1560. Most Visited Sector in a Circular Track

Given an integer n and an integer array rounds. We have a circular track which consists of n sectors labeled from 1 to n. A marathon will be held on this track, the marathon consists of m rounds. The ith round starts at sector rounds[i - 1] and ends at sector rounds[i]. For example, round 1 starts at sector rounds[0] and ends at sector rounds[1]

Return an array of the most visited sectors sorted in ascending order.

Notice that you circulate the track in ascending order of sector numbers in the counter-clockwise direction (See the first example).

 

Example 1:


Input: n = 4, rounds = [1,3,1,2]
Output: [1,2]
Explanation: The marathon starts at sector 1. The order of the visited sectors is as follows:
1 --> 2 --> 3 (end of round 1) --> 4 --> 1 (end of round 2) --> 2 (end of round 3 and the marathon)
We can see that both sectors 1 and 2 are visited twice and they are the most visited sectors. Sectors 3 and 4 are visited only once.
Example 2:

Input: n = 2, rounds = [2,1,2,1,2,1,2,1,2]
Output: [2]
Example 3:

Input: n = 7, rounds = [1,3,5,7]
Output: [1,2,3,4,5,6,7]
 

Constraints:

2 <= n <= 100
1 <= m <= 100
rounds.length == m + 1
1 <= rounds[i] <= n
rounds[i] != rounds[i + 1] for 0 <= i < m

这题类似于一个脑筋急转弯，无论如何跑，覆盖最大的区域永远是起点到终点经过的区域。

class MostVisitedSectorInACircularTrack : public Solution {
public:
    void Exec() {
        
    }
    vector<int> mostVisited(int n, const vector<int>& rounds) {
        vector<int> res;
        int curSector = rounds[0];
        res.push_back(curSector);

        while (curSector != rounds.back()) {
            curSector = curSector % n + 1;
            res.push_back(curSector);
        }
        std::sort(res.begin(), res.end());
        return res;
    }
};