Regular expression Matching DP

This topic, I from the night before (August 6 night) debugging to now (August 8 16:21), too sad, not a good sum up, I am too sorry for myself! This is the topic:

Implement regular expression matching with support for ‘.‘ and ‘*‘ .

‘.‘ Matches any single character. ' * ' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). The function prototype should be:bool IsMatch (const char *s, const char *p) Some examples:ismatch ("AA", "a") →falseismatch ( "AA", "AA") →trueismatch ("AAA", "AA") →falseismatch ("AA", "A *") →trueismatch ("AA", ". *") →trueismatch ("AB", ". *") →true IsMatch ("AaB", "C*a*b") →true

dynamic planning, in fact, any algorithm is: core thinking + boundary conditions, in fact, a lot of boundary conditions or in the debugging process added up!! the topic, I refer to a number of online blogs, their own summary as follows: first, a two-tuple array is used to record the match of S and P (s represents the source string, p for the target string), e.g. Res[i][j] to true means s[0 ... i-1] and p[0 ... j-1] have been matched. Suppose the current need to determine whether s[0 ... i] and p[0 ... j] matches, that is, res[i+1][j+1] is true:

• If p[j]== ' * '

(1) if p[j-1]!= '. ' RES[I+1][J+1] is set to true, only one of the following 3 conditions is required: A. Res[i+1][j] is true (at this point, the character that precedes it is taken only 1 times)

0		i-1	i	 	 
0	Span style= "color: #4dce1d;" >.....	j-1	J       *

B. res[i+1][j-1] is true (at this point, the character that precedes it is not taken once, that is, the two are empty) c. res[i][j+1] is true and s[i]==s[i-1] and s[i-1]==p[j-1]. (This time ' * ' takes one of the preceding characters 2 times). and the C option can continue to go recursively. (2) if p[j-1]== '. ' because '. * ' represents 0 or more '. ', which can match any character, so as long as any one of res[i+1][j-1] or res[i+1][j] is true, the remaining res[i+1][j+1],res [I+2] [J+1]......res[s.length ()][j+1] can all be set to true.

• If p[j]!= ' * '

when (S[i]==p[j] or p[j]== '. ' ) and Res[i][j] is true, res[i+1][j+1] is set to true.   last return Res[s.lenth ()][p.length ()]Here are my main references, with my own code:

1 classSolution {2  Public:3      BOOLIsMatch (stringSstringp) {4           //constexpr int len1 = Static_cast<const int> (s.length ()), len2 = P.length ();5           //bool Res[len1][len2] = {0}; Here itself originally wanted to use the array directly, but the array subscript is required constant expression, anxious Ah, now also did not solve6           intLen1 = s.length () +1, len2 = p.length () +1;7vector<vector<BOOL>> Res (LEN1, vector<BOOL> (Len2,false));8res[0][0] =true;9            for(inti =1; I < len2;i++)//without these 3 sentences, "AaB", "c*a*b" will not passTen           if(P[i-1] =='*') Oneres[0][i] = res[0][i-2]; A  -            for(intj =0; J < len2-1; J + +) -           { the  -                if(P[j] = ='*') -                { -  +                     if(j>0&&p[j-1] !='.') -                     { +                           for(inti =0; I < len1-1; ++i) A                          if(Res[i +1][j-1] || Res[i +1][J] | | I>0&& S[i-1] = = S[i] && s[i-1] = = P[j-1]&& (Res[i][j +1]|| RES[I][J]))//This place must pay attention to the specific conditions to add the restrictions on the good, do not go to the previous for loop from i=0 to I=1 atRes[i +1][j +1] =true; -                     } -                     Else -                     { -                          inti =0; -                     //For (; i < len1;) in                          //if (!res[i+1][j-1] &&!res[i][j]) -                          //++i; This place actually wrote a dead loop. to                           while(j>0&&i < len1-1&&!res[i +1][j-1] &&!res[i+1][j]) +++i; -                           for(; I < len1-1; ++i) theRes[i +1][j +1] =true; *  $                     }Panax Notoginseng                } -                Else  the                { +                      A                      for(inti =0; I < len1-1;++i) the                     if((s[i] = = P[j] | | p[j] = ='.') &&Res[i][j]) +Res[i +1][j +1] =true; -                } $           } $           returnRES[LEN1-1][len2-1]; -      } -};

Regular expression Matching DP