LeetCode in Kotlin
10. Regular Expression Matching
Hard
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:
1 <= s.length <= 201 <= p.length <= 30scontains only lowercase English letters.pcontains only lowercase English letters,'.', and'*'.- It is guaranteed for each appearance of the character
'*', there will be a previous valid character to match.
Solution
class Solution {
fun isMatch(s: String, p: String): Boolean {
val n = s.length
val m = p.length
return solve(n - 1, m - 1, s, p)
}
private fun solve(i: Int, j: Int, s: String, p: String): Boolean {
if (j < 0) {
return i < 0
}
if (i < 0) {
return p[j] == '*' && solve(i, j - 2, s, p)
}
// simple char matching
// if s char matchs with p char or it can be '.'
if (s[i] == p[j] || p[j] == '.') {
return solve(i - 1, j - 1, s, p)
}
return if (p[j] == '*') {
// if s char matches with p char or it can be '.'
if (s[i] == p[j - 1] || p[j - 1] == '.') {
solve(i - 1, j, s, p) || solve(i, j - 2, s, p)
} else {
solve(
i,
j - 2,
s,
p
)
}
} else {
false
}
}
}