Regular Expression Matching & Wildcard Matching

Regular Expression Matching

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)

  Example

IsMatch ("AA", "a") →falseismatch ("AA", "AA") →trueismatch ("AAA", "AA") →falseismatch ("AA", "A *") →trueismatch ("AA", ". *") ") →trueismatch (" AB ",". * ") →trueismatch (" AaB "," C*a*b ") →true

Analysis: If I did not review the interview when the problem, and then the interviewer to test the problem, I will be in the letter to the interviewer wrote two words: Fuck you! This guy obviously doesn't want me to. It took me more than one hours to understand the meaning of the question. 1.' * ' Matches zero or more of the preceding element.　　It doesn't mean that for a *, it can only match a, AA, AAA, infinitely multiple A. It can also match "" (an empty string). Lie down, do you want to get it, do you want it, do you really want it? You can't imagine it! 2. ". *" means that you can match any string, such as "ddd", "DDA", "abc". 3. For "A*b", it matches "B", but does not match "a". 4. If p starts with "*", you have to get rid of it. Well, if you understand the above section, we can solve the problem with DP. We first create a two-dimensional array to hold the intermediate variables.

int m = P.length ();
int n = s.length ();
boolean[][] Match = new Boolean[m + 1][n + 1]; (p is horizontal axis, S is longitudinal)

MATCH[I][J] Indicates whether the first i-1 characters of P match the first j-1 characters of S.

Here are a few things:

If P.chartat (i) is "." or P.charat (i-1) = = S.charat (j-1), then we have:

MATCH[I][J] = match[i-1][j-1];

If P.chartat (i) is not "." and P.charat (i-1)! = S.charat (j-1), then we have:

MATCH[I][J] = false;

OK, the key point is, if p.chartat (i) = = ' * ', then how to do?

First, if P.charat (i-2) = = '. ' | | P.charat (i-2) = = S.charat (j-1)

Then we can not take match[i-1][j-1] (because P.charat (i-1) = = S.charat (j-1) If the above conditions are established), or Match[i-2][j] ("x*" directly into "), or Match[i ][j-1] ("x*" becomes "x*x") | | MATCH[I-1][J] ("x*" becomes "X");

So, we have: match[i][j] = match[i-1][j-1] | | MATCH[I-2][J] | | Match[i][j-1] | | MATCH[I-1][J];

If P.charat (i-2)! = S.charat (j-1), we have only one way:

MATCH[I][J] = Match[i-2][j];

1  Public classSolution {2     /**3 * @param s:a string4 * @param p:a string includes "." and "*"5 * @return: A boolean6      */7      PublicBoolean IsMatch (string s, String p) {8         if(s = =NULL|| p = =NULL)return false;9         Ten          while(P.length () >=1&& P.charat (0) =='*') { Onep = p.substring (1); A         } -         intm =p.length (); -         intn =s.length (); theboolean[][] Match =NewBoolean[m +1][n +1]; -match[0][0] =true; -          for(inti =1; I <= m; i++) { -             if(P.charat (I-1) =='*') { +match[i][0] = Match[i-2][0]; -             } +         } A          at          for(inti =1; I <= m; i++) { -              for(intj =1; J <= N; J + +) { -                 if(P.charat (I-1) = = S.charat (J-1) || P.charat (I-1) =='.') { -MATCH[I][J] = match[i-1][j-1];  -}Else if(P.charat (I-1) =='*') { -                     if(P.charat (I-2) =='.'|| P.charat (I-2) = = S.charat (J-1)) { inMATCH[I][J] = match[i-1][j-1] || Match[i-2][J] | | Match[i][j-1] || Match[i-1][j]; -}Else { toMATCH[I][J] = match[i-2][j]; +                     } -}Else { theMATCH[I][J] =false; *                 } $             }Panax Notoginseng         } -         returnMatch[m][n]; the     } +      A}

Wildcard Matching

Implement wildcard pattern matching with the support for and ‘?‘ ‘*‘ .

• ‘?‘Matches any single character.
• ‘*‘Matches any sequence of characters (including the empty sequence).

The matching should cover the entire input string (not partial).

Has you met this question in a real interview?Yes Example

isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

1  Public classSolution {2     /**3 * @param s:a string4 * @param p:a string includes "?" and "*"5 * @return: A boolean6      */7      PublicBoolean IsMatch (string s, String p) {8         if(s = =NULL|| p = =NULL)9             return false;Tenboolean[][] Match =NewBoolean[p.length () +1][s.length () +1]; Onematch[0][0] =true; A          for(inti =1; I < match[0].length; i++) { -match[0][i] =false; -         } the          -          for(inti =1; i < match.length; i++) { -             if(P.charat (I-1) =='*') { -match[i][0] = Match[i-1][0]; +             }  -         } +          A          for(inti =1; i < match.length; i++) { at              for(intj =1; J < match[0].length; J + +) { -                 if(P.charat (I-1) = = S.charat (J-1) || P.charat (I-1) =='?') { -MATCH[I][J] = match[i-1][j-1]; -}Else if(P.charat (I-1) =='*') { -MATCH[I][J] = match[i-1][j-1] || Match[i-1][J] | | Match[i][j-1]; -                 } in             } -         } to          +         returnmatch[p.length ()][s.length ()]; -     } the}

Regular Expression Matching & Wildcard Matching