/*

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];

}

}