【leetcode】1560. Most Visited Sector in a Circular Track

题目如下：

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

解题思路：笨方法，把每个sector经过的次数算出来取最大值即可。

代码如下：

class Solution(object):
    def mostVisited(self, n, rounds):
        """
        :type n: int
        :type rounds: List[int]
        :rtype: List[int]
        """
        count = [0] * (n+1)
        for i in range(len(rounds)-1):
            start,end = rounds[i],rounds[i+1]
            if end > start:
                for inx in range(start,end):
                    count[inx] += 1
            else:
                for inx in range(start,n+1):
                    count[inx] += 1
                for inx in range(1,end):
                    count[inx] += 1
        count[rounds[-1]] += 1

        max_val = max(count)
        res = []
        for i,v in enumerate(count):
            if v == max_val:
                res.append(i)

        return res