[LeetCode] 10. Regular Expression Matching

Problem

Given an input string (s) and a pattern (p), 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).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

Solution

class Solution {
    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 = 1; 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++) {
                // '.' or same char
                if (p.charAt(j) == '.' || p.charAt(j) == s.charAt(i)) {
                    dp[i+1][j+1] = dp[i][j];
                } else if (j != 0 && 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 {
                        //multiple s.charAt(i) or none s.charAt(i)
                        dp[i+1][j+1] = dp[i][j+1] || dp[i+1][j-1]; 
                    }
                }
            }
        }
        return dp[s.length()][p.length()];
    }
}