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)。
So I have n operations log n times, n log n there we go, n log n. Took us a long time to get there, but it's a nice algorithm to have.
所以我log，n遍的n次操作，就得到了,虽然花了不少时间得到了这个结论。
As I mentioned in the overview section, random character access on a rope with many internal nodes is approximately o (log n), so traversal is o (n log n).
就像我在概述一节中提到过的，在拥有许多内部节点的rope上随机访问字符的时间大约为o (log n)，所以遍历时间为o (nlog n)。
Since the redo log group is inactive and is archived, just clear the redo log - alter database clear logfile group N (where N is the group # of the lost redo log)
由于重做日志组是不活动和归档的，所以只需清除重做日志即可 ——alterdatabaseclear logfilegroupN（其中 N 是丢失的重做日志的组 #）。
The JDK documentation indicates that this method gives you a modified mergesort with guaranteed N*log(N) performance.
JDK文档说明，该方法采用修正的合并排序法，保证N*log(N) 的效率。
Using this notation, you should know that search through a list is O(n) and binary search (through a sorted list) is log(n).
看到这个标记你就应该知道搜索链表的复杂度为O(n)，进行二进制搜索时（已排序）的空间复杂度为log(n)。
Record all the action and P/N and S/N of the removed/installed component as well as ATA chapter in the flight log book after any corrective maintenance action has been taken.
在采取任何纠正维修措施后，要将所作的所有工作以及拆下/安装部件的件号、序号，以及涉及到的ATA章节号，都记录在飞行记录本上。
There are two sorted arrays A and B of size mand n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
这个题是求两个有序数组的中间数，其实这个中间数，如果是奇数，就是中间的那个数，如果是偶数，就是中间的两个数的和的平均值。
There are two sorted arrays A and B of size mand n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
这个题是求两个有序数组的中间数，其实这个中间数，如果是奇数，就是中间的那个数，如果是偶数，就是中间的两个数的和的平均值。
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