208. Implement Trie (Prefix Tree)

Implement a trie with insert, search, and startsWith methods.

题目含义：实现字典树的insert, search, 和startsWith 方法

字典树概念参考http://blog.csdn.net/u012501459/article/details/46741267

Trie树的基本性质如下：

根节点不包括字符，除根节点外每个节点包括一个字符。
从根节点到某一节点，路径上经过的字符连接起来，即为对应的字符串。
每个节点的所有子节点包含的字符串各不相同。

本题中的Trie树可以简单实现如下：

Trie树节点数据结构定义如下：

val表示该节点对应的字符
子节点简单用一个数组表示，这样实现比较简单，但比较耗费内存
isWord标记从根节点到该节点是否表示一个单词，还是另一单词的前缀。

 1 class Trie {
 2     
 3     class TrieNode {
 4         public char val;
 5         public boolean isWord;
 6         public TrieNode[] children = new TrieNode[26];
 7         public TrieNode() {}
 8         TrieNode(char c){
 9             val = c;
10         }
11     }
12     
13     private TrieNode root;
14     /** Initialize your data structure here. */
15     public Trie() {
16         root = new TrieNode();
17         root.val = ' ';        
18     }
19     
20     /** Inserts a word into the trie. */
21     public void insert(String word) {
22         TrieNode ws = root;
23         for(int i = 0; i < word.length(); i++){
24             char c = word.charAt(i);
25             if(ws.children[c - 'a'] == null){
26                 ws.children[c - 'a'] = new TrieNode(c);
27             }
28             ws = ws.children[c - 'a'];
29         }
30         ws.isWord = true;        
31     }
32     
33     /** Returns if the word is in the trie. */
34     public boolean search(String word) {
35         TrieNode ws = root;
36         for(int i = 0; i < word.length(); i++){
37             char c = word.charAt(i);
38             if(ws.children[c - 'a'] == null) return false;
39             ws = ws.children[c - 'a'];
40         }
41         return ws.isWord;        
42     }
43     
44     /** Returns if there is any word in the trie that starts with the given prefix. */
45     public boolean startsWith(String prefix) {
46         TrieNode ws = root;
47         for(int i = 0; i < prefix.length(); i++){
48             char c = prefix.charAt(i);
49             if(ws.children[c - 'a'] == null) return false;
50             ws = ws.children[c - 'a'];
51         }
52         return true;        
53     }
54 }