Regular Expression Matching

描述
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
分析
这是一道很有挑战的题

''匹配任何单个字符。
“*”匹配前一个元素的零个或多个

代码

public class RegularExpressionMatching {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        System.out.println(isMatch("aa","a*"));
    }
    public static boolean isMatch(String s, String p) {
        if (p.length() == 0)
            return s.length() == 0;

        // length == 1 is the case that is easy to forget.
        // as p is subtracted 2 each time, so if original
        // p is odd, then finally it will face the length 1
        if (p.length() == 1)
            return (s.length() == 1) && (p.charAt(0) == s.charAt(0) || p.charAt(0) == '.');

        // next char is not '*': must match current character
        if (p.charAt(1) != '*') {
            if (s.length() == 0)
                return false;
            else
                return (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.') && isMatch(s.substring(1), p.substring(1));
        } else {
            // next char is *
            while (s.length() > 0 && (p.charAt(0) == s.charAt(0) || p.charAt(0) == '.')) {
                if (isMatch(s, p.substring(2)))
                    return true;
                s = s.substring(1);
            }
            return isMatch(s, p.substring(2));
        }
    }

}