10. Regular Expression Matching

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

 

AC code:

class Solution {
public:
    bool isMatch(string s, string p) {
        int len1 = s.length(), len2 = p.length();
        // 初始化
        vector<vector<bool> > dp(len1+1, vector<bool>(len2+1, false));
        dp[0][0] = true;
        // "a"
        // "aa*"   --->  "a"
        for (int i = 1; i <= len2; ++i) {
            if (p[i-1] == '*' && dp[0][i-2])
                dp[0][i] = true;
        }
        // DP
        for (int i = 1; i <= len1; ++i) {
            for (int j = 1; j <= len2; ++j) {
                if (s[i-1] == p[j-1] || p[j-1] == '.') {
                    dp[i][j] = dp[i-1][j-1];
                } else if (p[j-1] == '*') {
                    if (p[j-2] != '.' && p[j-2] != s[i-1])
                        dp[i][j] = dp[i][j-2];                                     
                    else {
                        dp[i][j] = (dp[i][j-2] || dp[i-1][j] || dp[i][j-1]);
                    }
                }
            }
        }
        return dp[len1][len2];
    }
}; 

　　

Runtime: 12 ms, faster than 51.36% of C++ online submissions for Regular Expression Matching.