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
struct Solution;
impl Solution {
fn check(nums: Vec<i32>) -> bool {
let n = nums.len();
let mut count = 0;
for i in 0..n {
if nums[i] > nums[(i + 1) % n] {
count += 1;
}
}
count < 2
}
}
#[test]
fn test() {
let nums = vec![3, 4, 5, 1, 2];
let res = true;
assert_eq!(Solution::check(nums), res);
let nums = vec![2, 1, 3, 4];
let res = false;
assert_eq!(Solution::check(nums), res);
let nums = vec![1, 2, 3];
let res = true;
assert_eq!(Solution::check(nums), res);
let nums = vec![1, 1, 1];
let res = true;
assert_eq!(Solution::check(nums), res);
let nums = vec![2, 1];
let res = true;
assert_eq!(Solution::check(nums), res);
}