Regular Expression Matching(正则表达式匹配) leetcode10

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:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like . or *.

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 precedeng 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

 public boolean isMatch(String s, String p) {
        //return s.matches(p);
        Boolean[][] dp = new Boolean[s.length() + 1][p.length() + 1];
        return isMatch(s, 0, p, 0, dp);
    }
    private boolean isMatch(String s, int i, String p, int j, Boolean[][] dp) {
        if (j == p.length()) {
            return i == s.length();
        }
        if (dp[i][j] != null) {
            return dp[i][j];
        }
        boolean ans;
        boolean matchFirstChar = i < s.length() && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.');
        if (j + 1 < p.length() && p.charAt(j + 1) == '*') {
            boolean matchZero = isMatch(s, i, p, j + 2, dp);
            boolean matchOne = matchFirstChar && isMatch(s, i + 1, p, j, dp);
            ans = matchZero || matchOne;
        } else {
            ans = matchFirstChar && isMatch(s, i + 1, p, j + 1, dp);
        }
        dp[i][j] = ans;
        return ans;
    
    }