Wildcard Matching leetcode java

描述
Implement wildcard paern matching with support for '?' and '*'.
'?' Matches any single character. '*' Matches any sequence of characters (including the empty
sequence).
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", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

分析
跟上一题很类似。
主要是'*' 的匹配问题。p 每遇到一个'*'，就保留住当前'*' 的坐标和 s 的坐标，然后 s 从前
往后扫描，如果不成功，则 s++，重新扫描

代码

 1 public class WildcardMatch {
 2 
 3     public static void main(String[] args) {
 4         // TODO Auto-generated method stu 
 5         String s= "aab";
 6         String p="*";
 7         System.out.println(isMatch(s,p));
 8     }
 9 //    递归
10         public static boolean isMatch(String s, String p) {
11         if (p.length() == 0)
12                 return s.length() == 0;
13             
14         if (p.charAt(0) == '*') {    
15                 
16         while (p!=null&&p.startsWith("*")) {
17             
18             p=p.substring(1); //跳过*
19         }    
20         if (p==null) 
21             return true;
22         while (s!= null && !isMatch(s, p)) 
23             s=s.substring(1);
24         return s != null;
25       }
26       
27         else if (p.charAt(0) == '\0' || s.charAt(0) == '\0') return p.charAt(0) == s.charAt(0);
28         else if (p.charAt(0) == s.charAt(0) || p.charAt(0) == '?') {
29             
30             return isMatch(s.substring(1), p.substring(1));
31         }
32         else return false;
33         }




34 //迭代版    
35     public static boolean isMatch2(String s, String p) {  
36         int i = 0;  
37         int j = 0;  
38         int star = -1;  
39         int mark = -1;  
40         while (i < s.length()) {  
41             if (j < p.length()  
42                     && (p.charAt(j) == '?' || p.charAt(j) == s.charAt(i))) {  
43                 ++i;  
44                 ++j;  
45             } else if (j < p.length() && p.charAt(j) == '*') {  
46                 star = j++;  
47                 mark = i;  
48             } else if (star != -1) {  
49                 j = star + 1;  
50                 i = ++mark;  
51             } else {  
52                 return false;  
53             }  
54         }  
55         while (j < p.length() && p.charAt(j) == '*') {  
56             ++j;  
57         }  
58         return j == p.length();  
59     }
60 }