1752. Check if Array Is Sorted and Rotated

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

这题踩了不少坑。首先我这里的判断条件是，找到第一个单调变化的点，也就是第一个递减的点，然后判断后面的子串是否是非减序列，并且每个元素必须小于等于整个串的第一个元素。

class CheckIfArrayIsSortedAndRotated : public Solution {
public:
    void Exec() {
    }
    bool check(vector<int>& nums) {
        if (nums.size() <= 2) return true;
        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] >= nums[i - 1]) {
                continue;
            }
            if (nums[i] > nums[0])
                return false;
            for (int j = i + 1; j < nums.size(); j++) {
                if (nums[j] < nums[j - 1] || nums[j] > nums[0]) {
                    return false;
                }
            }
            return true;
        }
        return true;
    }
};