Suppose that we have a list of 8 numbers. If we trace the migration of one value, it will be a member of the following sizes of lists: eight, four, two. The number of calls of mergeSort
needed to sort one value into its final resting spot is log2N. If N = 8, then it will take three calls of the algorithm for one value to find its final resting spot.
We must apply log2N steps to sort N elements in the list. The order of recursive Mergesort
is O(N * log2N) or O(N * log N).
What about the cost of merging the fragments of the list? The merge algorithm is a linear one, so when combined with the mergeSort
routine, we have a O (N * log N) + O(N), which remains in the category of an O(N * log N) algorithm.
Last modified: December 12, 2022
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