[Leetcode] Wildcard Matching

Implement wildcard pattern 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

Solution 1:
贪心算法:
http://blog.unieagle.net/2012/11/07/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Awildcard-matching/

只需要依据连续的’*’,将p分割成一些不带’*’的子串。然后在s中依次匹配就好了,只不过需要特殊照顾一下首尾两个子串:
1.对于开头不是’*’的p,第一个子串必须从s[0]开始匹配
2.对于结尾不是’*’的p,最后一个子串必须在s的尾巴上匹配

 1 package POJ;
 2 
 3 import java.util.ArrayList;
 4 import java.util.HashMap;
 5 import java.util.List;
 6 
 7 public class Main {
 8 
 9     public static void main(String[] args) {
10         Main so = new Main();
11         String s="AABCDEFGHIJKLMNOPQ";
12         String p="AAB*Q**Q";
13         System.out.println(so.isMatch(s, p));
14     }
15 
16     public boolean isMatch(String s, String p) {
17         if(s == null || p == null)
18             return false;
19         int front = p.indexOf("*");  //得到第一个*的位置,也即得到第一个*前边儿的字符个数
20         int back = p.length()-1-p.lastIndexOf("*");   //得到最后一个*后边儿还有多少个位数
21         if(front == -1){
22             //p中没有*
23             return (s.length()==p.length())&&(iMatch(s,p));
24         }
25         //p中有*
26         //首先,确定首尾是能满足条件的
27         if(!( (front+back<=s.length())&&(iMatch(s.substring(0, front),p.substring(0, front)))&&(iMatch(s.substring(s.length()-back),p.substring(p.length()-back)))))
28             return false;
29         
30         int i1=0;
31         int i2=0;
32         
33         //现在来确定首尾的两个*中间的部分,还是以*来作为分割,一段一段地看
34         while(true){
35             while((i2<p.length())&&(p.charAt(i2)=='*'))
36                 ++i2;
37             if(i2==p.length())
38                 break;
39             int st=i2;
40             while((i2<p.length())&&(p.charAt(i2)!='*'))
41                 ++i2;
42             String piece=p.substring(st,i2);  //找到被*分割的片段
43             while(((i1+piece.length())<=s.length())&&!iMatch(s.substring(i1, i1+piece.length()),piece))
44                 i1++;
45             if(i1+piece.length()>s.length())
46                 return false;
47             i1=i1+piece.length();           
48         }
49         return true;
50     }
51 
52     private boolean iMatch(String s, String p) {
53         // TODO Auto-generated method stub
54         for(int i=0;i<s.length();++i){
55             if(!((s.charAt(i)==p.charAt(i))||(p.charAt(i)=='?')))
56                 return false;
57         }
58         return true;
59     }
60 }

 



posted @ 2014-10-02 13:49  Phoebe815  阅读(187)  评论(0编辑  收藏  举报