Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

思想：利用动态规划的思想求解，distance[i][j]表示word1前i个字符和word2前j个字符所需要的最小编辑次数；而distance[i][j]肯定由distance[i-1][j] 和distance[i][j-1]以及distance[i-1][j-1]转变来的。从而可以得到

distance[i][j] = min { distance[i-1][j] + 1,distance[i][j-1] + 1,distance[i-1][j-1] + (word1[i-2]==word2[j-2]?0:1) } 

这里要注意＋ 与 ＝＝ （？：） 之间的优先级，要形成良好的编码习惯，根据思考流程加入括号。

1. public int minDistance(String word1, String word2) {
2. int wlen1 = word1.length();
3. int wlen2 = word2.length();
4. int[][] distance = new int[wlen1+1][wlen2+1]; //多申请一个长度
5. distance[0][0] = 0;
6. for(int i=1;i<=wlen1;i++) {
7. distance[i][0] = i;
8. }
9. for(int i=1;i<=wlen2;i++) {
10. distance[0][i] = i;
11. }
12. for(int i=1;i<=wlen1;i++) {
13. for(int j=1;j<=wlen2;j++) {
14. int cost = (word1.charAt(i-1) == word2.charAt(j-1) ? 0:1); //注意括号，保证正确优先级
15. distance[i][j] = Math.min(distance[i-1][j]+1,Math.min(distance[i][j-1]+1,distance[i-1][j-1]+cost));
16. }
17. }
18. return distance[wlen1][wlen2];
19. }

 

改进：可以backward的思想得到每次的操作，由于可能存在多条路径，可以今选择一条。

代码如下：

1. int dp[100][100];
2. typedef struct {
3. int left;
4. int down;
5. int slope;
6. } Direct;
7. Direct direct[100];
8. int minimalEditDistance(string s1,string s2) {
9. int s1len = s1.size();
10. int s2len = s2.size();
11. for(int i=1;i<=s2len;i++) {
12. dp[0][i] = i;
13. }
14. for(int i=1;i<=s1len;i++) {
15. dp[i][0] = i;
16. }
17. dp[0][0] = 0;
18. for(int i=1;i<=s1len;i++) {
19. for(int j=1;j<=s2len;j++) {
20. dp[i][j] = min(dp[i-1][j]+1,dp[i][j-1]+1);
21. dp[i][j] = min(dp[i][j],dp[i-1][j-1]+(s1[i-1]==s2[j-1]?0:1));
22. cout<<i<<'\t'<<j<<'\t'<<dp[i][j]<<endl;
23. int cnt=0;
24. if(dp[i][j]==dp[i-1][j]+1) {
25. direct[i].left = 1;
26. } else if(dp[i][j]==dp[i][j-1]) {
27. direct[i].down = 1;
28. } else {
29. direct[i].slope = 1;
30. }
31. }
32. }
33. return dp[s1len][s2len];
34. }
35. int main() {
36. //input the string s1 and s2
37. string s1 = "abc";
38. string s2 = "a";
39. memset(direct,0,sizeof(direct));
40. cout<< minimalEditDistance(s1,s2) <<endl;
41. for(int i=s1.length();i>=1;i--) {
42. if(direct[i].left == 1) {
43. cout<<"deletion"<<endl;
44. } else if(direct[i].down == 1) {
45. cout<<"insert"<<endl;
46. } else {
47. cout<<"substitution"<<endl;
48. }
49. }
50. }