Leetcode -- Regular Expression matching

Implement Regular Expression matching with support'.'And'*'.

'. 'Matches any single character.' * 'matches zero or more of the preceding element. The matching shoshould coverEntireInput string (not partial ). the function prototype shoshould 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 ",". * ") → trueismatch (" AAB "," C * a * B ") → true
 

Implement regular expressions that support '.' and.

'.' Matches any single character.

'*' Matches 0 or multiple forward elements.

Use recursion for determination. In general, there are two possible cases: one is starting with '*', and the other is not.

Public class regularexpressionmatching {public static void main (string [] ARGs) {system. out. println (ismatch ("AA", "A"); system. out. println (ismatch ("AA", "AA"); system. out. println (ismatch ("AAA", "AA"); system. out. println (ismatch ("AA", "A *"); system. out. println (ismatch ("AA ",". * "); system. out. println (ismatch ("AB ",". * "); system. out. println (ismatch ("AAB", "C * a * B");} public static Boolean ismatch (string s, String p) {If (P. length () = 0) return S. length () = 0; If (P. length () = 1 | P. charat (1 )! = '*') {If (S. Length () <1 | (P. charat (0 )! = '.' & S. charat (0 )! = P. charat (0) return false; return ismatch (S. substring (1), p. substring (1);} else {int I =-1; while (I <S. length () & (I <0 | P. charat (0) = '. '| P. charat (0) = S. charat (I) {If (ismatch (S. substring (I + 1), p. substring (2) return true; I ++;} return false ;}}}
 

reference: The http://www.programcreek.com/2012/12/leetcode-regular-expression-matching-in-java/