LeetCodeOJ——10. Regular Expression Matching

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”) → false
isMatch(“aa”,”aa”) → true
isMatch(“aaa”,”aa”) → false
isMatch(“aa”, “a*”) → true
isMatch(“aa”, “.*”) → true
isMatch(“ab”, “.*”) → true
isMatch(“aab”, “c*a*b”) → true

解题思路：
首先注意题目给出的例子isMatch(“aab”, “c*a*b”) → true，之所以aab”和”c*a*b”匹配，是因为c*代表0个或多个c而不是1个或多个c。其次，一开始我用基本的线性解法来做，想了半天没有结果，其实当碰到子问题和原问题类似的情况时，应该要想到递归，最后在网上搜到了答案http://blog.youkuaiyun.com/hopeztm/article/details/7992253。自己也按照提示也了一遍：

class Solution {
public:
    bool isMatch(string s, string p) {
        int i=0,j=0;
        return recurMatch(s,p,i,j);
    }

    bool recurMatch(string &s,string &p,int i,int j){
        if(j==p.length())
            return i==s.length();
        if(j+1<p.length()&&p[j+1]=='*'){
            while(i<s.length()&&(p[j]==s[i]||p[j]=='.')){
                if(recurMatch(s,p,i,j+2))
                    return true;
                i++;
            }
            return recurMatch(s,p,i,j+2);
        }else{
            if(s[i]==p[j]||(p[j]=='.'&&i<s.length())){
                if(recurMatch(s,p,i+1,j+1))
                    return true;
            }
            return false;
        }
    }
};