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];
        
        Stack<Integer> stack = new Stack<>();
        for (int i = n - 1; i >= 0; i--) {
            if (stack.isEmpty()) {
                result[i] = getValue(nums, i, i);
            } else {
                if (nums[i] >= stack.peek()) {
                    while (!stack.isEmpty() && stack.peek() <= nums[i]) {
                        stack.pop();
                    }
                    if (stack.isEmpty()) {
                        result[i] = getValue(nums, n - 1, i);
                        stack.push(nums[i]);
                        continue;
                    }
                }
                result[i] = stack.peek();
            }
            stack.push(nums[i]);
        }
        return result;
    }
    
    private int getValue(int[] nums, int start, int i) {
        int n = nums.length, index = (start + 1) % n;
        while (index != i) {
            if (nums[index] > nums[i]) return nums[index];
            index = (index + 1) % n;
        }
        return -1;
    }
}