1752. Check if Array Is Sorted and Rotated #

题目 #

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

题目大意 #

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 。

解题思路 #

• 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。

代码 #

package leetcode

func check(nums []int) bool {
	count := 0
	for i := 0; i < len(nums)-1; i++ {
		if nums[i] > nums[i+1] {
			count++
			if count > 1 || nums[0] < nums[len(nums)-1] {
				return false
			}
		}
	}
	return true
}