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Leetcode010-regularExpressionMatching

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Solution Report of LeetCode Acceptted

Description

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

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

Solution

Using DFS

Considering s =’’ or p=’’

Considering if there is a ‘*’

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class Solution:
    def isEndOfStar(self, p):
        while p!='':
            if len(p) ==1 or (len(p)>1 and p[1]!='*'):
                return False
            p = p[2:]
        return True
    
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        if p =='':
            return s == ''
        
        if s =='':
            return self.isEndOfStar(p)
        
        if len(p) == 1 or (len(p) > 1 and p[1]!='*'):
#       without *
            if s[0]!=p[0] and p[0] !='.':
                return False
            else:
                return self.isMatch(s[1:],p[1:])
        else:
#       with *
            if self.isMatch(s, p[2:]):
                return True
            while s[0] == p[0] or p[0] == '.':
                s = s[1:]
                
                if self.isMatch(s, p[2:]):
                    return True
                if s =='':
                    return self.isEndOfStar(p)
            return False