Programming-the beauty of string Similarity Calculation

ManyProgramA large number of strings are used. For different strings, we hope to be able to determine their similarity. We have defined a set of operation methods to make two different strings the same. The specific operation method is:

1. modify a character (for example, replace "A" with "B ");

2. Add a character (for example, change "abdd" to "aebdd ");

3. delete a character (for example, change "traveling" to "traveling ").

For example, for the strings "abcdefg" and "abcdef", we think we can increase/decrease a "G" to achieve the goal. The preceding two solutions only require one operation. The number of times required for this operation is defined as the distance between the two strings, and the similarity is equal to the reciprocal of "distance + 1. That is to say, the distance between "abcdefg" and "abcdef" is 1, and the similarity is 1/2 = 0.5.

Can you write a given string?AlgorithmTo calculate their similarity?

Using the idea of LCS:

Suppose two strings a = {A1, A2, A3,...}, B = {B1, B2, B3 ,...}. Use the DP idea similar to LCS. Set C [I] [J] to the string A1... AI, B1... BJ distance, if AI = BJ, then C [I] [J] = C [I-1] [J-1];

If Ai! = BJ, C [I] [J] = min (C [I-1] [J] + 1, C [I] [J-1] + 1, c [I-1] [J-1] + 1 );

CodeAs follows:

 
Int editdistance (char * dststr, char * srcstr) {char * tmpdststr = dststr; char * tmpsrcstr = srcstr; int C [N] [m]; int I = 0; int J = 0; while (* dststr ++! = '\ 0') I ++; while (* srcstr ++! = '\ 0') J ++; For (INT m = 0; m <j; ++ m) C [0] [m] = m; for (INT n = 0; n <I; ++ N) C [N] [0] = N; For (int m = 1; m <= I; ++ m) for (INT n = 1; n <= J; ++ N) if (tmpdststr [m] = tmpsrcstr [N]) c [m] [N] = M-1] [n-1]; else {C [m] [N] = min (C [M-1] [n-1] + 1, c M-1] [N] + 1); C [m] [N] = min (C [m] [N], c [m] [n-1] + 1);} return C [I] [J];}
 

Previously calculated: When the longest incrementing sub-sequence in the array is used, the array is sorted first, and then the LCS in the two arrays is located. This can also be used to find LCS, and then max (srcstr, dststr)-LCs. It cannot be seen that LCS is so useful.