[LeetCode] 853. Car Fleet 车队

2021年09月15日 阅读数:3
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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。
一辆车永远不会超过前面的另外一辆车,但它能够追上去,并与前车以相同的速度紧接着行驶。此时,咱们会忽略这两辆车之间的距离,也就是说,它们被假定处于相同的位置。车队是一些由行驶在相同位置、具备相同速度的车组成的非空集合。注意,一辆车也能够是一个车队。即使一辆车在目的地才遇上了一个车队,它们仍然会被视做是同一个车队。求会有多少车队到达目的地?code

解法:先把车按照位置进行排序,而后计算出每一个车在无阻拦的状况下到达终点的时间,若是后面的车到达终点所用的时间比前面车小,那么说明后车会比前面的车先到,因为后车不能超过前车,因此这种状况下就会合并成一个车队。用栈来存,对时间进行遍历,对于那些应该合并的车不进栈就好了,最后返回栈的长度。或者直接用一个变量存最近前车到达时间,用另外一变量记录车队的数量,若是循环的时间大于记录的前车时间,则当前的车不会比以前的车先到达,为一个新车队,更新变量。

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][2];
        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[0], b[0]));
        double cur = 0;

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

  

  

  

 

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