Python Classic Solution
This is what I supose to be the "classic" solution. \nSince the end-size of the list is known, we would ideally want to allocate 2*n space at the beginning and 
This is what I supose to be the "classic" solution. \nSince the end-size of the list is known, we would ideally want to allocate 2*n space at the beginning and avoid the dynamic resizing of the list during the ```append()``` method execution (see [Python-FAQ](https://docs.python.org/3/faq/design.html#how-are-lists-implemented-in-cpython) for details about python-list implementations)\n```\n    def getConcatenation(self, nums: List[int]) -> List[int]:\n        ans = []\n        n = len(nums)\n        for i in range(2*n):\n            ans.append(nums[i%n])\n        return ans\n```\n\nThe solutions with built-in python concat functions of lists\n```\n    def getConcatenation(self, nums: List[int]) -> List[int]:\n        return nums*2\n```\n```\n    def getConcatenation(self, nums: List[int]) -> List[int]:\n        return nums+nums\n```\n\nAs for the executiuon-time, all of the presented solutions fluctuated somewhere between 80 and 210 ms depending on the mighty Random.
['Python']