Leetcode Regular Expression Matching

The original title link is here: https://leetcode.com/problems/regular-expression-matching/

Regular expressions It is very unrealistic to expect a character-by-character match to be expected. And the "match" issue is very easy to convert to "match a part" and the entire match does not match.

So look for "the rest of the match" situation. It's a good thing to turn a big problem into a smaller problem: recursion

If pattern is a "x*" type, then the pattern is reduced by two two. Otherwise, it is a reduction of one. No matter how you reduce it, make sure there are so many pattern. For example, S.substring (n), where n is the largest and s.length ()

The terminating condition of the recursion is no? The length of P is 0 to see if s is 0, or P is 1 to see the last character of S and P RP the last char on or P is '. '.

Time Complexity:o (2^n).

AC Java:

1  Public classSolution {2      Public BooleanIsMatch (String s, String p) {3         if(p.length () = = 0){4             returnS.length () = = 0;5         }6         if(p.length () = = 1){7             returnS.length () = = 1 && (s.charat (0) = = P.charat (0) | | P.charat (0) = = '. ');8         }9         Ten         //after the above two if, the length of P is already greater than or equal to 2 One         if(P.charat (1)! = ' * ') {//The second character of P is not ' * ' A             returnS.length () >0 -&& (S.charat (0) = = P.charat (0) | | P.charat (0) = = '. ')//whether the first character is equal or the first character of the pattern is '. ' -&& IsMatch (s.substring (1), p.substring (1));//whether the remaining strings match the}Else{//P The second character is ' * ' -             if(IsMatch (S, p.substring (2))){ -                 return true; -             } +             returnS.length () >0//string length needs to be greater than 0 -&& (S.charat (0) = = P.charat (0) | | P.charat (0) = = '. ')//the current string is equal or the first character of a pattern is '. ' +&& IsMatch (s.substring (1), p);//Remove the first character of S A         } at     } -}

Ref:http://harrifeng.github.io/algo/leetcode/regular-expression-matching.html

Method two is to use DP, in fact, is a train of thought.

DP[I][J] Represents the length of I s and the length of J P can match, and finally return Dp[s.length ()][p.length ()].

So the DP generation should be set to a size of new Boolean[s.length () +1][p.length () +1].

AC Java:

1  Public classSolution {2      Public BooleanIsMatch (String s, String p) {3         //Method 2 DP4         Boolean[] DP =New Boolean[S.length () +1] [P.length () +1];5Dp[0][0] =true;6          for(inti = 0; I<=s.length (); i++){7              for(intj = 1; J<=p.length (); J + +){8                 if(i!=0){9                     Charc = P.charat (j-1);Ten                     if(j>1 && c = = ' * ') {//the P-tail character is ' * ' OneDP[I][J] = dp[i][j-2]//' * ' Match 0 A|| DP[I][J-1]//' * ' match 1 -|| (Issame (S.charat (i-1), P.charat (j-2)) && dp[i-1][j]);//' * ' Match multiple -}Else{//P-tail character is not ' * ' the                         if(Issame (S.charat (i-1), c) && dp[i-1][j-1]){ -DP[I][J] =true; -                         } -                     } +}Else{//S is an empty string -                     if(j>=2 && p.charat (j-1) = = ' * ' && dp[0][j-2]){ +DP[0][J] =true; A                     } at                 } -             } -         } -         returndp[s.length ()][p.length ()]; -     } -     Private BooleanIssame (CharSCharp) { in         returnp = = '. ' | | p = =s; -     } to}

Ref:http://www.cnblogs.com/icecreamdeqinw/p/4317183.html

Leetcode Regular Expression Matching