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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
function isMatch(s: string, p: string): boolean { // NOSONAR
const result = new Array(s.length + 1)
for (let i = 0; i < result.length; i++) {
result[i] = new Array(p.length + 1).fill(false)
}
result[0][0] = true
for (let j = 1; j < result[0].length; j++) {
const pChar = p[j - 1]
if (pChar === '*') {
result[0][j] = result[0][j - 2]
}
}
for (let i = 1; i < result.length; i++) {
for (let j = 1; j < result[0].length; j++) {
const sChar = s[i - 1]
const pChar = p[j - 1]
if (sChar === pChar || pChar === '.') {
result[i][j] = result[i - 1][j - 1]
} else if (pChar === '*') {
const prevCharMatch = sChar === p[j - 2] || p[j - 2] === '.' ? result[i - 1][j] : false
result[i][j] = result[i][j - 2] || prevCharMatch
}
}
}
return result[result.length - 1][result[0].length - 1]
}
export { isMatch }