# [leetcode题解]10. Regular Expression Matching

```'.' 匹配任意单个字符。
'*' 匹配零个或多个前面的元素。
```

• s 可能为空，且只包含从 a-z 的小写字母。
• p 可能为空，且只包含从 a-z 的小写字母，以及字符 . 和 *。

```输入: s = "aa"
p = "a" 输出: false 解释: "a" 无法匹配 "aa" 整个字符串。
```

```输入: s = "aa"
p = "a*" 输出: true 解释: '*' 代表可匹配零个或多个前面的元素, 即可以匹配 'a' 。因此, 重复 'a' 一次, 字符串可变为 "aa"。
```

```输入: s = "ab"
p = ".*" 输出: true 解释: ".*" 表示可匹配零个或多个('*')任意字符('.')。
```

```输入: s = "aab"
p = "c*a*b" 输出: true 解释: 'c' 可以不被重复, 'a' 可以被重复一次。因此可以匹配字符串 "aab"。
```

```输入: s = "mississippi"
p = "mis*is*p*." 输出: false```

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 precedeng 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
```

```bool isMatch(char* s, char* p)
{
if (*p == '\0')return *s == '\0';
if (*(p + 1) == '*')
{
while (*p == *s || (*s != '\0' && *p == '.'))
{
if (isMatch(s++, p + 2))return true;
}
return isMatch(s, p + 2);
}
else
{
if (*p == *s || (*s != '\0' && *p == '.'))return isMatch(++s, ++p);
return false;
}
}```