Problem
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.
https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/
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 byx = 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 byx = 0
positions (i.e. no rotation) to make nums.
Constraints:
1 <= nums.length <= 100
1 <= nums[i] <= 100
Test Cases
1 | class Solution: |
1 | import pytest |
Thoughts
遍历数组,统计数字变小的次数,如果超过一次就不是循环有序数组。要注意数组的第一个数字和最后一个数字也要比较。
时间复杂度 O(n)
,空间复杂度 O(1)
。
Code
1 | class Solution: |