10. 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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

这个题目参考DISCUSS中的解法，动态规划，主要难在各个状态的分析

小tips:对于字符串匹配的解法都是类似的，涉及字符串 动态规划数组的大小一般都是size+1 因为0表示空字符串

本题中i从0开始，j从1开始，因为题意是p中存在可以匹配s的，所以当s为0(空字符串)时，p存在1个(*)即可匹配

9-lines 16ms C++ DP Solutions with Explanations

This problem has a typical solution using Dynamic Programming. We define the state P[i][j] to be true if s[0..i) matches p[0..j) and false otherwise. Then the state equations are:

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.

Putting these together, we will have the following code.

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];    }};