leetcode 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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

解：主要考虑一些匹配出现的可能性，这一题主要是参考现有的一些解法，有递归，有循环遍历，发现递归的时间要长很多。

代码：递归：

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

循环：

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