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

Solution in python:

class Solution:
    def mostVisited(self, n: int, rounds: List[int]) -> List[int]:
        adic = dict()
        for i in range(n):
            adic[i] = 0
        adic[rounds[0]-1] = 1
        for i in range(len(rounds)-1):
            start = rounds[i]-1
            end = rounds[i+1]-1
            j = 1
            index = (start+j) % n
            while index != end:
                adic[index] += 1
                j += 1
                index = (start+j) % n
            adic[end] += 1
        max_value = max(adic.values())
        result = [key for key in adic.keys() if adic[key]==max_value]
        result.sort()
        return [i+1 for i in result]
最后修改日期: 2021年3月17日

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