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I have been looking at the application of merge sort to linked lists. Some of the articles I have looked at tout that merge sort is the best algorithm for sorting a linked list. It makes sense for the conquer part in the divide and conquer strategy where you merge the two sorted linked lists as you end up saving on required memory (compared to an array). But, what I don't understand is the time complexity of the divide step in the algorithm.

For an array, this step is constant time by leveraging random access and splitting the array into smaller chunks. But, for a linked list isn't this going to take an additional O(n)? I have seen Floyd's algorithm (tortoise-hare) used to find the mid-point of a linked list and divide the problem into smaller chunks. I did some analysis on the divide step. Suppose the linked list is of size n, then the # of operations involved in just dividing the problem is as follows,

n/2 + n/4 * 2 + n/8 * 4 + ... = n/2 * log(n)

From the above, it looks like compared to the array case, an additional factor of "n" appears from Floyd's algorithm. So, the final time complexity would be O(n^2 * log(n)). Can someone please explain the discrepancy?

Edit: based on @Yves comment, I identified the mistake,

I multiplied the work while merging back the sorted blocks from bottom to top when it should be added. So, the net time would be: nlogn/2 + nlogn = O(nlogn),

This is probably is most valid answer to the above question; other answers are a bit indirect/ provide no explanation

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