/*
 Author:     King, wangjingui@outlook.com
 Date:       Oct 26, 2014
 Problem:    Regular Expression Matching
 Difficulty: Hard
 Source:     https://oj.leetcode.com/problems/regular-expression-matching/
 Notes:
 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

 Solution: 1. Recursion.
           2. DP.
*/

public class Solution {
    public boolean isMatch_1(String s, String p) {
        if (p.length() == 0) return s.length() == 0;
        if (p.length() == 1) {
            if (s.length() != 1) return false;
            return (s.charAt(0) == p.charAt(0)) || (p.charAt(0) == '.');
        }
        if (s.length() != 0 && (p.charAt(0) == s.charAt(0) || (p.charAt(0) == '.'))) {
            if (p.charAt(1) == '*')
                return isMatch(s.substring(1),p) || isMatch(s, p.substring(2));
            return isMatch(s.substring(1), p.substring(1));
        }
        return p.charAt(1) == '*' && isMatch(s, p.substring(2));
    }
    public boolean isMatch_2(String s, String p) {
        if (p.length() == 0) return s.length() == 0;
        int sLen = s.length(), pLen = p.length();
        boolean[][] dp = new boolean[sLen + 1][pLen + 1];
        dp[0][0] = true;
        for (int i = 2; i <= pLen; ++i) {
            dp[0][i] = dp[0][i-2] && p.charAt(i-1) == '*';
        }
        for (int i = 1; i <= sLen; ++i) {
            for (int j = 1; j <= pLen; ++j) {
                char ch1 = s.charAt(i-1), ch2 = p.charAt(j-1);
                if (ch2 != '*') dp[i][j] = dp[i-1][j-1] && (ch1 == ch2 || ch2 == '.');
                else {
                    dp[i][j] = dp[i][j-2];
                    if (ch1 == p.charAt(j-2) || p.charAt(j-2) == '.')
                        dp[i][j] = dp[i][j] | dp[i-1][j];
                }
            }
        }
        return dp[sLen][pLen];
    }
}