Leetcode 853 Car Fleet

发表于 |

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.

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.

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:
0 <= N <= 10 ^ 4
0 < target <= 10 ^ 6
0 < speed[i] <= 10 ^ 6
0 <= position[i] < target
All initial positions are different.

分析：

  1. 这道题还有点意思，在不同的点有一系列的车，每辆车有不同的速度，都朝同一个target前进
  2. 无法超车，也就是初始位置在前面的车永远在前面
  3. 若追赶上前面的车则认为这些车属于一个集群，问到达终点后有多少个集群

思路：

  1. 显然我们先根据车的位置进行排序，从前面的车开始分析，如有A,B,C,D,E 5辆车，p[a] > p[b] > p[c] >…
  2. 我们先判断A到达终点前是否能被B追上，若追不上显然A自身是一个集群，再从B开始判断，若能追上，则判断C能不能追上，依次类推
  3. 假设B,C都能追上A，D追不上，那么显然D也不可能追上C，我们就把D当做此时最前面的车开始分析就行了，所以是O(n),one pass，不会重复计算，但是排序本身是O(nlogn)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
import collections
def carFleet(target, p, s):
	# 做的时候还没有note的第5条，所以处理了下初始position相同的情况
	# d = collections.defaultdict(lambda: 10**6)
	n = len(p)
	# for i in range(n):
	# 	d[p[i]] = min(d[p[i]],s[i])
	# 有第5条之后直接这样既可
	d = {a:b for a,b in zip(p,s)}
	p.sort(reverse=True)
	res = n
	i = 0
	while i + 1 < n:
		j = i+1
		while j < n and (target - p[i]) / float(d[p[i]]) * d[p[j]] + p[j] >= target:
			j += 1
			res -= 1
		i = j
	return res

O(nlogn) time, O(n) space
44 / 44 test cases passed.
difficulty: medium
Runtime: 174 ms