10. Regular Expression Matching

/*
 * 10. Regular Expression Matching  
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Total Accepted: 86843
Total Submissions: 387233
Difficulty: Hard
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
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 */
package dp;

public class RegularExpressMatch {

 public boolean isMatch(String s, String p) {
  // 参考视频 https://www.youtube.com/watch?v=l3hda49XcDE
  int M = s.length(), N = p.length();
  boolean[][] T = new boolean[M + 1][N + 1];

  T[0][0] = true;
  char[] pattern = p.toCharArray();
  char[] str = s.toCharArray();

  // deal with patterns like a*, a*b, a*b*c
  for (int j = 1; j < T[0].length; j++) {
   if (pattern[j - 1] == '*') {
    T[0][j] = T[0][j - 2];
   }
  }

  for (int i = 1; i < T.length; i++) {
   for (int j = 1; j < T[0].length; j++) {
    if (pattern[j - 1] == '.' || str[i - 1] == pattern[j - 1]) {
     T[i][j] = T[i - 1][j - 1];
    } else if (pattern[j - 1] == '*') {
     T[i][j] = T[i][j - 2];
     if (pattern[j - 2] == '.' || pattern[j - 2] == str[i - 1]) {
      T[i][j] = T[i][j] | T[i - 1][j];
     }
    } else {
     T[i][j] = false;
    }
   }
  }
  return T[M][N];
 }

 public static void main(String[] args) {
  RegularExpressMatch rem = new RegularExpressMatch();
  String str = "aab", pattern = "c*a*b";
  System.out.println(rem.isMatch(str, pattern)); // true

 }

}