10. Regular Expression Matching

题目：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”, “.*”) → false
isMatch(“aab”, “c*a*b”) → true

思路： 这道题目让我们去判断正则表达式是否匹配，其中‘.’表示可以匹配任何一个字符，’ * ‘表示它前面的一个字符可以有0个、1个或者多个，比如A *可以表示成：”、’A’、’AA’等等。利用动态规划的思想，令dp[i][j]表示s前i个字符与p的前j个字符是否匹配，状态转移方程为：
　　当p[i-1] != ‘*’时，dp[i][j] = (s[i-1]==p[j-1] || p[j-1]==’.’)&&dp[i-1][j-1]；
　　当p[i-1]==’*’时，dp[i][j] = (s[i-1]==p[j-2] || p[j-2]==’.’)&&dp[i-1][j] || dp[i][j-2];

代码如下：

class Solution {
public:
    bool isMatch(string s, string p) {
        if (s.size()==0 && p.size()== 0) return true;
        int m = s.size(), n = p.size();
        vector<vector<bool> > dp(m+1, vector<bool>(n+1, false));

        //init
        dp[0][0] = true;
        for (int j = 1; j <= n; j++) dp[0][j] = (p[j-1]=='*') && dp[0][j-2];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (p[j-1] != '*') dp[i][j] = (s[i-1]==p[j-1] || p[j-1]=='.')&&dp[i-1][j-1];
                else dp[i][j] = (s[i-1]==p[j-2] || p[j-2]=='.')&&dp[i-1][j] || dp[i][j-2];
            }
        }
        return dp[m][n];
    }
};