Here you can find the source of levenshteinDistanceRatio(CharSequence lhs, CharSequence rhs)
| Parameter | Description |
|---|---|
| lhs | a parameter |
| rhs | a parameter |
public static double levenshteinDistanceRatio(CharSequence lhs, CharSequence rhs)
//package com.java2s; //License from project: Open Source License public class Main { /**/* ww w . java 2s. c o m*/ * Returns string similarity as a ratio. This is calculated as a ratio of * levenshstein distance to max possible distance (which is the length of * the longer string). The more dissimilar the strings are, the lower * the value returned. Identical strings return 1.0. * @param lhs * @param rhs * @return */ public static double levenshteinDistanceRatio(CharSequence lhs, CharSequence rhs) { int maxDist = lhs.length(); if (rhs.length() > maxDist) { maxDist = rhs.length(); } if (maxDist == 0) { return 1; } return (maxDist - Double.valueOf(levenshteinDistance(lhs, rhs))) / Double.valueOf(maxDist); } /** * Shamelessly copied from this website with only a tiny modification: * https://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Levenshtein_distance#Java */ public static int levenshteinDistance(CharSequence lhs, CharSequence rhs) { // if strings are the same, dist is 0 if (lhs.equals(rhs)) { return 0; } int len0 = lhs.length() + 1; int len1 = rhs.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 = (lhs.charAt(i - 1) == rhs.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]; } }