Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.
'.' Matches any single character.
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
Note:
s could be empty and contains only lowercase letters a-z. p could be empty and contains only lowercase letters a-z, and characters like . or *.
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 precedeng 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
Solution:
ONE
Cheating with real RegExp matching.
/**
* @param {string} s
* @param {string} p
* @return {boolean}
*/
let isMatch = function (s, p) {
if (p[0] === "*") {
return false;
}
return new RegExp(`^${p}$`).test(s);
};
TWO
Let f(i, j) be the matching result of s[0...i) and p[0...j).
f(0, j) =
j == 0 || // empty
p[j-1] == '*' && f(i, j-2) // matches 0 time, which matches empty string
f(i, 0) = false // pattern must cover the entire input string
f(i, j) =
if p[j-1] == '.'
f(i-1, j-1)
else if p[j-1] == '*'
f(i, j-2) || // matches 0 time
f(i-1, j) && (s[i-1] == p[j-2] || p[j-2] == '.') // matches 1 or multiple times
else
f(i-1, j-1) && s[i-1] == p[j-1]
