# Leetcode Regular Expression Matching Java Solution

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 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 = "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```

Solution:

Explanation:

1, init dp && dp[j], to avoid corner cases;
2, transform function:
a, when p == ‘‘:
compare char[i-1][j-1];
if same, “char
” can count as 0, 1, multiple;
Important: if multiple, dp[i+ 1][j+ 1] = dp[i][j+1];
think backwards, not forwards.

Or simplify:
dp[i][j+1] represents 1/multiple, think previous step, represents 0 or multiple
dp[i+ 1][j+ 1] = dp[i + 1][j – 1] || dp[i][j] || dp[i][j+1]; // count 0, 1, multiple
here, counting 1 is redundant
So we can just use
dp[i+ 1][j+ 1] = dp[i + 1][j – 1] || dp[i][j+1]

``````class HackerHeap {
public boolean isMatch(String s, String p) {
int m = s.length(), n = p.length();
if (n == 0) return m == 0;
boolean[][] dp = new boolean[m + 1][n + 1];
dp = true;
for (int i = 1; i < n; i+=2) {
if (p.charAt(i) != '*') break;
else dp[i + 1] = true;
}

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
char a = s.charAt(i), b = p.charAt(j);
if (b == '.' || a == b) dp[i + 1][j + 1] = dp[i][j];
else if (b == '*') dp[i + 1][j + 1] = dp[i + 1][j - 1]  // zero
|| ((s.charAt(i) == p.charAt(j - 1) || p.charAt(j - 1) == '.') &&               dp[i][j + 1]); // 1 or more;  // consider ".*"
}
}

return dp[m][n];
}
}``````