10. Regular Expression Matching

Given an input string (s) and a pattern (p), 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).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

方法一：首先考虑递归解法。

如果p为空，那么表示已经匹配完成，如果此时s还未匹配完则一定无法匹配，反之则匹配成功。

如果p不为空，判断p的第二位是否为‘*’，如果是，分两种情况，第一种是忽略‘*’，s不变，p往后跳2格，递归调用函数。第二种是匹配一格，s前进一格，p不变。

如果p的第二位不为‘*’，则可以直接判断当前位是否相同，然后s、p各进一格递归调用。

注意每次都要判断s是否为空，若为空则返回false。

class Solution {
public:
    bool isMatch(string s, string p) {
        if(p.empty())
            return s.empty();
        if(p[1]=='*')
            return isMatch(s,p.substr(2)) || ( !s.empty() && (s[0]==p[0] || p[0] == '.') && isMatch(s.substr(1),p) ); 
        else
            return !s.empty() && (s[0]==p[0] || p[0] == '.') && isMatch(s.substr(1),p.substr(1));
    } 
};

方法二：动态规划。

其实和方法一思想基本相同，只是把递归转换为了dp数组。

class Solution {
public:
    bool isMatch(string s, string p) {
        if(p.empty())
            return s.empty();
        int len1=s.size(),len2=p.size();
        vector<vector<bool>> dp(len1+1,vector<bool>(len2+1,false));
        dp[0][0]=true;
        for(int i=0;i<=len1;i++)
        {
            for(int j=1;j<=len2;j++)
            {
                if(p[j-1]=='*')
                    dp[i][j] = dp[i][j-2] || ( i>0 && dp[i-1][j] && (s[i-1]==p[j-2] || p[j-2]=='.') );
                else
                    dp[i][j] = i>0 && dp[i-1][j-1] && (s[i-1]==p[j-1] || p[j-1]=='.');
            }
        }
        return dp[len1][len2];
    } 
};