您可以构建集合中所有可能单词的语料库的后缀树（查看此链接）
使用此数据结构，您的复杂性将包括O（n）花费一倍的时间来构建树，其中n是您所有单词的长度之和。

You could build a suffix tree of the corpus of all possible words in your set (see this link) Using this data structure your complexity would include a one time cost of O(n) to build the tree, where n is the sum of the lengths of all your words.

建立树后，查找字符串是否匹配应仅取O（n），其中n为长度的字符串。

Once the tree is built finding if a string matches should take just O(n) where n is length of the string.