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)。

youdao

Before giving an overview of the SOA patterns it is helpful to think of an n-tier architecture as outlined below.

在讨论SOA模式的概况前，将n层体系结构视为以下所给出的情况。

youdao