leet code 853,car fleet

链接：https://www.lintcode.com/problem/car-fleet/description

题意：有好多车和它们的位置，速度，如果后边车在终点前追上前边的，则可以形成一个车队，问最终到达终点的有几个车队；

直接想比较难，算一下它们到终点的时间，如果后边车的时间小于等于前边车的时间则一定可以追上，否则它自己就形成一个车队；

1，用个TreeMap建立位置和时间之间的映射，位置由大到小，所以TreeMap的first存-position；

code：

class Solution {
public:
    /**
     * @param target: the target
     * @param position: the initial position
     * @param speed: the speed
     * @return: How many car fleets will arrive at the destination
     */
    int carFleet(int target, vector<int> &position, vector<int> &speed) {
        // Write your code here
        //sort 
        int res=0; double cur=0;
        map<int ,double> pos2time;
        for(int i=0; i<position.size(); ++i)
        {
            pos2time[-position[i]]=(double)(target-position[i])/speed[i];
        }
        for(auto a : pos2time)
        {
            if(a.second <= cur) continue;
            cur = a.second;
            ++res;
        }
        return res;
    }
};

2，用优先队列直接维护；

code；

class Solution {
public:
    /**
     * @param target: the target
     * @param position: the initial position
     * @param speed: the speed
     * @return: How many car fleets will arrive at the destination
     */
    int carFleet(int target, vector<int> &position, vector<int> &speed) {
        // Write your code here
        //sort 
        int res=0; double cur=0;
        priority_queue<pair<int ,int> > q;
        for(int i=0; i<position.size(); i++)
        {
            q.push({position[i],speed[i]});
        }
        while(!q.empty()){
            auto t=q.top();
            q.pop();
            double pos2time=(double) (target-t.first)/t.second;
            if(pos2time<=cur) continue;
            cur= pos2time;
            ++res;
        }
        return res;
    }
};