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

1.p為空，那s必然也為空

2.s為空，看p的第二字符是否為*，為*的話 則為isMatch(s,p.substring(2));

3.都不為空，p的第二字符為*，一種情況是直接跳過*部分isMatch(s,p.substring(2))，一種情況是首字符相等isMatch(s.substring(1),p)，一種情況是p首字符是.

4.都不為空，p的第二字符不為*，看首字符是否相等

 1 public class Solution {
 2     public boolean isMatch(String s, String p) {
 3          if ("".equals(p))
 
 4             return "".equals(s);
 5         else if ("".equals(s)){
 6             if (p.length()>1&&p.charAt(1)==‘*‘){
 7                 return isMatch(s,p.substring(2));
 8             }else
 9                 return false;
10         }
11         else if (p.length()>1&&p.charAt(1)==‘*‘){
 
12             if (isMatch(s,p.substring(2)))//skip .*
13                 return true;
14             else if(s.charAt(0)==p.charAt(0)||p.charAt(0)==‘.‘){
15                 return isMatch(s.substring(1),p);
16             }else {
17                 return  false;
18             }
19         }else {
20             if (p.charAt(0)==s.charAt(0)||p.charAt(0)==‘.‘)
21                 return isMatch(s.substring(1),p.substring(1));
22             else
23                 return false;
24         }
25     }
26 }

leetcode 10. Regular Expression Matching