检查是否给定的字符串遵循特定的模式 [英] Check if the given string follows the given pattern

问题描述

我的一个朋友刚做了他的采访在谷歌，得到了拒绝，因为他不能给解决这个问题。

A friend of mine just had his interview at Google and got rejected because he couldn't give a solution to this question.

我有几天我自己的面试，似乎无法想出一个办法来解决它。

I have my own interview in a couple of days and can't seem to figure out a way to solve it.

下面的问题：

正在给定的图案，如[A B A B]。您也给予了   字符串，例如redblueredblue。我需要编写一个程序，告诉   是否该字符串的给定模式或没有。

You are given a pattern, such as [a b a b]. You are also given a string, example "redblueredblue". I need to write a program that tells whether the string follows the given pattern or not.

举几个例子：

模式：[利群]字符串：catdogdogcat返回1

Pattern: [a b b a] String: catdogdogcat returns 1

模式：ABAB]字符串：redblueredblue返回1

Pattern: [a b a b] String: redblueredblue returns 1

模式：[利群]字符串：redblueredblue返回0

Pattern: [a b b a] String: redblueredblue returns 0

我想到了几个办法，想获得在图案独特的字符数，然后找到该字符串的许多独特的子然后使用一个HashMap的模式相比较。但是，原来是一个问题，如果一个子串为b的一部分。

I thought of a few approaches, like getting the number of unique characters in the pattern and then finding that many unique substrings of the string then comparing with the pattern using a hashmap. However, that turns out to be a problem if the substring of a is a part of b.

这将会是真正伟大如果你们可以帮助我吧。 ：）

It'd be really great if any of you could help me out with it. :)

更新：

新增信息：可以有任意数量的模式（亚利桑那州）的字符。两个字符不会再present相同的子字符串。此外，一个字符不能再present一个空字符串。

Added Info: There can be any number of characters in the pattern (a-z). Two characters won't represent the same substring. Also, a character can't represent an empty string.

推荐答案

难道你只需要翻译使用反向引用的模式应用到正则表达式，即是这样的（Python 3中与装载的重模块）：

Don't you just need to translate the pattern to a regexp using backreferences, i.e. something like this (Python 3 with the "re" module loaded):

>>> print(re.match('(.+)(.+)\\2\\1', 'catdogdogcat'))
<_sre.SRE_Match object; span=(0, 12), match='catdogdogcat'>

>>> print(re.match('(.+)(.+)\\1\\2', 'redblueredblue'))
<_sre.SRE_Match object; span=(0, 14), match='redblueredblue'>

>>> print(re.match('(.+)(.+)\\2\\1', 'redblueredblue'))
None

的正则表达式看起来pretty的微不足道的产生。如果需要支持超过9 backrefs，你可以使用命名组 - 见href="https://docs.python.org/3.4/library/re.html"> Python的正则表达式文档的

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