[Leetcode questions and Notes] Regular Expression matching

Implement Regular Expression matching with support‘.‘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)Some examples:isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

Question: see the recursive solution. This question has been handed in many times, mainly for details. Note the following points:

• When S is 0, if p is 1, false; otherwise, if the 1st bits of P are '*', we need to evaluate the elements behind P, so we call ismatch (S, p. substring (2). In this example, S = "", P = "C *"; if the first of P is not 1, then S and P certainly do not match.
• When the length of P is 1, it must be processed separately, because at this time, we do not need to judge whether the 1st bits of P are. The code for processing This section is as follows:

1. if(p.length() == 1){        if(p.charAt(0) == ‘.‘ && s.length() == 1)            return true;        return s.equals(p);}

• In other cases, the score is based on whether the 1st bits of P are. If it is not '*', The 0th bits of P and S must match (equal or P 0th bits are '. '), otherwise P and S do not match. In this example, S = "AB", P = "C * B" is similar ". If it is '*', we will match 0 bits, 1 bits, 2 bits ..... similar examples include S = "abbc", P = "AB * BBC". At this time, '*' does not match any character in S, then S = "AA", P = "A *", '*' matches two a s.

The Code is as follows:

 1 public class Solution { 2     public boolean isMatch(String s, String p) { 3          if(p.length() == 0) 4             return s.length() == 0; 5          6         if(s.length() == 0){ 7             if(p.length() == 1) 8                 return false; 9             if(p.charAt(1) == ‘*‘)10                 return isMatch(s, p.substring(2));11             return false;12         }13         14         15         if(p.length() == 1){16             if(p.charAt(0) == ‘.‘ && s.length() == 1)17                 return true;18             return s.equals(p);19         }20             21         if(p.length() >= 2 && p.charAt(1) != ‘*‘){22             //if p(1) is not *, we need p(0) equals to s(0) or p(0) equals ‘.‘23             if(p.charAt(0) == s.charAt(0) || p.charAt(0) == ‘.‘ && s.length() != 0)24                 //check if the left is also match25                 return isMatch(s.substring(1), p.substring(1));26             return false;27         }28         else{29             //if p(1) is ‘*‘,we check how many we can match from this * by trying30             int i = 0;31             char now = p.charAt(0);32             while(i<s.length() && (s.charAt(i) == now || now == ‘.‘)){33                 if(isMatch(s.substring(i+1),p.substring(2)))34                         return true;35                 i++;36             }37             //this means we don‘t use this ‘*‘ to match any character, just skip it38             return isMatch(s, p.substring(2));39         }40     }41 }