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 = "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
```

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.

Idea:

Backtracking / Depth-first search (DFS)

There are two normal cases without consideration of `*`:

`1. s === p2. s !== p, but p === '.' and s is any character`

And then let’s consider how to handle the pattern `*`:

If we have a `x*` in the pattern(x stands for any character), we may ignore this part of
the pattern, or delete a matching character in the text

```// no match: aa , c*                || match: aaa , a*
(isMatch(s, p.substring(2)) || (first_match && isMatch(s.substring(1), p)));```

According to these analyses, we can construct a depth-first search algorithm, it’s a recursive algorithm.

Solution:

``````/**
* @param {string} s
* @param {string} p
* @return {boolean}
*/
var isMatch = function(s, p) {
let sLen = s.length;
let pLen = p.length;
if (pLen === 0) return sLen === 0;
let first_match = (sLen !== 0 && (p[0] === s[0] || p[0] === '.'));

// If a star is present in the pattern, it will be in the
// second position p[1]. Then, we may ignore this part of
// the pattern, or delete a matching character in the text.
// If we have a match on the remaining strings after any of
// these operations, then the initial inputs matched.
if (pLen >= 2 && p[1] === '*') {
// no match:  aa , c*       ||      match:       aaa , a*
return (isMatch(s, p.substring(2)) || (first_match && isMatch(s.substring(1), p)));
} else {
return first_match && isMatch(s.substring(1), p.substring(1));
}
};
``````