Regular Expression Matching

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:

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 preceding 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) {
        // dp[i][j] = p[0..i) matches s[0..j)
        // dp[i][j] = if p[i - 1] == s[j - 1] || p[j - 1] == '.'
        //               dp[i - 1][j - 1]
        //            if p[j - 1] == '*'
        //               dp[i - 2][j] || dp[i][j - 1] && s[i - 1] == p[i - 2] || p[i - 2] == '.'
        //          * matches 0 elements   * matches s[i - 1]                
        // dp[0][0] = true
        // dp[i][0] = true if dp[i - 2][0] and p[i - 1] == '*'
        // dp[m][n]
        int m = p.length();
        int n = s.length();
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;
        for (int i = 2; i <= m; i ++) {
            if (p.charAt(i - 1) == '*' && dp[i - 2][0]) {
                dp[i][0] = true;
            }
        }
        for (int i = 1; i <= m; i ++) {
            char r = p.charAt(i - 1);
            for (int j = 1; j <= n; j ++) {
                char c = s.charAt(j - 1);
                if (c == r || r == '.') {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (r == '*') {
                    dp[i][j] = dp[i - 2][j] || dp[i][j - 1] && (c == p.charAt(i - 2) || p.charAt(i - 2) == '.');
                }
            }
        }
        return dp[m][n];
    }
}