算法代码实现:

package com.util;

public class SimFeatureUtil {

	private static int min(int one, int two, int three) {
		int min = one;
		if (two < min) {
			min = two;
		}
		if (three < min) {
			min = three;
		}
		return min;
	}

	public static int ld(String str1, String str2) {
		int d[][]; // 矩阵
		int n = str1.length();
		int m = str2.length();
		int i; // 遍历str1的
		int j; // 遍历str2的
		char ch1; // str1的
		char ch2; // str2的
		int temp; // 记录相同字符,在某个矩阵位置值的增量,不是0就是1
		if (n == 0) {
			return m;
		}
		if (m == 0) {
			return n;
		}
		d = new int[n + 1][m + 1];
		for (i = 0; i <= n; i++) { // 初始化第一列
			d[i][0] = i;
		}
		for (j = 0; j <= m; j++) { // 初始化第一行
			d[0][j] = j;
		}
		for (i = 1; i <= n; i++) { // 遍历str1
			ch1 = str1.charAt(i - 1);
			// 去匹配str2
			for (j = 1; j <= m; j++) {
				ch2 = str2.charAt(j - 1);
				if (ch1 == ch2) {
					temp = 0;
				} else {
					temp = 1;
				}
				// 左边+1,上边+1, 左上角+temp取最小
				d[i][j] = min(d[i - 1][j] + 1, d[i][j - 1] + 1, d[i - 1][j - 1]+ temp);
			}
		}
		return d[n][m];
	}
	public static double sim(String str1, String str2) {
		try {
			double ld = (double)ld(str1, str2);
			return (1-ld/(double)Math.max(str1.length(), str2.length()));
		} catch (Exception e) {
			return 0.1;
		}
	}

	public static void main(String[] args) {
		String str1 = "测试12";
		String str2 = "测试123";
		System.out.println("ld=" + ld(str1, str2));
		System.out.println("sim=" + sim(str1, str2));
	}
}

算法介绍:

算法原理:

设我们可以使用d[ i , j ]个步骤（可以使用一个二维数组保存这个值），表示将串s[ 1…i ] 转换为 串t [ 1…j ]所需要的最少步骤个数，那么，在最基本的情况下，即在i等于0时，也就是说串s为空，那么对应的d[0,j] 就是 增加j个字符，使得s转化为t，在j等于0时，也就是说串t为空，那么对应的d[i,0] 就是 减少 i个字符，使得s转化为t。

  然后我们考虑一般情况，加一点动态规划的想法，我们要想得到将s[1..i]经过最少次数的增加，删除，或者替换操作就转变为t[1..j]，那么我们就必须在之前可以以最少次数的增加，删除，或者替换操作，使得现在串s和串t只需要再做一次操作或者不做就可以完成s[1..i]到t[1..j]的转换。所谓的“之前”分为下面三种情况：

1）我们可以在k个操作内将 s[1…i] 转换为 t[1…j-1]

2）我们可以在k个操作里面将s[1..i-1]转换为t[1..j]

3）我们可以在k个步骤里面将 s[1…i-1] 转换为 t [1…j-1]

针对第1种情况，我们只需要在最后将 t[j] 加上s[1..i]就完成了匹配，这样总共就需要k+1个操作。

针对第2种情况，我们只需要在最后将s[i]移除，然后再做这k个操作，所以总共需要k+1个操作。

针对第3种情况，我们只需要在最后将s[i]替换为 t[j]，使得满足s[1..i] == t[1..j]，这样总共也需要k+1个操作。而如果在第3种情况下，s[i]刚好等于t[j]，那我们就可以仅仅使用k个操作就完成这个过程。

  最后，为了保证得到的操作次数总是最少的，我们可以从上面三种情况中选择消耗最少的一种最为将s[1..i]转换为t[1..j]所需要的最小操作次数。

算法实现步骤:

算法步骤详解:

本小节将演示如何计算"GUMBO"和"GAMBOL"两个字符串的Levenshtein距离。

步骤1、2

初始化完了之后重点是理解步骤6.

步骤3-6，当 i = 1

我们算V1中的值：以红色的0所在的格子为例

步骤3-6，当 i = 2

步骤3-6，当 i = 3

步骤3-6，当 i = 4

步骤3-6，当 i = 5

步骤7

编辑距离就是矩阵右下角的值，v1[m] == 2。由"GUMBO"变换为"GAMBOL"的过程对于我来说是很只管的，即通过将"A"替换为"U"，并在末尾追加"L"这样子（实际上替换的过程是由移除和插入两个操作组合而成的）。

我们得到最小编辑距离为2

那么它们的相似度为   (1-ld/(double)Math.max(str1.length(), str2.length()));

1 - 2/6=0.6666666666666667

参考链接:

http://www.cnblogs.com/ymind/archive/2012/03/27/fast-memory-efficient-Levenshtein-algorithm.html

http://teiraisan.blog.163.com/blog/static/12278141420098685835372/

http://teiraisan.blog.163.com/blog/static/12278141420098685835372/

其他语言的代码实现:

c++

In C++, the size of an array must be a constant, and this code fragment causes an error at compile time:   
  
int sz = 5;   
int arr[sz];   
  
This limitation makes the following C++ code slightly more complicated than it would be if the matrix could simply be declared as a two-dimensional array, with a size determined at run-time.   
  
In C++ it's more idiomatic to use the System Template Library's vector class, as Anders Sewerin Johansen has done in an alternative C++ implementation.   
  
Here is the definition of the class (distance.h):   
  
class Distance   
{   
  public:   
    int LD (char const *s, char const *t);   
  private:   
    int Minimum (int a, int b, int c);   
    int *GetCellPointer (int *pOrigin, int col, int row, int nCols);   
    int GetAt (int *pOrigin, int col, int row, int nCols);   
    void PutAt (int *pOrigin, int col, int row, int nCols, int x);   
};    
  
Here is the implementation of the class (distance.cpp):   
  
#include "distance.h"  
#include <string.h>   
#include <malloc.h>   
  
//****************************   
// Get minimum of three values   
//****************************   
  
int Distance::Minimum (int a, int b, int c)   
{   
int mi;   
  
  mi = a;   
  if (b < mi) {   
    mi = b;   
  }   
  if (c < mi) {   
    mi = c;   
  }   
  return mi;   
  
}   
  
//**************************************************   
// Get a pointer to the specified cell of the matrix   
//**************************************************    
  
int *Distance::GetCellPointer (int *pOrigin, int col, int row, int nCols)   
{   
  return pOrigin + col + (row * (nCols + 1));   
}   
  
//*****************************************************   
// Get the contents of the specified cell in the matrix    
//*****************************************************   
  
int Distance::GetAt (int *pOrigin, int col, int row, int nCols)   
{   
int *pCell;   
  
  pCell = GetCellPointer (pOrigin, col, row, nCols);   
  return *pCell;   
  
}   
  
//*******************************************************   
// Fill the specified cell in the matrix with the value x   
//*******************************************************   
  
void Distance::PutAt (int *pOrigin, int col, int row, int nCols, int x)   
{   
int *pCell;   
  
  pCell = GetCellPointer (pOrigin, col, row, nCols);   
  *pCell = x;   
  
}   
  
//*****************************   
// Compute Levenshtein distance   
//*****************************   
  
int Distance::LD (char const *s, char const *t)   
{   
int *d; // pointer to 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   
int result; // result   
int cell; // contents of target cell   
int above; // contents of cell immediately above   
int left; // contents of cell immediately to left   
int diag; // contents of cell immediately above and to left   
int sz; // number of cells in matrix   
  
  // Step 1    
  
  n = strlen (s);   
  m = strlen (t);   
  if (n == 0) {   
    return m;   
  }   
  if (m == 0) {   
    return n;   
  }   
  sz = (n+1) * (m+1) * sizeof (int);   
  d = (int *) malloc (sz);   
  
  // Step 2   
  
  for (i = 0; i <= n; i++) {   
    PutAt (d, i, 0, n, i);   
  }   
  
  for (j = 0; j <= m; j++) {   
    PutAt (d, 0, j, n, j);   
  }   
  
  // Step 3   
  
  for (i = 1; i <= n; i++) {   
  
    s_i = s[i-1];   
  
    // Step 4   
  
    for (j = 1; j <= m; j++) {   
  
      t_j = t[j-1];   
  
      // Step 5   
  
      if (s_i == t_j) {   
        cost = 0;   
      }   
      else {   
        cost = 1;   
      }   
  
      // Step 6    
  
      above = GetAt (d,i-1,j, n);   
      left = GetAt (d,i, j-1, n);   
      diag = GetAt (d, i-1,j-1, n);   
      cell = Minimum (above + 1, left + 1, diag + cost);   
      PutAt (d, i, j, n, cell);   
    }   
  }   
  
  // Step 7   
  
  result = GetAt (d, n, m, n);   
  free (d);   
  return result;   
       
}   

'*******************************   
'*** Get minimum of three values   
'*******************************   
  
Private Function Minimum(ByVal a As Integer, _   
                         ByVal b As Integer, _   
                         ByVal c As Integer) As Integer   
Dim mi As Integer   
                             
  mi = a   
  If b < mi Then   
    mi = b   
  End If   
  If c < mi Then   
    mi = c   
  End If   
     
  Minimum = mi   
                             
End Function   
  
'********************************   
'*** Compute Levenshtein Distance   
'********************************   
  
Public Function LD(ByVal s As String, ByVal t As String) As Integer   
Dim d() As Integer ' matrix   
Dim m As Integer ' length of t   
Dim n As Integer ' length of s   
Dim i As Integer ' iterates through s   
Dim j As Integer ' iterates through t   
Dim s_i As String ' ith character of s   
Dim t_j As String ' jth character of t   
Dim cost As Integer ' cost   
     
  ' Step 1  
     
  n = Len(s)   
  m = Len(t)   
  If n = 0 Then   
    LD = m   
    Exit Function   
  End If    
  If m = 0 Then   
    LD = n   
    Exit Function   
  End If    
  ReDim d(0 To n, 0 To m) As Integer   
     
  ' Step 2  
     
  For i = 0 To n   
    d(i, 0) = i   
  Next i   
     
  For j = 0 To m   
    d(0, j) = j   
  Next j   
  
  ' Step 3  
  
  For i = 1 To n   
       
    s_i = Mid$(s, i, 1)   
       
    ' Step 4  
       
    For j = 1 To m   
         
      t_j = Mid$(t, j, 1)   
         
      ' Step 5  
         
      If s_i = t_j Then   
        cost = 0  
      Else   
        cost = 1  
      End If   
         
      ' Step 6  
         
      d(i, j) = Minimum(d(i - 1, j) + 1, d(i, j - 1) + 1, d(i - 1, j - 1) + cost)   
       
    Next j   
       
  Next i   
     
  ' Step 7  
     
  LD = d(n, m)   
  Erase d   
  
End Function  

Python代码 

#!/user/bin/env python   
# -*- coding: utf-8 -*-   
  
class arithmetic():   
       
    def __init__(self):   
        pass  
    ''''' 【编辑距离算法】 【levenshtein distance】 【字符串相似度算法】 '''  
    def levenshtein(self,first,second):   
        if len(first) > len(second):   
            first,second = second,first   
        if len(first) == 0:   
            return len(second)   
        if len(second) == 0:   
            return len(first)   
        first_length = len(first) + 1  
        second_length = len(second) + 1  
        distance_matrix = [range(second_length) for x in range(first_length)]    
        #print distance_matrix   
        for i in range(1,first_length):   
            for j in range(1,second_length):   
                deletion = distance_matrix[i-1][j] + 1  
                insertion = distance_matrix[i][j-1] + 1  
                substitution = distance_matrix[i-1][j-1]   
                if first[i-1] != second[j-1]:   
                    substitution += 1  
                distance_matrix[i][j] = min(insertion,deletion,substitution)   
        print distance_matrix   
        return distance_matrix[first_length-1][second_length-1]   
       
if __name__ == "__main__":   
    arith = arithmetic()   
    print arith.levenshtein('GUMBOsdafsadfdsafsafsadfasfadsfasdfasdfs','GAMBOL00000000000dfasfasfdafsafasfasdfdsa'