0503. Next Greater Element II

https://leetcode.com/problems/next-greater-element-ii

Description

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]

Constraints:

• 1 <= nums.length <= 104

• -109 <= nums[i] <= 109

ac

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        // edge cases
        if (nums == null || nums.length == 0) return new int[0];

        int n = nums.length;
        int[] res = new int[n];
        Arrays.fill(res, -1);
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < n * 2; i++) {
            while (!stack.isEmpty() && nums[i%n] > nums[stack.peek()]) {
                int tmp = stack.pop();
                res[tmp] = nums[i%n];
            }
            stack.push(i%n);
        }

        return res;
    }
}