If I'm running at nanosecond speed, 1000 n, the size of the problem, whatever it is, is 1000, and I've got a log algorithm, it takes 10 nanoseconds to complete.
如果这个问题的规模，也就是n，是，如果这个问题是对数级的，这将会占据10纳秒的时间，你一眨眼的时间。
The red-black tree is O(log n) in time for most operations, while the previous scheduler employed O(1), using a priority array with a fixed number of priorities.
对于大多数操作，红黑树的执行时间为O(log n)，而以前的调度程序通过具有固定优先级的优先级数组使用 O(1)。
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)。
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。
This will allow a log (n) look up of packets referenced by SACK options.
这允许对SACK选项指向的包进行log (N)查找。
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)。
Recall that returning a single character at an arbitrary position within a rope is an o (log n) operation.
返回rope内任意位置字符的操作是个o (log n)操作。
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。
Unlike other self-balancing binary search trees that provide worst case O(log n) lookup time, scapegoat trees have no additional per-node overhead compared to a regular binary search tree.
和其它的提供了最坏情况O（log n）查找时间的自平衡二分查找树不同，替罪羊树与普通的二分查找树相比，并没有对每个节点增加额外的开销。
The circuit of the log is actually a typical measuring circuit which can be used as the secondary circuit of some n...
该计程仪的电路为一典型的测量电路，可给导航仪器作二次转换电路。
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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