[Leetcode] edit distance

Edit distance

Given two wordsWord1AndWord2, Find the minimum number of steps required to convertWord1ToWord2. (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

The original question in the beauty of programming, but the answer in the book is Recursive Implementation, but I don't know whether the original author gives the answer is not tested?

Very small cases will time out, for example: "trinitrophenylmethylnitramine", "dinitrophenylhydrazine"

Recursive version:

 1 public class Solution { 2     public int minDistance(String word1, String word2) { 3         if(word1 == null || word1.length() == 0) return word2.length(); 4         if(word2 == null || word2.length() == 0) return word1.length(); 5         if(word1.length() > word2.length()) return minDistance(word2,word1);//assume word1 is shorter than 2 6         if(word1.charAt(0) == word2.charAt(0)){ 7           return minDistance(word1.substring(1),word2.substring(1));   8         }else { 9             int delete = minDistance(word1,word2.substring(1)) + 1;10             int change = minDistance(word1.substring(1),word2.substring(1)) + 1;11             return Math.min(delete,change);12         }13     }14 }   

View code

 

This question is explained in detail in Wikipedia. The specific implementation uses DP:

DP algorithm:

Maintain a two-dimensional matrix to record the distance status:
Dinstance [I] [J] represents the string word1 [0 ~ I] and word2 [0 ~ J] distance
Here we need to open distance to [word1.length () + 1] [word2.length () + 1]

[0] [0] indicates that when both are empty strings, distance is obviously 0.
When I = 0, distance [0] [J] = J (where 1 <= j <= word2.length (), similarly
When J = 0, distance [I] [0] = I (1 <= I <= word1.length () and distance [I] [J]
When word1.charat (I) = word2.charat (J,
Apparently distance [I] [J] = distance [I-1] [J-1];
When word1.charat (I )! = Word2.charat (J,
We need to check that distance [I-1] [J-1], distance [I] [J-1], and distance [I-1] [J] correspond to three situations: change word1 [I] To word2 [J], delete word2 [J], delete word1 [I], and find the smallest number of the three, then + 1 (indicates the delete operation or modify Operation)

The Code is as follows:

 1 public class Solution { 2     public int minDistance(String word1, String word2) { 3         if(word1 == null || word1.length() == 0) return word2.length(); 4         if(word2 == null || word2.length() == 0) return word1.length(); 5         if(word1.length() > word2.length()) return minDistance(word2,word1);//assume word1 is shorter than 2 6         int height = word1.length() + 1,width = word2.length() + 1; 7         int[][] dp = new int[height][width]; 8         for(int i = 0; i < width;i++){ 9             if(i < height){10                 dp[i][0] = i;11             }12             dp[0][i] = i;13         }14         for(int i = 1; i < height ; i++){15             for(int j = 1; j < width ; j++){16                 if(word1.charAt(i - 1) == word2.charAt(j - 1)){17                     dp[i][j] = dp[i - 1][j - 1];18                 }else{19                     dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j])) + 1;20                 }21             }22         }23         return dp[word1.length()][word2.length()];24     }25 }