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
func isMatch(s string, p string) bool {
m, n := len(s), len(p)
dp := make([][]bool, m+1)
for i := 0; i <= m; i++ {
dp[i] = make([]bool, n+1)
}
dp[0][0] = true
for i := 1; i <= n; i++ {
if p[i-1] == '*' {
dp[0][i] = dp[0][i-2]
}
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
switch {
case p[j-1] == '.' || p[j-1] == s[i-1]:
dp[i][j] = dp[i-1][j-1]
case p[j-1] == '*':
dp[i][j] = dp[i][j-2]
if s[i-1] == p[j-2] || p[j-2] == '.' {
dp[i][j] = dp[i][j] || dp[i-1][j]
}
}
}
}
return dp[m][n]
}