LetCode 10. 正则表达式匹配

static auto x = [ ] ( ) {
    std :: ios :: sync_with_stdio ( false );
    cin.tie ( NULL );
    return 0;
}();


class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.length(), n = p.length();
        return backtracking(s, m, p, n);
    }

    bool backtracking(string& s, int i, string& p, int j) {
        if (i == 0 && j == 0) return true;
        if (i != 0 && j == 0) return false;
        if (i == 0 && j != 0) {
            //in this case only p == "c*c*c*" this pattern can match null string
            if (p[j-1] == '*') {
                return backtracking(s, i, p, j-2);
            }
            return false;
        }
        //now both i and j are not null
        if (s[i-1] == p[j-1] || p[j-1] == '.') {
            return backtracking(s, i - 1, p, j - 1);
        } else if (p[j-1] == '*') {
            //two cases: determines on whether p[j-2] == s[i-1]
            //first p[j-2]* matches zero characters of p
            if (backtracking(s, i, p, j - 2)) return true;
            //second consider whether p[j-2] == s[i-1], if true, then s[i-1] is matched, move to backtracking(i - 1, j)
            if (p[j-2] == s[i-1] || p[j-2] == '.') {
                return backtracking(s, i - 1, p, j);
            }
            return false;
        }
        return false;
    }
};