Regular Expression Matching

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

'.' Matches any single character.

'*' Matches zero or more of the preceding element. "a/star" means zero to any number of a

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") → false
• isMatch("aa","aa") → true
• isMatch("aaa","aa") → false
• isMatch("aa", "a*") → true
• isMatch("aa", ".*") → true
• isMatch("ab", ".*") → true
• isMatch("aab", "cab") → true

https://leetcode.com/problems/regular-expression-matching/

Solution I:

public boolean isMatch(String s, String p) {
    // base case
    if (p.length() == 0) {
        return s.length() == 0;
    }

    // special case
    if (p.length() == 1) {

        // if the length of s is 0, return false
        if (s.length() < 1) {
            return false;
        }

        //if the first does not match, return false
        else if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != '.')) {
            return false;
        }

        // otherwise, compare the rest of the string of s and p.
        else {
            return isMatch(s.substring(1), p.substring(1));
        }
    }

    // case 1: when the second char of p is not '*'
    if (p.charAt(1) != '*') {
        if (s.length() < 1) {
            return false;
        }
        if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != '.')) {
            return false;
        } else {
            return isMatch(s.substring(1), p.substring(1));
        }
    }

    // case 2: when the second char of p is '*', complex case.
    else {
        //case 2.1: a char & '*' can stand for 0 element
        if (isMatch(s, p.substring(2))) {
            return true;
        }

        //case 2.2: a char & '*' can stand for 1 or more preceding element, 
        //so try every sub string
        int i = 0;
        while (i<s.length() && (s.charAt(i)==p.charAt(0) || p.charAt(0)=='.')){
            if (isMatch(s.substring(i + 1), p.substring(2))) {
                return true;
            }
            i++;
        }
        return false;
    }
}

Solution II: concise version

public class Solution {
    public boolean isMatch(String s, String p) {

        if(p.length() == 0)
            return s.length() == 0;

        //p's length 1 is special case    
        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 len = s.length();

            int i = -1; 
            while(i<len && (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;
        } 
    }
}