leetcode 10 - Regular Expression Matching - hard

Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

• '.' Matches any single character.​​​​
• '*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

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 preceding 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 (.)".

Constraints:

• 1 <= s.length <= 20
• 1 <= p.length <= 30
• s contains only lowercase English letters.
• p contains only lowercase English letters, '.', and '*'.
• It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.

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 preceding 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 (.)".

Constraints:

• 1 <= s.length <= 20
• 1 <= p.length <= 30
• s contains only lowercase English letters.
• p contains only lowercase English letters, '.', and '*'.
• It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.

涉及两个string match的问题，一般都是动态规划。

State: dp[i][j] i，j是字符数，总共[len(s)+1][len(p)+1]。s中取前i个字符(s[i-1])和p中前j个字符是否match

初始化：p为空，s只有是空字符才是true；s为空，dp[0][0]=T, dp[0][1]=F, 往下每个, 因为可以引入*了，取决于dp[0][j-2]

转移方程：1. if s[i-1] match p[j-1], 当前值等于左上角的值. 2. 当前*，从上往下取i-2的值（abcd, abcde*），从左往右，取i-1的值(abcddd, abcdddd => abcd*)

Result: dp[len(s)][len(p)]

实现：

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;
        
        for (int j = 1; j <= p.length(); j++) {
            if (p.charAt(j-1) == '*') {
                dp[0][j] = dp[0][j-2];
            }
        }

        for (int i = 1; i <= s.length(); i++) {
            for (int j = 1; j<=p.length(); j++) {
                if (p.charAt(j-1) == '.' || p.charAt(j-1) == s.charAt(i-1)) {
                    dp[i][j] = dp[i-1][j-1];
                } else if (p.charAt(j-1) == '*') {
                    if (p.charAt(j-2) == s.charAt(i-1) || p.charAt(j-2) == '.') {
                        dp[i][j] = dp[i][j-2] || dp[i-1][j];
                    } else {
                        dp[i][j] = dp[i][j-2];
                    }
                }
            }
        }
        
        return dp[s.length()][p.length()];

    }
}