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
public class Solution {
public IList<int> MostVisited(int n, int[] rounds) {
int start = rounds[0];
int end = rounds[rounds.Length-1];
var res = new List<int>();
if(start <= end) {
while(start<=end){
res.Add(start++);
}
} else {
int i=1;
while(i<=end) {
res.Add(i++);
}
while(start <= n) {
res.Add(start++);
}
}
return res;
}
}Time Complexity: O(n)
Space Complexity: O(n)
