LeetCode--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

方法一：动态规划。
如果s[0..i]和p[0..j]匹配，则状态dp[i][j]为ture，反之为false。状态方程如下:
dp[i][j] = dp[i - 1][j - 1], if p[j - 1] != ‘*’ && (s[i - 1] == p[j - 1] || p[j - 1] == ‘.’);
dp[i][j] = dp[i][j - 2], if p[j - 1] == ‘*’ ;（字符出现0次）
dp[i][j] = dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == ‘.’), if p[j - 1] == ‘*’ （字符出现至少一次）

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];    }};

方法二：递归法。
按照下一个字符是否为’*’的两种情况进行讨论，分别递归求解，但是耗时较长，推荐使用动态规划，可以用空间换时间。

class Solution {public:    bool isMatch(string s, string p) {        return isMatch(s.c_str(),p.c_str());    }private:    bool isMatch(const char*s,const char*p)    {        if(*p=='\0') return *s=='\0';        if(*(p+1)!='*')        {            if(*p==*s||(*p=='.'&&*s!='\0'))                return isMatch(s+1,p+1);            else                 return false;        }        else        {            while(*p==*s||(*p=='.'&&*s!='\0'))            {                if(isMatch(s,p+2))                    return true;                s++;            }            return isMatch(s,p+2);        }    }};