使用Java中的通配符返回字符串列表的最快方法 [英] Quickest way to return list of Strings by using wildcard from collection in Java

问题描述

我设置了100000字符串。例如，我想得到所有的字符串从JO开始。那么最好的解决方案是什么？

我在想 Aho-Corasick但是我的实现不支持通配符。

解决方案

p>如果你希望所有的字符串以一个序列开始，你可以将所有的字符串添加到一个NavigableSet像TreeSet，并得到 subSet（text，text +'\\\￿'）将提供以 text开头的所有条目这个查找是O（log n）

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如果你想让所有的字符串以一个序列结尾，你可以做一个类似的事情，除非你必须反转String。在这种情况下，从反向String到正向String的TreeMap将是一个更好的结构。

如果你想要x * z，你可以搜索第一个集合，

如果您想要包含 x ，您可以使用导航< String，Set< String >>其中键是每个String从第一个，第二个，第三个char等开始的值该值是一个Set，因为你可以得到重复。你可以像结构的开始一样进行搜索。

I have set of 100000 String. And for example I want to get all strings starting with "JO" from that set. What would be the best solution for that?

I was thinking Aho-Corasick but the implementation I have does not support wild cards.

解决方案

If you want all the strings starting with a sequence you can add all the String into a NavigableSet like TreeSet and get the subSet(text, text+'\uFFFF') will give you all the entries starting with text This lookup is O(log n)

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If you want all the Strings with end with a sequence, you can do a similar thing, except you have to reverse the String. In this case a TreeMap from reversed String to forward String would be a better structure.

If you want "x*z" you can do a search with the first set and take a union with the values of the Map.

if you want contains "x", you can use a Navigable<String, Set<String>> where the key is each String starting from the first, second, third char etc The value is a Set as you can get duplicates. You can do a search like the starts with structure.

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