Implementing a simple Trie for efficient Levenshtein Distance calculation - Java

Question

UPDATE 3

Done. Below is the code that finally passed all of my tests. Again, this is modeled after Murilo Vasconcelo\'s modified version of Steve

Answer 1

Well, here's how I did it a long time ago. I stored the dictionary as a trie, which is simply a finite-state-machine restricted to the form of a tree. You can enhance it by not making that restriction. For example, common suffixes can simply be a shared subtree. You could even have loops, to capture stuff like "nation", "national", "nationalize", "nationalization", ...

Keep the trie as absolutely simple as possible. Don't go stuffing strings in it.

Remember, you don't do this to find the distance between two given strings. You use it to find the strings in the dictionary that are closest to one given string. The time it takes depends on how much levenshtein distance you can tolerate. For distance zero, it is simply O(n) where n is the word length. For arbitrary distance, it is O(N) where N is the number of words in the dictionary.