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

思路：对于字符串匹配，s[i]和p[j].此题的关键就在于p[j+1]是否为'*'.递归思路解题。

如果p[j+1]!='*',s[i]==p[j]，则匹配下一位(i+1,j+1)，如果s[i]!=p[j]，匹配失败

如果p[j+1]=='*',s[i]==p[j]，则匹配(i,j+2)或者(i+1,j+2)，如果s[i]!=p[j]，则匹配(i,j+2)

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

 使用动态规划解题，思路和上述很相似，dp[j][i]表示s[0..j]和p[0...j]是否匹配；

class Solution {
public:
    bool isMatch(const char *s, const char *p) {
        int slen=strlen(s);
    int plen=strlen(p);
    vector<vector<bool> > dp(slen+1,vector<bool>(plen+1,false));
    dp[0][0]=true;
    for(int i=1;i<=plen;++i)
    {
        if(p[i]!='*')
        {
            for(int j=1;j<=slen;++j)
            {
                dp[j][i]=(dp[j-1][i-1] && (s[j-1]==p[i-1]||p[i-1]=='.'));
            }
        }
        else
        {
            dp[0][i]=dp[0][i+1]=dp[0][i-1];
            for(int j=1;j<=slen;++j)
            {
                dp[j][i]=dp[j][i+1]=(dp[j-1][i]&&(s[j-1]==p[i-1] || p[i-1]=='.')||dp[j][i-1]);
            }
            ++i;
        }
    }
    return dp[slen][plen];
    }
};