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

╰+哭是因爲堅強的太久メ 2022-05-24 01:44 207阅读 0赞

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;
        
        }