10. Regular Expression Matching

Given an input string (`s`) and a pattern (`p`), implement regular expression matching with support for `'.'` and `'*'`.

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'.' 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:**
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Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
**Example 2:**
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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:**
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Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
**Example 4:**
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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:**
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Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

这道题有两种方法,一种是动态法,另外一种是递归方法,递归速度比较慢,代码如下:

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class Solution {
public:
bool isMatch(string s, string p) {
if (p.empty()) return s.empty();
if (p.size() == 1) {
return (s.size() == 1 && (s[0] == p[0] || p[0] == '.'));
}
if (p[1] != '*') {
if (s.empty()) return false;
return (s[0] == p[0] || p[0] == '.') && isMatch(s.substr(1), p.substr(1));
}
while (!s.empty() && (s[0] == p[0] || p[0] == '.')) {
if (isMatch(s, p.substr(2))) return true;
s = s.substr(1);
}
return isMatch(s, p.substr(2));
}
};

4ms, Ranking 27%