10. Regular Expression Matching

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一道一开始看题解都看不懂的题，原来是看错题了，*代表某一个字符的0次或者n次，而不是任意字符（一开始就是看错了任意字符，所以看不懂题解）
1. P[i][j] = P[i - 1][j - 1], if p[j - 1] != ‘*’ && (s[i - 1] == p[j - 1] || p[j - 1] == ‘.’);
2. P[i][j] = P[i][j - 2], if p[j - 1] == ‘*’ and the pattern repeats for 0 times;
3. P[i][j] = P[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == ‘.’), if p[j - 1] == ‘*’ and the pattern repeats for at least 1 times.
好题，要2刷，其实是容易明白的这题，不等于*的时候超级容易懂，就是2，3要看看画图看明白才行。
Ps，注意|| 和&&这些符号的运行顺序，否则容易runwrong。

class Solution {public:    bool isMatch(string s, string p) {        int n = s.length();        int m = p.length();        vector<vector<bool>>dp(n + 1, vector<bool>(m + 1, false));        dp[0][0] = true;        for(int i = 0; i <= n; ++ i)            for(int j = 1; j <= m; ++ 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[n][m];    }};

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