DAY 3
0
Software Development

## 題目

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
``````

## 解法

``````class Solution {
public boolean isMatch(String text, String pattern) {
if (pattern.isEmpty()) return text.isEmpty();
boolean first_match = (!text.isEmpty() &&
(pattern.charAt(0) == text.charAt(0) || pattern.charAt(0) == '.'));

if (pattern.length() >= 2 && pattern.charAt(1) == '*'){
return (isMatch(text, pattern.substring(2)) ||
(first_match && isMatch(text.substring(1), pattern)));
} else {
return first_match && isMatch(text.substring(1), pattern.substring(1));
}
}
}
``````

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