leetcode.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.

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.

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.

   
   分析：先对car按照与target距离的由小到大排序，然后计算每辆car在可超车情况下的到达时间。由题目可知：（1）若车i的到达时间大于车j（排序后：i<j），则车i与车j不会合并成一个车队；若车i的到达时间小于等于于车j（排序后：i<j），则车i与车j将合并车一个车队，并按车j的速度行驶。因此可以利用栈的特点来实现。按距离由近到远来遍历car的时间，若栈顶car的时间大于当前的car的时间，则将这个car进栈，比较下一个car；若栈顶car的时间小于当前元素，则这两个car可以合并成一个，弹出栈顶car，循环比较栈顶car与当前的car，直至栈为空或者栈顶car的时间大于当前的car的时间。遍历完，栈的大小就是到达的车队数量。

class Solution {
public:
    int carFleet(int target, vector<int>& position, vector<int>& speed) {
        if(position.size() == 0)
            return 0;
        if(position.size() == 1)
            return 1;
        sortCar(position,speed);
        stack<int> sd;
        vector<double> time;
        //计算每辆car可超车时的到达target时间
        for(int i = 0; i < position.size(); ++i){
            double len_i = target - position[i];
            double time_i = len_i / speed[i];
            time.push_back(time_i);
        }
        sd.push(0);
        for(int j = 1; j < time.size(); ++j){
            while(!sd.empty() && time[sd.top()]<=time[j]){
                sd.pop();
            }
            sd.push(j);
        }
        return sd.size();
    }
    
//对car按离target的距离由小到大排序
    void sortCar(vector<int>& position, vector<int>& speed){
        for(int i = 0; i < position.size(); ++i){
            for(int j = 0; j < position.size()-i-1; ++j){
                if(position[j] > position[j+1]){
                    swap(position[j],position[j+1]);
                    swap(speed[j],speed[j+1]);
                }
            }
        }
    }
};