一、題目

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 (.)”.

Constraints:

1 <= s.length <= 20
1 <= p.length <= 20
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.

二、題解

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