How would you sort a linked list using merge sort, and what is the time complexity?

• A. Recursively split into halves, sort each half, then merge; Time O(n log n); Space O(log n) due to recursion (or O(1) bottom-up).
• B. Use quicksort; Time O(n log n) average; Space O(1).
• C. In-place selection sort; Time O(n^2).
• D. Bubble sort; Time O(n^2).

Answer: (A)

Sorting a linked list with merge sort uses divide-and-conquer: split the list into two roughly equal halves, sort each half recursively, then merge the two sorted halves back together. To split, use a slow and a fast pointer to find the middle, cut the list there, and apply the same process to the two halves. The merge step builds the sorted list by repeatedly linking the smaller head from either half, advancing that list’s pointer each time, until both halves are exhausted. This merge runs in linear time with respect to the total number of nodes.

Because the list is halved at every level, the number of levels is logarithmic in n, so the total time is O(n log n). The usual recursive implementation uses stack space proportional to the recursion depth, which is O(log n); with a bottom-up iterative (non-recursive) approach you can achieve O(1) extra space. This method suits linked lists well since splitting is easy with pointers and merging is done by pointer adjustments without random access.