010 Regular Expression Matching 正则表达式匹配

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).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

详见：https://leetcode.com/problems/regular-expression-matching/description/

方法一：

class Solution {
public:
    bool isMatch(string s, string p) {
        if(p.empty())
        {
            return s.empty();
        }
        if(p.size()==1)
        {
            return (s.size()==1&&(s[0]==p[0]||p[0]=='.'));
        }
        if(p[1]!='*')
        {
            if(s.empty())
            {
                return false;
            }
            return (s[0]==p[0]||p[0]=='.')&&isMatch(s.substr(1),p.substr(1));
        }
        while(!s.empty()&&(s[0]==p[0]||p[0]=='.'))
        {
            if(isMatch(s,p.substr(2)))
            {
                return true;
            }
            s=s.substr(1);
        }
        return isMatch(s,p.substr(2));
    }
};

 方法二：动态规划，dp[i][j] 表示 s[0..i] 和 p[0..j] 是否 match

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size(), n = p.size();
        vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));

        dp[0][0] = true;
        for (int i = 1; i <= m; i++)
        {  
            dp[i][0] = false;
        }
        for (int j = 1; j <= n; j++)
        {
            dp[0][j] = j > 1 && '*' == p[j - 1] && dp[0][j - 2];
        }

        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (p[j - 1] != '*')
                {
                    dp[i][j] = dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);
                }
                else
                {
                    dp[i][j] = dp[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && dp[i - 1][j];
                }
            }
        }

        return dp[m][n];
    }
};

参考：http://www.cnblogs.com/grandyang/p/4461713.html