LeetCode-in-All
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
impl Solution {
pub fn is_match(s: String, p: String) -> bool {
is_match_dp(s.as_bytes(), p.as_bytes())
}
}
pub fn is_match_dp(s: &[u8], p: &[u8]) -> bool {
let mut grid = vec![vec![false; s.len() + 1]; p.len() + 1];
grid[p.len()][s.len()] = true;
let mut pi = p.len();
while pi > 0 {
pi -= 1;
let mstar = pi > 0 && p[pi] == b'*';
if mstar {
pi -= 1;
}
for si in 0..=s.len() {
let is_match = if mstar {
let mut si2 = si;
loop {
if grid[pi + 2][si2] {
break true;
}
if si2 < s.len() && (p[pi] == b'.' || s[si2] == p[pi]) {
si2 += 1;
continue;
}
break false;
}
} else {
si < s.len() && (p[pi] == b'.' || s[si] == p[pi]) && grid[pi + 1][si + 1]
};
grid[pi][si] = is_match;
}
}
grid[0][0]
}