1 package com.algorithm.dynamicprogramming;
  2
  3 /**
  4  * 动态规划——字符串的编辑距离
  5  * s1 = "abc", s2 = "def"
  6  * 计算公式：
  7  *          | 0                                           i = 0, j = 0
  8  *          | j                                           i = 0, j > 0
  9  * d[i,j] = | i                                           i > 0, j = 0
 10  *          | min(d[i,j-1]+1, d[i-1,j]+1, d[i-1,j-1])    s1(i) = s2(j)
 11  *          | min(d[i,j-1]+1, d[i-1,j]+1, d[i-1,j-1]+1)  s1(i) ≠ s2(j)
 12  * 定义二维数组[4][4]：
 13  *      d e f            d e f
 14  *   |x|x|x|x|        |0|1|2|3|
 15  * a |x|x|x|x|  =>  a |1|1|2|3|  => 编辑距离d = [3][3] = 3
 16  * b |x|x|x|x|      b |2|2|2|3|
 17  * c |x|x|x|x|      c |3|3|3|3|
 18  *
 19  * Created by yulinfeng on 6/29/17.
 20  */
 21 public class Levenshtein {
 22
 23     public static void main(String[] args) {
 24         String s1 = "abc";
 25         String s2 = "def";
 26         int editDistance = levenshtein(s1, s2);
 27         System.out.println("s1=" + s1 + "与s2=" + s2 + "的编辑距离为：" + editDistance);
 28     }
 29
 30     /**
 31      * 编辑距离求解
 32      * @param s1 字符串s1
 33      * @param s2 字符串s2
 34      * @return 编辑距离
 35      */
 36     private static int levenshtein(String s1, String s2) {
 37         int i = 0;  //s1字符串中的字符下标
 38         int j = 0;  //s2字符串中的字符下标
 39         char s1i = 0;   //s1字符串第i个字符
 40         char s2j = 0;   //s2字符串第j个字符
 41         int m = s1.length();    //s1字符串长度
 42         int n = s2.length();    //s2字符串长度
 43         if (m == 0) {   //s1字符串长度为0，此时的编辑距离就是s2字符串长度
 44             return n;
 45         }
 46         if (n == 0) {
 47             return m;   //s2字符串长度为0，此时的编辑距离就是s1字符串长度
 48         }
 49         int[][] solutionMatrix = new int[m + 1][n + 1];     //求解矩阵
 50         /**
 51          *      d e f
 52          *   |0|x|x|x|
 53          * a |1|x|x|x|
 54          * b |2|x|x|x|
 55          * c |3|x|x|x|
 56          */
 57         for (i = 0; i < m + 1; i++) {
 58             solutionMatrix[i][0] = i;
 59         }
 60         /**
 61          *      d e f
 62          *   |0|1|2|3|
 63          * a |x|x|x|x|
 64          * b |x|x|x|x|
 65          * c |x|x|x|x|
 66          */
 67         for (j = 0; j < n + 1; j++) {
 68             solutionMatrix[0][j] = j;
 69         }
 70         /**
 71          * 上面两个操作后，求解矩阵变为
 72          *      d e f
 73          *   |0|1|2|3|
 74          * a |1|x|x|x|
 75          * b |2|x|x|x|
 76          * c |3|x|x|x|
 77          * 接下来就是填充剩余表格
 78          */
 79         for (i = 1; i < m + 1; i++) {   //i = 1,j = 1, 2, 3，以行开始填充
 80             s1i = s1.charAt(i - 1);
 81             for (j = 1; j < n + 1; j++) {
 82                 s2j = s2.charAt(j - 1);
 83                 int flag = (s1i == s2j) ? 0 : 1;    //根据公式，如果s1[i] = s2[j]，则d[i,j]=d[i-1,j-1]，如果s1[i] ≠ s2[j]，则其中一个公式为d[i,j]=d[i-1,j-1]+1
 84                 solutionMatrix[i][j] = min(solutionMatrix[i][j-1] + 1, solutionMatrix[i-1][j] + 1, solutionMatrix[i-1][j-1] + flag);
 85             }
 86         }
 87         return solutionMatrix[m][n];
 88     }
 89
 90     /**
 91      * 根据公式求解编辑距离
 92      * @param insert s1插入操作
 93      * @param delete s1删除操作
 94      * @param edit s1修改操作
 95      * @return 编辑距离
 96      */
 97     private static int min(int insert, int delete, int edit) {
 98         int tmp = insert < delete ? insert : delete;
 99         return tmp < edit ? tmp : edit;
100     }
101 }

 1 '''
 2     动态规划——字符串的编辑距离
 3     s1 = "abc", s2 = "def"
 4     计算公式：
 5              | 0                                           i = 0, j = 0
 6              | j                                           i = 0, j > 0
 7     d[i,j] = | i                                           i > 0, j = 0
 8              | min(d[i,j-1]+1, d[i-1,j]+1, d[i-1,j-1])    s1(i) = s2(j)
 9              | min(d[i,j-1]+1, d[i-1,j]+1, d[i-1,j-1]+1)  s1(i) ≠ s2(j)
10     定义二维数组[4][4]：
11         d e f            d e f
12     |x|x|x|x|        |0|1|2|3|
13     a |x|x|x|x|  =>  a |1|1|2|3|  => 编辑距离d = [4][4] = 3
14     b |x|x|x|x|      b |2|2|2|3|
15     c |x|x|x|x|      c |3|3|3|3|
16 '''
17 def levenshtein(s1, s2):
18     i = 0   #s1字符串中的字符下标
19     j = 0   #s2字符串中的字符下标
20     s1i = ""    #s1字符串第i个字符
21     s2j = ""    #s2字符串第j个字符
22     m = len(s1) #s1字符串长度
23     n = len(s2) #s2字符串长度
24     if m == 0:
25         return n    #s1字符串长度为0，此时的编辑距离就是s2字符串长度
26     if n == 0:
27         return m    #s2字符串长度为0，此时的编辑距离就是s1字符串长度
28     solutionMatrix = [[0 for col in range(n + 1)] for row in range(m + 1)]  #长为m+1，宽为n+1的矩阵
29     '''
30              d e f
31           |0|x|x|x|
32         a |1|x|x|x|
33         b |2|x|x|x|
34         c |3|x|x|x|
35     '''
36     for i in range(m + 1):
37         solutionMatrix[i][0] = i
38     '''
39              d e f
40           |0|1|2|3|
41         a |x|x|x|x|
42         b |x|x|x|x|
43         c |x|x|x|x|
44
45     '''
46     for j in range(n + 1):
47         solutionMatrix[0][j] = j
48     '''
49         上面两个操作后，求解矩阵变为
50              d e f
51           |0|1|2|3|
52         a |1|x|x|x|
53         b |2|x|x|x|
54         c |3|x|x|x|
55         接下来就是填充剩余表格
56     '''
57     for x in range(1, m + 1):
58         s1i = s1[x - 1]
59         for y in range(1, n + 1):
60             s2j = s2[y - 1]
61             flag = 0 if s1i == s2j  else 1
62             solutionMatrix[x][y] = min(solutionMatrix[x][y-1] + 1, solutionMatrix[x-1][y] + 1, solutionMatrix[x-1][y-1] + flag)
63
64     return solutionMatrix[m][n]
65
66 def min(insert, delete, edit):
67     tmp = insert if insert < delete else delete
68     return tmp if tmp < edit else edit
69
70 s1 = "abc"
71 s2 = "def"
72 distance = levenshtein(s1, s2)
73 print(distance)