struct carData {
int position;
double time;
carData(int position, double time) : position(position), time(time) {}
};
class Solution {
public:
int carFleet(int target, vector<int> &position, vector<int> &speed) {
vector<carData> cars;
for (int i = 0; i < position.size(); ++i) {
cars.push_back(carData(position[i], static_cast<double>(target - position[i]) / speed[i]));
}
sort(cars.begin(), cars.end(), [](carData &car1, carData &car2){ return car1.position < car2.position; });
int carFleetCnt = 1;
for (int curr = cars.size() - 1; curr >= 1; --curr) {
if (cars[curr - 1].time <= cars[curr].time) {
cars[curr - 1].time = cars[curr].time;
continue;
}
++carFleetCnt;
}
return carFleetCnt;
}
};
My Solution Solution 1: DFS (recursion) The Key Idea for Solving This Coding Question C++ Code /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right;
Jun 6, 2025MyCircularQueueO(k)
Apr 20, 2025O(m)
Mar 4, 2025O(n)n is the length of nums.
Feb 19, 2025or
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