10. Regular Expression Matching

Problem:

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 stringf(i, 0) = false // pattern must cover the entire input stringf(i, j) =if p[j-1] == '.'f(i-1, j-1)else if p[j-1] == '*'f(i, j-2) || // matches 0 timef(i-1, j) && (s[i-1] == p[j-2] || p[j-2] == '.') // matches 1 or multiple timeselsef(i-1, j-1) && s[i-1] == p[j-1]
///**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */let isMatch = function (s, p) {if (p[0] === '*') {return false;}const dp = [[true]];for (let j = 2; j <= p.length; j++) {
        dp[0][j] = p[j - 1] === '*' && dp[0][j - 2];}for (let i = 1; i <= s.length; i++) {
        dp[i] = [];for (let j = 1; j <= p.length; j++) {switch (p[j - 1]) {case '.':
                    dp[i][j] = dp[i - 1][j - 1];break;case '*':
                    dp[i][j] = dp[i][j - 2] || (dp[i - 1][j] && (p[j - 2] === '.' || s[i - 1] === p[j - 2]));break;default:
                    dp[i][j] = dp[i - 1][j - 1] && s[i - 1] === p[j - 1];}}}return !!dp[s.length][p.length];};

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