CODEWITHSUNDEEP
pythonpython-3.x
Weining Peng
Weining Peng

Reputation: 13

Regular expression matching (if not ... return not...)

Here's the solution to leetcode problem: Regular expression matching.

I'm looking at this recursive approach but don't quite understand the logic of the first case. "if not pattern: return not text" what does that mean?

def isMatch(self, text, pattern):
    if not pattern:
        return not text

    first_match = bool(text) and pattern[0] in {text[0], '.'}

    if len(pattern) >= 2 and pattern[1] == '*':
        return (self.isMatch(text, pattern[2:]) or
                first_match and self.isMatch(text[1:], pattern))
    else:
        return first_match and self.isMatch(text[1:], pattern[1:])

I only want someone explain the first logic. Thanks.

Upvotes: 0

Views: 196

Answers (2)

Aaron_ab
Aaron_ab

Reputation: 3758

when you put something into "if" statement, is should be evaluated to boolean.

if not pattern

means pattern should be evaluated to false for this "not pattern" to be "true". Then, return "not text", which again, evaluated to boolean

Some examples:

str = "bla"
>> not str
>> false 

str = ""
>> not str
>> true

The pattern "if":

pattern = "some_pattern"
if not pattern:
    print("no pattern")  # nothing be printed

pattern = ""
if not pattern:
    print("no pattern")
>> no pattern

Upvotes: 0

L3viathan
L3viathan

Reputation: 27283

not pattern evaluates to True if the pattern is empty, False otherwise.

not text evaluates to True if the text is empty, False otherwise.

if not pattern: return not text therefore means: If the pattern is empty, return True if the text is empty, too. Otherwise return False.

A confusion matrix:

|Pattern \ Text  |  empty   | non-empty |
|---------------------------------------|
|empty           |  True    |  False    |
|---------------------------------------|
|non-empty       | (rest of the code)   |
|---------------------------------------|

Upvotes: 1

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