Created
February 4, 2015 19:07
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The Levenshtein distance between two strings is defined as the minimum number of edits needed to transform one string into the other, with the allowable edit operations being insertion, deletion, or substitution of a single character.
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using System; | |
/// <summary> | |
/// Contains approximate string matching | |
/// </summary> | |
static class LevenshteinDistance | |
{ | |
/// <summary> | |
/// Compute the distance between two strings. | |
/// </summary> | |
public static int Compute(string s, string t) | |
{ | |
int n = s.Length; | |
int m = t.Length; | |
int[,] d = new int[n + 1, m + 1]; | |
// Step 1 | |
if (n == 0) | |
{ | |
return m; | |
} | |
if (m == 0) | |
{ | |
return n; | |
} | |
// Step 2 | |
for (int i = 0; i <= n; d[i, 0] = i++) | |
{ | |
} | |
for (int j = 0; j <= m; d[0, j] = j++) | |
{ | |
} | |
// Step 3 | |
for (int i = 1; i <= n; i++) | |
{ | |
//Step 4 | |
for (int j = 1; j <= m; j++) | |
{ | |
// Step 5 | |
int cost = (t[j - 1] == s[i - 1]) ? 0 : 1; | |
// Step 6 | |
d[i, j] = Math.Min( | |
Math.Min(d[i - 1, j] + 1, d[i, j - 1] + 1), | |
d[i - 1, j - 1] + cost); | |
} | |
} | |
// Step 7 | |
return d[n, m]; | |
} | |
} | |
class Program | |
{ | |
static void Main() | |
{ | |
Console.WriteLine(LevenshteinDistance.Compute("aunt", "ant")); | |
Console.WriteLine(LevenshteinDistance.Compute("Sam", "Samantha")); | |
Console.WriteLine(LevenshteinDistance.Compute("flomax", "volmax")); | |
} | |
} |
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