Find the Levenshtein distance between two Strings.

Introduction

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).
 Code:
 /*
  * Copyright 2013 Guidewire Software, Inc.
  */
 /**
  * This class is based, in part, on org.apache.commons.lang.StringUtils and is intended
  * to break the dependency on that project.
  *
  * @author <a href="http://jakarta.apache.org/turbine/">Apache Jakarta Turbine</a>
  * @author <a href="mailto:jon@latchkey.com">Jon S. Stevens</a>
  * @author Daniel L. Rall
  * @author <a href="mailto:gcoladonato@yahoo.com">Greg Coladonato</a>
  * @author <a href="mailto:ed@apache.org">Ed Korthof</a>
  * @author <a href="mailto:rand_mcneely@yahoo.com">Rand McNeely</a>
  * @author Stephen Colebourne
  * @author <a href="mailto:fredrik@westermarck.com">Fredrik Westermarck</a>
  * @author Holger Krauth
  * @author <a href="mailto:alex@purpletech.com">Alexander Day Chaffee</a>
  * @author <a href="mailto:hps@intermeta.de">Henning P. Schmiedehausen</a>
  * @author Arun Mammen Thomas
  * @author Gary Gregory
  * @author Phil Steitz
  * @author Al Chou
  * @author Michael Davey
  * @author Reuben Sivan
  * @author Chris Hyzer
  *  Johnson
                          
  */
 //package com.java2s;
                          
 public class Main {
     public static void main(String[] argv) {
         String s = "java2s.com";
         String t = "java2s.com1";
         System.out.println(getLevenshteinDistance(s, t));
     }
                          
     /**
      * <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>
      * getLevenshteinDistance(null, *)             = IllegalArgumentException
      * getLevenshteinDistance(*, null)             = IllegalArgumentException
      * getLevenshteinDistance("","")               = 0
      * getLevenshteinDistance("","a")              = 1
      * getLevenshteinDistance("aaapppp", "")       = 7
      * getLevenshteinDistance("frog", "fog")       = 1
      * getLevenshteinDistance("fly", "ant")        = 3
      * getLevenshteinDistance("elephant", "hippo") = 7
      * getLevenshteinDistance("hippo", "elephant") = 7
      * getLevenshteinDistance("hippo", "zzzzzzzz") = 8
      * 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>
      */
     public static int getLevenshteinDistance(String s, String t) {
         if (s == null || t == null) {
             throw new IllegalArgumentException("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.
          */
                          
         int n = s.length(); // length of s
         int m = t.length(); // length of t
                          
         if (n == 0) {
             return m;
         } else if (m == 0) {
             return n;
         }
                          
         if (n > m) {
             // swap the input strings to consume less memory
             String tmp = s;
             s = t;
             t = tmp;
             n = m;
             m = t.length();
         }
                          
         int p[] = new int[n + 1]; //'previous' cost array, horizontally
         int d[] = new int[n + 1]; // cost array, horizontally
         int _d[]; //placeholder to assist in swapping p and d
                          
         // indexes into strings s and t
         int i; // iterates through s
         int j; // iterates through t
                          
         char t_j; // jth character of t
                          
         int cost; // cost
                          
         for (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 counts
         return p[n];
     }
                          
     /**
      * Gets a String's length or <code>0</code> if the String is <code>null</code>.
      *
      * @param str
      *            a String or <code>null</code>
      * @return String length or <code>0</code> if the String is <code>null</code>.
      * @since 2.4
      */
     public static int length(String str) {
         return str == null ? 0 : str.length();
     }
 }
Find the Levenshtein distance between two Strings. - Java java.lang

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