15 thinking question editing distance

 

/* Description: Set A and B to two strings. Use the minimum number of characters to convert string a to string B. The character operations mentioned here include: (1) deleting one character; (2) Inserting one character; (3) changing one character to another. The minimum number of characters used to convert string a to string B is the distance from string a to string B and is recorded as D (A, B ). Try to design an effective Algorithm For any two strings a and B, calculate their editing distance D (A, B ). Requirement: input: the first line is string a, and the second line is string B. Output: string a and B editing distance D (A, B) idea: open a two-dimensional array d [I] [J] to record the editing distance between a0-ai and b0-bj, to implement recursion, you need to consider the overhead of the delete, insert, and replace operations on one of the strings, and find out a minimum overhead that is the specific algorithm: first, specify the first row and the first column, and then calculate each value D [I, j] as follows: d [I] [J] = min (d [I-1] [J] + 1, D [I] [J-1] + 1, d [I-1] [J-1] + (S1 [I] = S2 [J]? ); In the last row, the value in the last column is the minimum editing distance */# include <iostream> using namespace STD; char S1 [100], S2 [100]; int DP [101] [101]; int min (int A, int B, int c) {int temp = (a <B )? A: B; Return (temp <c )? Temp: C;} void levenshteindistance (INT len1, int len2) {for (INT I = 0; I <= len1; I ++) DP [I] [0] = I; for (Int J = 0; j <= len2; j ++) DP [0] [J] = J; for (INT I = 1; I <= len1; I ++) {for (Int J = 1; j <= len2; j ++) {int cost = (S1 [I-1] = S2 [J-1])? 0: 1; // note that this is not S1 [I] = S2 [J] int deletion = DP [I-1] [J] + 1; int insertion = DP [I] [J-1] + 1; int substitution = DP [I-1] [J-1] + cost; DP [I] [J] = min (deletion, insertion, substitution) ;}} cout <DP [len1] [len2] <Endl;} void main () {cout <"Enter string 1:" <Endl; cin> S1; cout <"Enter string 2:" <Endl; CIN> S2; levenshteindistance (strlen (S1), strlen (S2 ));}