Next Greater Element II

 Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

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, return -1 for this number.

 

Example 1:

Input: nums = [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.

Example 2:

Input: nums = [1,2,3,4,3]

Output: [2,3,4,-1,4]

 

import java.util.*;

class Solution {

    public int[] nextGreaterElements(int[] nums) {

        int n = nums.length;

        int[] result = new int[n];

        Arrays.fill(result, -1); // default to -1

        Stack<Integer> stack = new Stack<>(); // stores indices

        for (int i = 0; i < 2 * n; i++) {

            int curr = nums[i % n];

            // Resolve next greater for elements in stack

            while (!stack.isEmpty() && nums[stack.peek()] < curr) {

                int index = stack.pop();

                result[index] = curr;

            }

            // Only push index in first round (0 to n-1)

            if (i < n) {

                stack.push(i);

            }

        }

        return result;

    }

}