Find the Levenshtein distance between two Strings.

/* * Licensed to the Apache Software Foundation (ASF) under one or more * contributor license agreements. See the NOTICE file distributed with * this work for additional information regarding copyright ownership. * The ASF licenses this file to You under the Apache License, Version 2.0 * (the "License"); you may not use this file except in compliance with * the License. You may obtain a copy of the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. * See the License for the specific language governing permissions and * limitations under the License. */publicclassMain { // Misc //----------------------------------------------------------------------- /** * <p>Find the Levenshtein distance between two Strings.</p> * * <p>This is the number of changes needed to change one String into * another, where each change is a single character modification (deletion, * insertion or substitution).</p> * * <p>The previous implementation of the Levenshtein distance algorithm * was from <a href="http://www.merriampark.com/ld.htm">http://www.merriampark.com/ld.htm</a></p> * * <p>Chas Emerick has written an implementation in Java, which avoids an OutOfMemoryError * which can occur when my Java implementation is used with very large strings.<br> * This implementation of the Levenshtein distance algorithm * is from <a href="http://www.merriampark.com/ldjava.htm">http://www.merriampark.com/ldjava.htm</a></p> * * <pre> * StringUtils.getLevenshteinDistance(null, *) = IllegalArgumentException * StringUtils.getLevenshteinDistance(*, null) = IllegalArgumentException * StringUtils.getLevenshteinDistance("","") = 0 * StringUtils.getLevenshteinDistance("","a") = 1 * StringUtils.getLevenshteinDistance("aaapppp", "") = 7 * StringUtils.getLevenshteinDistance("frog", "fog") = 1 * StringUtils.getLevenshteinDistance("fly", "ant") = 3 * StringUtils.getLevenshteinDistance("elephant", "hippo") = 7 * StringUtils.getLevenshteinDistance("hippo", "elephant") = 7 * StringUtils.getLevenshteinDistance("hippo", "zzzzzzzz") = 8 * StringUtils.getLevenshteinDistance("hello", "hallo") = 1 * </pre> * * @param s the first String, must not be null * @param t the second String, must not be null * @return result distance * @throws IllegalArgumentException if either String input <code>null</code> */publicstaticintgetLevenshteinDistance(String s, String t) {if(s == null || t == null) {thrownewIllegalArgumentException("Strings must not be null"); } /* The difference between this impl. and the previous is that, rather than creating and retaining a matrix of size s.length()+1 by t.length()+1, we maintain two single-dimensional arrays of length s.length()+1. The first, d, is the 'current working' distance array that maintains the newest distance cost counts as we iterate through the characters of String s. Each time we increment the index of String t we are comparing, d is copied to p, the second int[]. Doing so allows us to retain the previous cost counts as required by the algorithm (taking the minimum of the cost count to the left, up one, and diagonally up and to the left of the current cost count being calculated). (Note that the arrays aren't really copied anymore, just switched...this is clearly much better than cloning an array or doing a System.arraycopy() each time through the outer loop.) Effectively, the difference between the two implementations is this one does not cause an out of memory condition when calculating the LD over two very large strings. */intn = s.length(); // length of sintm = t.length(); // length of tif(n == 0) {returnm; }elseif(m == 0) {returnn; }if(n > m) { // swap the input strings to consume less memory String tmp = s; s = t; t = tmp; n = m; m = t.length(); }intp[] =newint[n+1]; //'previous' cost array, horizontallyintd[] =newint[n+1]; // cost array, horizontallyint_d[]; //placeholder to assist in swapping p and d // indexes into strings s and tinti; // iterates through sintj; // iterates through tchart_j; // jth character of tintcost; // costfor(i = 0; i<=n; i++) { p[i] = i; }for(j = 1; j<=m; j++) { t_j = t.charAt(j-1); d[0] = j;for(i=1; i<=n; i++) { cost = s.charAt(i-1)==t_j ? 0 : 1; // minimum of cell to the left+1, to the top+1, diagonally left and up +cost d[i] = Math.min(Math.min(d[i-1]+1, p[i]+1), p[i-1]+cost); } // copy current distance counts to 'previous row' distance counts _d = p; p = d; d = _d; } // our last action in the above loop was to switch d and p, so p now // actually has the most recent cost countsreturnp[n]; } }