Description
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
分析
题目的意思是：给你一个数组，按照数组里面的数来转圈圈，问转圈圈过程中访问频次最多的数是哪些？这道题可以用模拟的方法求出来，但是比较笨，如果能够想到不管怎么转圈圈，访问最多频次只与起点位置和终点位置有关，如果能够想到这个，就好办了，判断并求一下起点位置到终点位置覆盖的数就是答案了。
代码
class Solution:
def mostVisited(self, n: int, rounds: List[int]) -> List[int]:
start=rounds[0]
end=rounds[-1]
m=[]
if(start>end):
for i in range(end):
m.append(i+1)
for i in range(start,n+1):
m.append(i)
else:
for i in range(start,end+1):
m.append(i)
return m
参考文献
[LeetCode]Python3 Faster than 99.67%
