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 <= 200 <= p.length <= 30s contains only lowercase English letters.p contains only lowercase English letters, '.', and '*'.'*', there will be a previous valid character to match.struct Solution;
impl Solution {
fn is_match(s: String, p: String) -> bool {
let n = s.len();
let m = p.len();
let s: Vec<char> = s.chars().collect();
let p: Vec<char> = p.chars().collect();
let mut memo: Vec<Vec<Option<bool>>> = vec![vec![None; m + 1]; n + 1];
Self::is_match_dp(n, m, &mut memo, &s, &p)
}
fn is_match_dp(
n: usize,
m: usize,
memo: &mut Vec<Vec<Option<bool>>>,
s: &[char],
p: &[char],
) -> bool {
if let Some(ans) = memo[n][m] {
ans
} else {
let res = if n == 0 && m == 0 {
true
} else if n != 0 && m == 0 {
false
} else if n == 0 && m != 0 {
if p[m - 1] == '*' {
Self::is_match_dp(n, m - 2, memo, s, p)
} else {
false
}
} else {
if s[n - 1] == p[m - 1] {
Self::is_match_dp(n - 1, m - 1, memo, s, p)
} else {
match p[m - 1] {
'*' => match p[m - 2] {
'*' => false,
'.' => {
Self::is_match_dp(n - 1, m, memo, s, p)
|| Self::is_match_dp(n, m - 2, memo, s, p)
}
_ => {
if s[n - 1] != p[m - 2] {
Self::is_match_dp(n, m - 2, memo, s, p)
} else {
Self::is_match_dp(n - 1, m, memo, s, p)
|| Self::is_match_dp(n, m - 2, memo, s, p)
}
}
},
'.' => Self::is_match_dp(n - 1, m - 1, memo, s, p),
_ => false,
}
}
};
memo[n][m] = Some(res);
res
}
}
}
#[test]
fn test() {
let s = "aa".to_string();
let p = "a".to_string();
let res = false;
assert_eq!(Solution::is_match(s, p), res);
let s = "aa".to_string();
let p = "a*".to_string();
let res = true;
assert_eq!(Solution::is_match(s, p), res);
let s = "ab".to_string();
let p = ".*".to_string();
let res = true;
assert_eq!(Solution::is_match(s, p), res);
let s = "aab".to_string();
let p = "c*a*b".to_string();
let res = true;
assert_eq!(Solution::is_match(s, p), res);
let s = "mississippi".to_string();
let p = "mis*is*p*.".to_string();
let res = false;
assert_eq!(Solution::is_match(s, p), res);
}