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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

题意：

分类：动态规划，回溯，贪心，字符串

解法1：

[java] view plain copy

1. public class Solution {  
2.     public boolean isMatch(String s, String p) {  
3.         int slen = s.length()+1;  
4.         int plen = p.length()+1;  
5.         boolean[][] dp = new boolean[slen][plen];  
6.         dp[0][0]=true;  
7.           
8.         for(int i=1; i<plen; i++){  
9.             if(p.charAt(i-1)=='*')  
10.                 dp[0][i] = true;  
11.             else  
12.                 break;  
13.         }  
14.           
15.         for(int i=1; i<slen; i++){  
16.             for(int j=1; j<plen; j++){  
17.                 if(p.charAt(j-1)=='*' && (dp[i-1][j] || dp[i][j-1] || dp[i-1][j-1]))  
18.                     dp[i][j] = true;  
19.                 else if(dp[i-1][j-1] && (p.charAt(j-1)=='?' || p.charAt(j-1)==s.charAt(i-1))){  
20.                     dp[i][j] = true;  
21.                 }  
22.             }  
23.         }  
24.         return dp[slen-1][plen-1];  
25.     }  
26. }  

解法2：

[java] view plain copy

1. if(p.length()==0)    
2.         return s.length()==0;    
3.     boolean[] res = new boolean[s.length()+1];    
4.     res[0] = true;    
5.     for(int j=0;j<p.length();j++)    
6.     {    
7.         if(p.charAt(j)!='*')    
8.         {    
9.             for(int i=s.length()-1;i>=0;i--)    
10.             {    
11.                 res[i+1] = res[i]&&(p.charAt(j)=='?'||s.charAt(i)==p.charAt(j));    
12.             }    
13.         }    
14.         else    
15.         {    
16.             int i = 0;    
17.             while(i<=s.length() && !res[i])    
18.                 i++;    
19.             for(;i<=s.length();i++)    
20.             {    
21.                 res[i] = true;    
22.             }    
23.         }    
24.         res[0] = res[0]&&p.charAt(j)=='*';    
25.     }    
26.     return res[s.length()];  

原文链接http://blog.csdn.net/crazy__chen/article/details/47359779