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第 12 屆 iThome 鐵人賽

DAY 11
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Software Development

刷刷題 or Alan Becker's game 製作 is a question 系列 第 11

(Hard?) 10. Regular Expression Matching

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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 = "cab"
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 = "misisp*."
Output: false


思路
懂 Python 的 Regular Expression 的 function re.fullmatch()
則 兩行程式碼搞定 !

解法 python3

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        result = re.fullmatch(p,s)
        return result

Result
https://ithelp.ithome.com.tw/upload/images/20200926/20111516UOYhQ9O0Sk.png


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(Medium) 6. ZigZag Conversion
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(Easy) 242. Valid Anagram
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刷刷題 or Alan Becker's game 製作 is a question 30
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