列表切片的Big-O [英] Big-O of list slicing

问题描述

说我有一些Python列表，my_list，其中包含N个元素.可以使用my_list[i_1]为单个元素建立索引，其中i_1是所需元素的索引.但是，Python列表也可能被索引为my_list[i_1:i_2]，其中需要从i_1到i_2的列表切片".切成大小为N的列表的Big-O(最坏情况)表示法是什么?

Say I have some Python list, my_list which contains N elements. Single elements may be indexed by using my_list[i_1], where i_1 is the index of the desired element. However, Python lists may also be indexed my_list[i_1:i_2] where a "slice" of the list from i_1 to i_2 is desired. What is the Big-O (worst-case) notation to slice a list of size N?

就我个人而言，如果我正在编码切片器"，我会从i_1迭代到i_2，生成一个新列表并返回它，暗示O(N)，这是Python的方式吗?

Personally, if I were coding the "slicer" I would iterate from i_1 to i_2, generate a new list and return it, implying O(N), is this how Python does it?

谢谢

推荐答案

获取切片是O(i_2 - i_1).这是因为Python的列表内部表示是一个数组，因此您可以从i_1开始并迭代到i_2.

Getting a slice is O(i_2 - i_1). This is because Python's internal representation of a list is an array, so you can start at i_1 and iterate to i_2.

有关更多信息，请参见Python 时间复杂度Wiki条目

For more information, see the Python time complexity wiki entry

您还可以在 CPython源代码<中查看实现/a>如果愿意.

You can also look at the implementation in the CPython source if you want to.

这篇关于列表切片的Big-O的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！