Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.

**Example 1:**

Input:[1,2,1]Output:[2,-1,2]Explanation:The first 1's next greater number is 2;

The number 2 can't find next greater number;

The second 1's next greater number needs to search circularly, which is also 2.

**Note:**
The length of given array won't exceed 10000.

```
struct Solution;
impl Solution {
fn next_greater_elements(nums: Vec<i32>) -> Vec<i32> {
let mut stack: Vec<usize> = vec![];
let n = nums.len();
let mut res = vec![-1; n];
for i in 0..2 * n {
let j = i % n;
let x = nums[j];
while let Some(top) = stack.pop() {
if nums[top] < x {
res[top] = x;
} else {
stack.push(top);
break;
}
}
stack.push(j);
}
res
}
}
#[test]
fn test() {
let nums = vec![1, 2, 1];
let res = vec![2, -1, 2];
assert_eq!(Solution::next_greater_elements(nums), res);
}
```