Created
December 18, 2012 05:39
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Match a regular expression (including '.' and '*' using dynamic programming
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bool canMatch(char a, char b) | |
{ | |
return (a == b || b == '.'); | |
} | |
bool isMatch(const char *s, const char *p) { | |
int lS = strlen(s); | |
int lP = strlen(p); | |
vector<vector<bool> > F(lS + 1, vector<bool>(lP + 1)); | |
F[0][0] = true; | |
for (int i = 0; i <= lS; i++) | |
{ | |
for (int j = 1; j <= lP; j++) | |
{ | |
if(i>0) | |
// matches one character, index of both string and pattern move forward by 1 | |
if (F[i-1][j-1] && canMatch(s[i-1], p[j-1])) | |
{ | |
F[i][j] = true; | |
continue; | |
} | |
if (i > 0 && j > 1) | |
// matches the situation when the next char in the string is the same as the char before the '*' in the pattern | |
// pre-match: string xxxCyyy, pattern mmmC*nnn, xxx matches mmmC* | |
// current match: xxxC matches mmmC* | |
if (F[i-1][j] && canMatch(s[i-1], p[j-2]) && p[j-1] == '*') | |
{ | |
F[i][j] = true; | |
continue; | |
} | |
if (j > 1) | |
// matches the situation when the next two chars in the pattern is in the form of C* | |
// pre-match: string xxxyyy, pattern mmmC*nnn, xxx matches mmm | |
// current match: xxx matches mmmC* | |
if (F[i][j-2] && p[j-1] == '*') | |
{ | |
F[i][j] = true; | |
continue; | |
} | |
} | |
} | |
return F[lS][lP]; | |
} | |
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Hey, can you please explain the code in detail. I am not able to understand why are you taking an array of size ls+1 and lp+1???