[leetcode] Regular Expression Matching

'.' 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

本题主要处理*号，先把代码贴上，C++,运行时间4ms

class Solution
{
public:
    bool isMatch(string s, string p)
    {
        int sSize = s.size(),pSize = p.size();
        int storeSize = (1+sSize)*(1+pSize);
        bool *flag = new bool[storeSize];
        flag[0]=true;
        for(int i=1;i<storeSize;i++) flag[i]=false;
        for(int i=0;i<pSize;i++)
        {
            if('*'==p[i])
            {
                for(int j=0;j<=sSize;j++)
                {
                    flag[(i+1)*(sSize+1)+j] =  flag[(i-1)*(sSize+1)+j] || flag[(i)*(sSize+1)+j] || (j-1>=0 && flag[(i+1)*(sSize+1)+j-1]&&(p[i-1]==s[j-1]|| p[i-1]=='.'));
                }
            }
            else
            {
                for(int j=1;j<=sSize;j++)
                {
                    flag[(i+1)*(sSize+1)+j] = flag[i*(sSize+1)+j-1] && (p[i]==s[j-1] || p[i]=='.');
                }
            }
        }
        bool resutl = flag[storeSize-1];
        delete []flag;
        return resutl;
    }
};

使用动态规划算法，有空再写写思路