10. Regular Expression Matching

Given an input string (s) and a pattern §, 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

Solution

C++

class Solution { public: bool isMatch(string s, string p) { int m = s.size(), n = p.size(); if(n && p[0]=='*') return false; vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false)); dp[0][0] = true; for (int i = 0; i <= m; i++) { for (int j = 1; j <= n; j++) { if (p[j - 1] == '*') { dp[i][j] = dp[i][j - 2] || (i && dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.')); } else { dp[i][j] = i && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.'); } } } return dp[m][n]; } };

Explanation dp: