0010. Regular Expression Matching.md

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;
        // This for loop deals with pattern like 'a*b*'
        for (int i = 1; i < dp[0].length; i++) {
            if (p.charAt(i - 1) == '*') {
                dp[0][i] = dp[0][i - 2];
            }
        }
        for (int i = 1; i < dp.length; i++) {
            for (int j = 1; j < dp[0].length; j++) {
                if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.') {  
                    dp[i][j] = dp[i - 1][j - 1];    // match, so whether regex match depends on status without these two 
                }
                else if (p.charAt(j - 1) == '*') {  // when meeting '*', dp thinking kicks in 
                    // dp[i][j - 2] checks if 0 occurence of last character in pattern could result in regex match.
                    // p.charAt(j - 2) == s.charAt(i - 1) checks if character before '*' in p matches current character in s
                    // or if character before '*' is a '.'. If either of them is true, we can eliminate character at i
                    // position and depends on status at i - 1 position.
                    dp[i][j] = dp[i][j - 2];
                    if (p.charAt(j - 2) == s.charAt(i - 1) || p.charAt(j - 2) == '.') {
                        dp[i][j] = dp[i][j] || dp[i - 1][j];
                    }
                }
                else {
                    dp[i][j] = false;
                }
            }
        }
        return dp[dp.length - 1][dp[0].length - 1];
    }
}

note

• time O(m * n) space O(m * n)
• The idea is to use dynammic programming to solve this problem. Notice that '.*' can represent 0 to many of any characters, so 'abxsdady' would match with 'a.*y'.
• There are couple of cases that we need to consider for this problem.
  1). Match.
  2). Meeting *
  i). check whether 0 occurence of character before * will result in regex matching
  ii). check whether character before * combined with * could result in regex matching. To achieve this, we check if character before * match with current character in s or if character before '*' is a '.'.
  3). If not conditions above, set the current matching to false.