Leetcode 211: Add and Search Word - Data structure design

Posted 2020-10-24 Keep walking

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.

 

 1 public class WordDictionary {
 2     public class TrieNode
 3     {
 4         public Dictionary<char, TrieNode> children = new Dictionary<char, TrieNode>();
 5         public bool isWord = false;
 6     }
 7     
 8     private TrieNode root = new TrieNode();
 9 
10     /** Initialize your data structure here. */
11     public WordDictionary() {
12     }
13     
14     /** Adds a word into the data structure. */
15     public void AddWord(string word) {
16         var node = root;
17         
18         for (int i = 0; i < word.Length; i++)
19         {
20             if (!node.children.ContainsKey(word[i]))
21             {
22                 node.children[word[i]] = new TrieNode();
23             }
24             
25             node = node.children[word[i]];
26         }
27         
28         node.isWord = true;
29     }
30     
31     /** Returns if the word is in the data structure. A word could contain the dot character ‘.‘ to represent any one letter. */
32     public bool Search(string word) {
33         return InternalSearch(word, root);
34     }
35     
36     private bool InternalSearch(string word, TrieNode node) {
37         for (int i = 0; i < word.Length; i++)
38         {
39             if (word[i] == ‘.‘)
40             {
41                 foreach (var c in node.children.Values)
42                 {
43                     if (InternalSearch(word.Substring(i + 1, word.Length - i - 1), c))
44                     {
45                         return true;
46                     }
47                 }
48                 
49                 return false;
50             }
51             else if (node.children.ContainsKey(word[i]))
52             {
53                 node = node.children[word[i]];
54             }
55             else
56             {
57                 return false;
58             }
59         }
60         
61         return node.isWord;
62     }
63 }
64 
65 /**
66  * Your WordDictionary object will be instantiated and called as such:
67  * WordDictionary obj = new WordDictionary();
68  * obj.AddWord(word);
69  * bool param_2 = obj.Search(word);
70  */