题目

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 (.)”.
Example 4:

Input: s = “aab”, p = “cab”
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 = “misisp*.”
Output: false

Constraints:

0 <= s.length <= 20
0 <= 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.

解法：DP

class Solution {
public:
    bool isMatch(string s, string p) {
        int n = s.size(), m = p.size();
        s = ' ' + s, p = ' ' + p;
        vector<vector<bool>> f(n + 1, vector<bool>(m + 1));
        f[0][0] = true;
        for (int i = 0; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (j + 1 <= m && p[j + 1] == '*') continue;
                if (p[j] == '*') 
                    f[i][j] = f[i][j - 2] || i && f[i - 1][j] && (s[i] == p[j - 1] || p[j - 1] == '.');
                else
                    f[i][j] = i && f[i - 1][j - 1] && (s[i] == p[j] || p[j] == '.');
            }
        }
        return f[n][m];
    }
};