# Next Greater Element II

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.

```
public class Solution {
public int[] nextGreaterElements(int[] nums) {
int n = nums.length;
int[] result = new int[n];
if (nums == null || n == 0) {
return result;
}
Arrays.fill(result, -1);
Stack<Integer> stack = new Stack<>();
for (int i = 0; i < n * 2; i++) {
int num = nums[i % n];
while (!stack.isEmpty() && nums[stack.peek()] < num) {
result[stack.pop()] = num;
}
if (i < n) {
stack.push(i);
}
}
return result;
}
}
```

Hope this helps,

Michael