Main.java Source code

Introduction

Here is the source code for Main.java

Source

//package com.java2s;

import android.text.TextUtils;

public class Main {
    /**
     * Author: Chas Emerick (source: http://mrfoo.de/archiv/1176-Levenshtein-Distance-in-Java.html)
     *
     * This method uses the LevenstheinDistance algorithm to compute the similarity of two strings.
     *
     * @return the minimum number of single-character edits required to change one of the given
     * strings into the other
     */
    public static int getLevenshteinDistance(String s, String t) {
        if (s == null || t == null) {
            throw new IllegalArgumentException("Strings must not be null");
        }
        if (TextUtils.isEmpty(s)) {
            return t.length();
        } else if (TextUtils.isEmpty(t)) {
            return s.length();
        }
        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;
        }

        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];
    }
}