Most efficient way to calculate Levenshtein distance

The wikipedia entry on Levenshtein distance has useful suggestions for optimizing the computation -- the most applicable one in your case is that if you can put a bound k on the maximum distance of interest (anything beyond that might as well be infinity!) you can reduce the computation to O(n times k) instead of O(n squared) (basically by giving up as soon as the minimum possible distance becomes > k).

Since you're looking for the closest match, you can progressively decrease k to the distance of the best match found so far -- this won't affect the worst case behavior (as the matches might be in decreasing order of distance, meaning you'll never bail out any sooner) but average case should improve.

I believe that, if you need to get substantially better performance, you may have to accept some strong compromise that computes a more approximate distance (and so gets "a reasonably good match" rather than necessarily the optimal one).

According to a comment on this blog, Speeding Up Levenshtein, you can use VP-Trees and achieve O(nlogn). Another comment on the same blog points to a python implementation of VP-Trees and Levenshtein. Please let us know if this works.