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
Solution: Counting and checking
Count how many turning points (nums[i] < nums[i – 1]) in the array. Return false if there are more than 1.
For the turning point r, (nums[r] < nums[r – 1), return true if both of the following conditions are satisfied:
1. nums[r – 1] is the largest number, e.g. nums[r – 1] >= nums[n – 1]
2. nums[r] is the smallest number, e.g. nums[r] <= nums[0]
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: bool check(vector<int>& nums) { int count = 0; int r = -1; for (size_t i = 1; i < nums.size(); ++i) if (nums[i] < nums[i - 1]) { ++count; r = i; } if (count == 0) return true; if (count > 1) return false; return nums[r] <= nums[0] && nums[r - 1] >= nums.back(); } }; |
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