10. Regular Expression Matching

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
解题思路：此题的难点在于 x*可以延伸为xx*，即s中xxxyzyz与p中x*yzyz匹配的时候，每次都能把s中的x去掉一个，于是可以转化成了xxyzyz与p中x*yzyz匹配。
dfs解法：

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

 dp解法：膜 https://discuss.leetcode.com/topic/6183/my-concise-recursive-and-dp-solutions-with-full-explanation-in-c/2

需要注意的是：初始化的时候,当p[j-1]='*'的时候，f[0][j]=f[0][j-2] 比如s="",p="a*b*",f[0][2]是可以为true的。

class Solution {
public:
    bool isMatch(string s, string p) {
        /**
         * f[i][j]: if s[0..i-1] matches p[0..j-1]
         * if p[j - 1] != '*'
         *      f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1]
         * if p[j - 1] == '*', denote p[j - 2] with x
         *      f[i][j] is true iff any of the following is true
         *      1) "x*" repeats 0 time and matches empty: f[i][j - 2]
         *      2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j]
         * '.' matches any single character
         */
        int n=s.length(),m=p.length();
        vector<vector<bool>> f(n + 1, vector<bool>(m + 1, false));
        f[0][0]=true;
        for(int i=1;i<=n;i++){
            f[i][0]=false;
        } 
        for(int j=1;j<=m;j++)
            f[0][j]=j>1&&p[j-1]=='*'&&f[0][j-2];
        
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++)
            if(p[j-1]!='*'){
                f[i][j]=f[i-1][j-1]&&(s[i-1]==p[j-1]||p[j-1]=='.');
            }
            else {
                f[i][j]=f[i][j-2]||(s[i-1]==p[j-2]||p[j-2]=='.')&&f[i-1][j];
            }
        }
        return f[n][m];
    }
};