Add and Search Word - Data structure design 解答

Question

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") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

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

Solution

This kind of problem (searching a String) can be solved by Trie.

We use DFS to implement search. Notice corner cases.

When we choose to implement it by recursion, think one step each time.

 1 class TrieNode {
 2     public char value;
 3     public boolean isLeaf;
 4     public HashMap<Character, TrieNode> children;
 5     
 6     public TrieNode(char c) {
 7         value = c;
 8         children = new HashMap<Character, TrieNode>();
 9         isLeaf = false;
10     }
11 }
12 
13 public class WordDictionary {
14     public TrieNode root;
15 
16     public WordDictionary() {
17         root = new TrieNode('!');
18     }
19 
20     // Adds a word into the data structure.
21     public void addWord(String word) {
22         TrieNode currentNode = root;
23         for (int i = 0; i < word.length(); i++) {
24             char tmp = word.charAt(i);
25             HashMap<Character, TrieNode> children = currentNode.children;
26             TrieNode nextNode;
27             if (children.containsKey(tmp)) {
28                 nextNode = children.get(tmp);
29             } else {
30                 nextNode = new TrieNode(tmp);
31                 children.put(tmp, nextNode);
32             }
33             currentNode = nextNode;
34             // Check whether it's the last character
35             if (i == word.length() - 1)
36                 currentNode.isLeaf = true;
37         }
38     }
39 
40     // Returns if the word is in the data structure. A word could
41     // contain the dot character '.' to represent any one letter.
42     public boolean search(String word) {
43         return dfsSearch(word, 0, root);
44     }
45     
46     private boolean dfsSearch(String word, int index, TrieNode prevNode) {
47         // If prevNode is null but word has not been completely traversed, return fasle
48         if (prevNode == null) {
49             return false;
50         }
51         
52         // If word has been completely traversed, check whether tree is at bottom
53         if (index == word.length()) {
54             return prevNode.isLeaf;
55         }
56         char target = word.charAt(index);
57         HashMap<Character, TrieNode> currentMap = prevNode.children;
58         
59         if (target != '.') {
60             if (!currentMap.containsKey(target))
61                 return false;
62             else
63                 return dfsSearch(word, index + 1, currentMap.get(target));
64         } else {
65             boolean result = false;
66             for (Character key : currentMap.keySet()) {
67                 if (dfsSearch(word, index + 1, currentMap.get(key))) {
68                     result = true;
69                 }
70             }
71             return result;
72         }
73     }
74 }
75 
76 // Your WordDictionary object will be instantiated and called as such:
77 // WordDictionary wordDictionary = new WordDictionary();
78 // wordDictionary.addWord("word");
79 // wordDictionary.search("pattern");

 

posted @ 2015-10-13 10:20  树獭君  阅读(125)  评论(0)  编辑  收藏  举报