So I always have to do a merge of n elements.

所以我始终需要做n个元素的合并。

youdao

Merge sort has o (n log n) worst-case and average-case performance.

合并排序的最差性能和平均性能为o (n log n)。

youdao

Use merge sort, in a binary search algorithm complexity: nlogn, n number two together, if can equal to a specific number m, the output yes, otherwise no output.

说明：运用归并排序，在用二分法查找，算法复杂度：nlogn，n个数两两相加，若能等于特定数m，则输出yes，否则输出no。

youdao

So in other words, every time I merge the point that I kept emphasizing verbally there and that I'm only touching each number once, means that we have to account for the amount of time it takes to merge N which is going to be just N. Now, this is again one of these cyclical answers.

换言之，之前在做合并时,我不停地强调,对每个数字我只碰了一次，这就是说,我们要记录合并所花的时间量,也就是这里的,这又是一种循环性的答案。

哈佛公开课 - 计算机科学课程节选

We said each of the merge operations O was of order n. But n is different. Right?

注意这里发生了什么,我们说过每一次合并操作的复杂度都是？

麻省理工公开课 - 计算机科学及编程导论课程节选

It at least does corroborate the claim that merge sort N*log N as we argue intuitively is in fact, N log N in running time.

但这至少证实了归并排序,的时间复杂度为。

哈佛公开课 - 计算机科学课程节选