[Eetcode 10] Regular Expression Matching

1 Topics:

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

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

2 Ideas:

OK, this question I started a comparison really kneeling, no recursion, consider various situations, various if-else, found still can't consider all the situation. See what others write, use recursion to write, the idea is very clear.

Https://leetcode.com/discuss/32424/clean-java-solution

Look at that C + +, want to turn Java, really kneeling, C + + with ' \ ' to represent the end of the string, and Java string is an array, without this one said, various cross-border plus timeout.

Https://leetcode.com/discuss/9405/the-shortest-ac-code

3 Code:

     Public BooleanIsMatch (String s, String p) {if(P.isempty ()) {returnS.isempty (); }        if(p.length () = = 1 | | P.charat (1)! = ' * ') {            if(S.isempty () | | (P.charat (0)! = '. ' && p.charat (0)! = S.charat (0))) {                return false; } Else {                returnIsMatch (s.substring (1), p.substring (1)); }        }        //p.length () >=2 && p.charat (1) = = ' * '//The first char is match         while(!s.isempty () && (S.charat (0) = = P.charat (0) | | P.charat (0) = = '. ')) {              if(IsMatch (S, p.substring (2))) {                 return true; }            //See if the next char of S is still matchs = s.substring (1); }        //First char not match, make * to 0        returnIsMatch (S, p.substring (2)); }

[Eetcode 10] Regular Expression Matching