10. Regular Expression Matching

Update time

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*' where:

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

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

Constraints:

0 <= s.length <= 20
0 <= p.length <= 30
s contains only lowercase English letters.
p contains only lowercase English letters, '.', and '*'.
It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.

# @lc code=start
using LeetCode

function is_match(s::AbstractString, p::AbstractString)::Bool
    if isempty(p)
        isempty(s)
    elseif length(p) >= 2 && p[2] == '*'
        if !isempty(s) && (s[1] == p[1] || p[1] == '.')
            # case 1: * means 1 or more of previous char, then the first char of `s` and `p` must match.
            is_match(SubString(s, 2), p)
        else
            # case 2: * means zero of previous char, ignore it and match the rest
            is_match(s, SubString(p, 3))
        end
    else
        if isempty(s)
            false  ## length mismatch
        elseif (s[1] == p[1] || p[1] == '.')
            is_match(SubString(s, 2), SubString(p, 2))
        else
            false  ## first char mismatch
        end
    end
end
# @lc code=end

is_match (generic function with 1 method)

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