LeetCode | 10. Regular Expression Matching

题目：

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

 

代码：

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