leetcode--Add and Search Word

Design a data structure that supports the following two operations:

void addWord(word)bool search(word)

search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.

For example:

addWord("bad")addWord("dad")addWord("mad")search("pad") -> falsesearch("bad") -> truesearch(".ad") -> truesearch("b..") -> true

Note:
You may assume that all words are consist of lowercase letters a-z.

click to show hint.

You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.

[java] view plain copy

1. class TrieNode {  
2.     // Initialize your data structure here.  
3.     public TrieNode() {}  
4.     Map<Character,TrieNode> next = new HashMap<Character,TrieNode>();  
5.     char c='\0';  
6.     boolean isEnd = false;  
7.     public TrieNode(char c) {  
8.         this.c = c;          
9.     }  
10. }  
11. public class WordDictionary {  
12.     private TrieNode root;  
13.   
14.     public WordDictionary() {  
15.         root = new TrieNode();  
16.     }  
17.       
18.     // Adds a word into the data structure.  
19.     public void addWord(String word) {  
20.         Map<Character, TrieNode> children = root.next;    
21.         for(int i=0; i<word.length(); i++) {    
22.             char c = word.charAt(i);    
23.             TrieNode t;    
24.             if(children.containsKey(c)) {    
25.                 t = children.get(c);    
26.             } else {    
27.                 t = new TrieNode(c);    
28.                 children.put(c, t);    
29.             }    
30.             children = t.next;    
31.             if(i==word.length()-1) t.isEnd=true;    
32.         }    
33.     }  
34.   
35.     // Returns if the word is in the data structure. A word could  
36.     // contain the dot character '.' to represent any one letter.  
37.     public boolean search(String word) {  
38.         return helper(word,root);  
39.     }  
40.       
41.     public boolean helper(String word,TrieNode tn){  
42.         if(tn==null) return false;    
43.         if(word.length() == 0 ) return tn.isEnd;    
44.             
45.         Map<Character, TrieNode> children = tn.next;    
46.         TrieNode t = null;    
47.         char c = word.charAt(0);    
48.         if(c=='.') {    
49.             for(char key : children.keySet() ) {    
50.                 if(helper(word.substring(1), children.get(key) )) return true;    
51.             }    
52.             return false;    
53.         } else if(!children.containsKey(c)) {    
54.             return false;    
55.         } else {    
56.             t = children.get(c);    
57.             return helper(word.substring(1), t);    
58.         }    
59.     }  
60. }  
61.   
62. // Your WordDictionary object will be instantiated and called as such:  
63. // WordDictionary wordDictionary = new WordDictionary();  
64. // wordDictionary.addWord("word");  
65. // wordDictionary.search("pattern");

原文链接http://blog.csdn.net/crazy__chen/article/details/46576277