503 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]

Constraints:

• 1 <= nums.length <= 104
• -109 <= nums[i] <= 109

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class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        stack = []
        res = [-1] * len(nums)

        for i, num in enumerate(nums):
            while stack and num > nums[stack[-1]]:
                res[stack.pop()] = num
            stack.append(i)
            
        for num in nums:
            while stack and num > nums[stack[-1]]:
                res[stack.pop()] = num
            if not stack:
                break
        return res