Leetcode simple question: check whether the array is sorted and rotated

subject

Give you an array nums .nums In the source array of , All elements are related to nums identical , But in non decreasing order .
If nums It can rotate several positions by the source array （ Include 0 A place ） obtain , Then return to true ; otherwise , return false .
There may be... In the source array Duplicate item .
Be careful ： We call it an array A In rotation x You get an array of the same length after two positions B , When they meet A[i] == B[(i+x) % A.length] , among % For the remainder operation .
Example 1：
Input ：nums = [3,4,5,1,2]
Output ：true
explain ：[1,2,3,4,5] For ordered source arrays .
It can rotate x = 3 A place , Make the new array from the value to 3 Element start of ：[3,4,5,1,2] .
Example 2：
Input ：nums = [2,1,3,4]
Output ：false
explain ： The source array cannot be rotated nums .
Example 3：
Input ：nums = [1,2,3]
Output ：true
explain ：[1,2,3] For ordered source arrays .
It can rotate x = 0 A place （ That is, it doesn't rotate ） obtain nums .
Example 4：
Input ：nums = [1,1,1]
Output ：true
explain ：[1,1,1] For ordered source arrays .
Rotation can be obtained at any position nums .
Example 5：
Input ：nums = [2,1]
Output ：true
explain ：[1,2] For ordered source arrays .
It can rotate x = 5 A place , Make the new array from the value to 2 Element start of ：[2,1] .
Tips ：
1 <= nums.length <= 100
1 <= nums[i] <= 100
source ： Power button （LeetCode）

Their thinking

   First find the subscript of the smallest element in the array . There may be more than one smallest element , We select an element as the beginning and loop through the entire array to see whether an increasing sequence can be formed if we start with the current minimum . If there is a beginning traversal, it can make the array in increasing order , Then it can be judged that the modified array is true .

class Solution:
    def check(self, nums: List[int]) -> bool:
        MIN=min(nums)
        temp=[]
        for i in range(len(nums)):
            if nums[i]==MIN:
                temp.append(i)
        def match(MIN):
            for i in range(MIN,MIN+len(nums)-1):
                if nums[i%len(nums)]>nums[(i+1)%len(nums)]:
                    return False
            return True
        return True if all([not match(i) for i in temp])==False else False