10. Regular Expression Matching

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题意：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

思路：这题有几个地方需要注意的，首先是声明DP数组时，可以多开一行一列的空间，但是要注意，第0行相当于是用空的字符串作为s，这个需要处理，第0列的话直接赋全0就好，另外的结果大致分为三类，根据Solutions的解释：

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.

注意其中p[j-1]代表的是当前项，所以对于1来说就是s和p的对应项相等，则状态由前一状态决定，对于p[j-1]==’*’时，分为两类，一类是s=ac, p=a *c，另一类是s=aab, p=a *b或 . *b即可，下面的代码也是Solutions里的。

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];    }};

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