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## Problem

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.

## Code

x a * b . c
T F F F F F F
x F T F T F F F
a F F T T F F F
a F F F T F F F
b F F F F T F F
y F F F F F T F
c F F F F F F T
class Solution {
public boolean isMatch(String s, String p) {
int m = s.length();
int n = p.length();

boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;

for(int i = 1; i <= n; i++){
if(p.charAt(i - 1) =='*' && dp[0][i - 2] == true){
dp[0][i] = true;
}
}

for(int i = 1; i <= m; i++){
for(int j = 1; j <= n; j++){
if(s.charAt(i - 1) == p.charAt(j - 1)){
dp[i][j] = dp[i - 1][j - 1];
}

if (p.charAt(j - 1) =='.'){
dp[i][j] = dp[i - 1][j - 1];
}

if(p.charAt(j - 1) =='*'){
// p的前一个字符不能变成s的当前字符
// -> 只能删除p的前一个字符和p的当前字符, 看是不是可以匹配
if(p.charAt(j - 2) != s.charAt(i - 1) && p.charAt(j - 2) != '.'){
dp[i][j] = dp[i][j - 2];
} else {
// p的前一个字符可以等于s的当前字符
dp[i][j] = dp[i - 1][j] || dp[i][j - 1] || dp[i][j - 2];
}
}
}
}

return dp[m][n];
}
}