Leetcode------Edit Distance

Title:	Edit Distance
Pass Rate:	26.1%
Difficulty:	Difficult

Given 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

The topic is the classic problem of dynamic programming, the formula is as follows:

1, S1.length==0,return s2.length;

2, S2.length==0,return s1.length;

3. If S1.charat (i) ==s2.charat (j), Res[i][j]=res[i-1][j-1]

else Res[i][j]=min (res[i-1][j],res[i][j-1],res[i-1][i-1]+1)

See the code specifically:

1  Public classSolution {2      Public intmindistance (String word1, String word2) {3 intLen1 =word1.length ();4     intLen2 =word2.length ();5  6     //len1+1, len2+1, because finally return DP[LEN1][LEN2]7     int[] DP =New int[Len1 + 1] [Len2 + 1];8  9      for(inti = 0; I <= len1; i++) TenDp[i][0] =i; One      A      for(intj = 0; J <= Len2; J + +)  -DP[0][J] =J; -      the   -     //iterate though, and check last char -      for(inti = 1; I <= len1; i++) { -         CharC1 = Word1.charat (i-1); +          for(intj = 1; J <= Len2; J + +) { -             CharC2 = Word2.charat (j-1); +   A             //if last chars equal at             if(C1 = =C2) { -                 //Update DP value for +1 length -DP[I][J] = dp[i-1][j-1]; -}Else { -                 intReplace = Dp[i-1][j-1] + 1; -                 intInsert = Dp[i-1][j] + 1; in                 intDelete = Dp[i][j-1] + 1; -   to                 intMin =math.min (replace, insert); +Min =math.min (min,delete); -DP[I][J] =min; the             } *         } $     }Panax Notoginseng   -     returnDp[len1][len2]; the     } +}

　　　　　

Leetcode------Edit Distance