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No10.regular_expression_matching.rb
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=begin
Difficulty:
Hard
Desc:
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:
Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
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".
Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
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".
Input:
s = "mississippi"
p = "mis*is*p*."
Output: false
=end
def match(s, p, i, j)
if i >= s.size && j >= p.size then
return true
end
if i < s.size && j >= p.size then
return false
end
if i >= s.size then
return is_empty(p, j)
end
s1 = s[i]
s2 = p[j]
if s1 == s2 || s2 == '.' then
if j < p.size - 1 && p[j + 1] == '*' then
result = match(s, p, i, j + 2)
return result unless result == false
i += 1
while i < s.size && (s2 == '.' || (s2 != '.' && s[i] == s2)) do
result = match(s, p, i, j + 2)
return result unless result == false
i += 1
end
return match(s, p, i, j + 2)
else
return match(s, p, i + 1, j + 1)
end
else
return false unless j < p.size - 1 && p[j + 1] == '*'
return match(s, p, i, j + 2)
end
end
def is_empty(p, j)
count = p.size - j
return false unless count % 2 == 0
while j < p.size do
return false unless p[j + 1] == '*'
j += 2
end
return true
end
# @param {String} s
# @param {String} p
# @return {Boolean}
def is_match(s, p)
return match(s, p, 0, 0)
end