Compute Levenshtein distance : String Compare « Data Type « Java





Compute Levenshtein distance

     
/*   Copyright 2004 The Apache Software Foundation
 *
 *   Licensed 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.
 */
//based on Apache technology http://www.apache.org
public class Levenshtein {
  // ****************************
  // Get minimum of three values
  // ****************************

  private static int minimum(int a, int b, int c) {
    int mi = a;
    if (b < mi)
      mi = b;
    if (c < mi)
      mi = c;
    return mi;
  }

  // *****************************
  // Compute Levenshtein distance
  // *****************************
  public static int distance(String s, String t) {
    int d[][]; // matrix
    int n; // length of s
    int m; // length of t
    int i; // iterates through s
    int j; // iterates through t
    char s_i; // ith character of s
    char t_j; // jth character of t
    int cost; // cost

    // Step 1
    n = s.length();
    m = t.length();
    if (n == 0)
      return m;
    if (m == 0)
      return n;
    d = new int[n + 1][m + 1];

    // Step 2
    for (i = 0; i <= n; i++)
      d[i][0] = i;
    for (j = 0; j <= m; j++)
      d[0][j] = j;

    // Step 3
    for (i = 1; i <= n; i++) {
      s_i = s.charAt(i - 1);

      // Step 4
      for (j = 1; j <= m; j++) {
        t_j = t.charAt(j - 1);

        // Step 5
        if (s_i == t_j)
          cost = 0;
        else
          cost = 1;

        // Step 6
        d[i][j] = minimum(d[i - 1][j] + 1, d[i][j - 1] + 1, d[i - 1][j - 1] + cost);
      }
    }

    // Step 7
    return d[n][m];
  }

}

   
    
    
    
    
  










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