Regular Expression Matching

Regular Expression Matching dynamic programming

Implement regular expression matching with support for ‘.’ 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") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

Regular Expression Matching Solution

public class RegularExpressionMatching {
    public boolean isMatch(String s, String p) {
        if(s == null || p == null) {
            return false;
        }

        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
        dp[0][0] = true;

        for(int i = 0; i < p.length(); i++) {
            if(p.charAt(i) == '*' && dp[0][i - 1]) {
                dp[0][i + 1] = true;
            }
        }

        for(int i = 0; i < s.length(); i++) {
            for(int j = 0; j < p.length(); j++) {
                if(p.charAt(j) == '.') {
                    dp[i + 1][j + 1] = dp[i][j];
                }

                if(p.charAt(j) == s.charAt(i)) {
                    dp[i + 1][j + 1] = dp[i][j];
                }

                if(p.charAt(j) == '*') {
                    if(p.charAt(j - 1) != s.charAt(i) && p.charAt(j - 1) != '.') {
                        dp[i + 1][j + 1] = dp[i + 1][j - 1];
                    } else {
                        dp[i + 1][j + 1] = (dp[i + 1][j] || dp[i][j + 1] || dp[i + 1][j - 1]);
                    }
                }
            }
        }

        return dp[s.length()][p.length()];
    }
}

Last modified October 4, 2020