[Leetcode] Edit Distance

Topic:

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

To find the string edit distance (edit Distance), the editing distance refers to the minimum number of edit operations required to turn from one two string to another. Permission edits include a) insert a character B) delete one character C) replace one character with another character such as: Tea->eat the minimum two steps.　　Analysis process: Assuming that two strings are S and T, enumeration string s and T last character S[i], T[j] corresponds to four cases: (character-character) (character-blank) (Blank-character) (blank-blank); Obviously, (blank-blank) is necessarily an extra edit operation. (1) 　　　s + blank    t + character x     Then, S becomes T, finally, insert "character x" dp[i,j] = dp[i,j-1] + 1  (2) s + character X T + character y at the end of S  s becomes T, finally, in X modified into y         DP[I,J] = Dp[i-1,j-1] + (x==y? 0:1)       (3) S + Character x    t + blank    s to t,x removed    dp[i,j] = Dp[i-1,j] + 1 with Dp[i][j] for first string s[0...i] and 2nd string t[0　　　　The shortest editing distance of ... j]. Then    if (S[i]==t[j])  dp[i][j] = min (dp[i-1][j-1],dp[i][j-1]+1,dp[i-1][j-1]+1) else                  DP[I][J] = min (dp[i-1][j],dp[i][j-1],dp[i-1][j-1]) +1 The final code is as follows:

Class Solution {public:    int mindistance (string word1, String word2) {int len1 = word1.length (); int len2 = Word2.length () if (len1 = = 0) return len2;if (Len2 = = 0) return Len1;int i,j;vector<vector<int> > dp (len1 + 1, vector<int > (len2 + 1)); for (i = 0; I <= len1; i++) {dp[i][0] = i;} for (j = 0; J <= Len2; j + +) {Dp[0][j] = j;} for (i = 1, i <= len1; i++) {for (j = 1; J <= Len2; j + +) {/*int cost = (word1[i-1] = = Word2[j-1])? 0:1;dp[i][j] = m In (Dp[i-1][j-1]+cost,min (dp[i-1][j]+1,dp[i][j-1]+1)); */if (word1[i-1] = word2[j-1]) {dp[i][j] = min (dp[i-1][j-1], Min (dp[i-1][j],dp[i][j-1]) + 1);} Else{dp[i][j] = min (dp[i-1][j-1],min (dp[i-1][j],dp[i][j-1])) + 1;}}} return dp[len1][len2];}    ;

　　

[Leetcode] Edit Distance