Given an array nums, return true if the array was originally sorted in non-decreasing order, then rotated some number of positions (including zero). Otherwise, return false.

There may be duplicates in the original array.

Note: An array A rotated by x positions results in an array B of the same length such that A[i] == B[(i+x) % A.length], where % is the modulo operation.

Example 1:

Input: nums = [3,4,5,1,2] Output: true Explanation: [1,2,3,4,5] is the original sorted array. You can rotate the array by x = 3 positions to begin on the the element of value 3: [3,4,5,1,2]. 

Example 2:

Input: nums = [2,1,3,4] Output: false Explanation: There is no sorted array once rotated that can make nums. 

Example 3:

Input: nums = [1,2,3] Output: true Explanation: [1,2,3] is the original sorted array. You can rotate the array by x = 0 positions (i.e. no rotation) to make nums. 

Example 4:

Input: nums = [1,1,1] Output: true Explanation: [1,1,1] is the original sorted array. You can rotate any number of positions to make nums. 

Example 5:

Input: nums = [2,1] Output: true Explanation: [1,2] is the original sorted array. You can rotate the array by x = 5 positions to begin on the element of value 2: [2,1]. 

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100

给你一个数组 nums 。nums 的源数组中，所有元素与 nums 相同，但按非递减顺序排列。如果 nums 能够由源数组轮转若干位置（包括 0 个位置）得到，则返回 true ；否则，返回 false 。源数组中可能存在 重复项 。

• 简单题。从头扫描一遍数组，找出相邻两个元素递减的数对。如果递减的数对只有 1 个，则有可能是轮转得来的，超过 1 个，则返回 false。题干里面还提到可能有多个重复元素，针对这一情况还需要判断一下 nums[0] 和 nums[len(nums)-1] 。如果是相同元素，nums[0] < nums[len(nums)-1]，并且数组中间还存在一对递减的数对，这时候也是 false。判断好上述这 2 种情况，本题得解。

package leetcode funccheck(nums []int) bool { count:=0fori:=0; i<len(nums)-1; i++ { ifnums[i] >nums[i+1] { count++ifcount>1||nums[0] <nums[len(nums)-1] { returnfalse } } } returntrue }