[Leetcode] Regular Expression Matching

Regular Expression Matching 

implement regular expression matching with support For ". '  and  ' * ' .

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

Problem Solving Ideas:

Test instructions to implement a regular expression match. which supports. and *

Divided into three different situations. The subscript for the current string and pattern is i,j, respectively.

1, if the current pattern is complete, that is, p[j]== ' ", if the end of the string, then return True, otherwise return false

2, if the next character of the pattern is not *, that is p[j+1]!= ' * '. Here is a discussion of the situation.

(1) If S[I]==P[J], recursive verification i+1, j+1

(2) if p[i]== '. ' and s[i]!= ' + ', recursive verification i+1, j+1

(3) Otherwise, return false

3, if the next character of the pattern is *, that is, p[j+1]== ' * ', then continuously through the recursive backtracking i+k,j+2 (k from 0 to Len (s)-i,j+2 means to cross the current character and *).

The code is as follows:

Class Solution {public:    bool IsMatch (string s, String p) {        return matchhelper (S, p, 0, 0);    }        BOOL Matchhelper (string& S, string& p, int i, int j) {        if (p[j]== ')            } {return s[i]== ';        }                if (p[j + 1]! = ' * ') {            return (s[i] = = P[j]) | | (P[j] = = '. ' && s[i]!= ') ") && Matchhelper (S, p, i + 1, j + 1);        }                while ((s[i] = = P[j]) | | (P[j] = = '. ' && s[i]!= ') ") {            if (Matchhelper (S, p, I, j+2)) return true;            i++;        }        Return Matchhelper (S, p, I, j+2);    }};

[Leetcode] Regular Expression Matching