Here you can find the source of levenshteinEquals(double threshold, String dom1, String dom2)
| Parameter | Description |
|---|---|
| threshold | a parameter |
| dom1 | a parameter |
| dom2 | a parameter |
public static boolean levenshteinEquals(double threshold, String dom1, String dom2)
//package com.java2s; //License from project: Apache License public class Main { /**//w w w. jav a2s. co m * This method implements the equality check between * two strings using the Levenshtein Distance algorithm * @param threshold * @param dom1 * @param dom2 * @return true/false */ public static boolean levenshteinEquals(double threshold, String dom1, String dom2) { double l = levenshteinDistance(dom1, dom2); if (l >= ((dom1.length() + dom2.length()) * (1.0 - threshold))) { return false; } return true; } /** * This method implements the Levenshtein Distance algorithm * between two strings * @param s0 first string to be compared * @param s1 second string to be compared * @return the cost to turn s0 into s1 */ public static int levenshteinDistance(String s0, String s1) { int len0 = s0.length() + 1; int len1 = s1.length() + 1; // the array of distances int[] cost = new int[len0]; int[] newcost = new int[len0]; // initial cost of skipping prefix in String s0 for (int i = 0; i < len0; i++) cost[i] = i; // dynamically computing the array of distances // transformation cost for each letter in s1 for (int j = 1; j < len1; j++) { // initial cost of skipping prefix in String s1 newcost[0] = j; // transformation cost for each letter in s0 for (int i = 1; i < len0; i++) { // matching current letters in both strings int match = (s0.charAt(i - 1) == s1.charAt(j - 1)) ? 0 : 1; // computing cost for each transformation int cost_replace = cost[i - 1] + match; int cost_insert = cost[i] + 1; int cost_delete = newcost[i - 1] + 1; // keep minimum cost newcost[i] = Math.min(Math.min(cost_insert, cost_delete), cost_replace); } // swap cost/newcost arrays int[] swap = cost; cost = newcost; newcost = swap; } // the distance is the cost for transforming all letters in both strings return cost[len0 - 1]; } }