Next Greater Element II

503. 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