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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

实现支持 ' . '和 ' * '的正則表達式。

' . ' 匹配不论什么单字符。

' * '匹配0或多个前向元素。

使用递归进行推断。

整体上能够分成两种情况，一种是以 ' * ‘开头的，还有一种不是。

public class RegularExpressionMatching {	public static void main(String[] args) {		System.out.println(isMatch("aa","a"));		System.out.println(isMatch("aa","aa"));		System.out.println(isMatch("aaa","aa"));		System.out.println(isMatch("aa", "a*"));		System.out.println(isMatch("aa", ".*"));		System.out.println(isMatch("ab", ".*"));		System.out.println(isMatch("aab", "c*a*b"));	}	public static boolean isMatch(String s,String p){		if(p.length() == 0)			return s.length() == 0;		if(p.length() == 1 || p.charAt(1) != '*'){			if(s.length()  < 1 || (p.charAt(0) != '.' && s.charAt(0) != p.charAt(0)))				return false;			return isMatch(s.substring(1),p.substring(1));		}else{			int i = -1;			while(i < s.length() && (i < 0 || p.charAt(0) == '.' || p.charAt(0) == s.charAt(i))){				if(isMatch(s.substring(i+1),p.substring(2)))					return true;				i++;			}			return false;		}			}}

Reference：http://www.programcreek.com/2012/12/leetcode-regular-expression-matching-in-java/