LeetCode-10-Regular Expression Matching

算法描述：

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

解題思路：用動態規劃求解。

1. dp[i][j] = dp[i-1][j-1] if(s[i-1]==p[j-1] || p[j-1]=='.') 字符相同或者字符爲'.' 

2. dp[i][j] = dp[i][j-2] if(p[j-1] == '*' )字符重複0次或者屢次。

3. dp[i][j] = dp[i-1][j] if(p[j-1] == '*' && s[i-1]==p[j-2] || p[j-2]=='.')字符重複至少1次。

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