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.

Constraints:

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

Solution:

class Solution {
    public boolean check(int[] n) {
        int l= n.length; // Calculating length of given array.
        int nRotation=0;// Declaring a variable to keep a count of number of rotations.
//Running a for loop to identify any rotation in given array.
        for(int i=0;i<l;i++){
            if(n[i]>n[(i+1)%l]){
                nRotation++;
            }
        }
//If number of rotations are greater than 1 return false.
        if(nRotation>1)
            return false;
//Else return True.
        return true;

    }
}

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原文链接：1752. Check if Array Is Sorted and Rotated