-pos[i]] = (double)(target - pos[i]) / speed[i];
int res = 0; double cur = 0;
for (auto it : m) if (it.second > cur) cur = it.second, res++;
return res;
}
```