[LeetCode] Add and Search Word - Data Structure Design

This problem is an application of the Trie data structure. In the following, it is assumed that you have solved Implement Trie (Prefix Tree).

Now, let's first look at the TrieNode class. I define it as follows.

1 class TrieNode {
2 public:
3     bool isKey;
4     TrieNode* children[26];
5     TrieNode(): isKey(false) {
6         memset(children, NULL, sizeof(TrieNode*) * 26); 
7     }
8 };

The field isKey is to label whether the string comprised of characters starting from root to the current node is a key (word that has been added). In this problem, only lower-case letters a - zneed to be considered, so each TrieNode has at most 26 children. I store it in an array ofTrieNode*: children[i] corresponds to letter 'a' + i. The remaining code defines the constructor of the TrieNode class.

Adding a word can be done in the same way as in Implement Trie (Prefix Tree). The basic idea is to create a TrieNode corresponding to each letter in the word. When we are done, label the last node to be a key (set isKey = true). The code is as follows.

1 void addWord(string word) {
2     TrieNode* run = root;
3     for (char c : word) {
4         if (!(run -> children[c - 'a']))
5             run -> children[c - 'a'] = new TrieNode();
6         run = run -> children[c - 'a']; 
7     }
8     run -> isKey = true;
9 }

By the way, root is defined as private data of WordDictionary:

1 private:
2     TrieNode* root;

And the WordDictionary class has a constructor to initialize root:

1 WordDictionary() {
2     root = new TrieNode();
3 }

Now we are left only with search. Let's do it. The basic idea is still the same as typical search operations in a Trie. The critical part is how to deal with the dots .. Well, my solution is very naive in this place. Each time when we reach a ., just traverse all the children of the current node and recursively search the remaining substring in word starting from that children. So I define a helper function query for search that takes in a string and a starting node. And the initial call to query is like query(word, root).

By the way, I pass a char* instead of string to query and it greatly speeds up the code. So the initial call to query is actually query(word.c_str(), root).

Now I put all the codes together below. Hope it to be useful!

 1 class TrieNode {
 2 public:
 3     bool isKey;
 4     TrieNode* children[26];
 5     TrieNode(): isKey(false) {
 6         memset(children, NULL, sizeof(TrieNode*) * 26);
 7     }
 8 };
 9 
10 class WordDictionary {
11 public:
12     WordDictionary() {
13         root = new TrieNode();
14     }
15 
16     // Adds a word into the data structure.
17     void addWord(string word) {
18         TrieNode* run = root;
19         for (char c : word) {
20             if (!(run -> children[c - 'a'])) 
21                 run -> children[c - 'a'] = new TrieNode();
22             run = run -> children[c - 'a'];
23         }
24         run -> isKey = true;
25     }
26 
27     // Returns if the word is in the data structure. A word could
28     // contain the dot character '.' to represent any one letter.
29     bool search(string word) {
30         return query(word.c_str(), root);
31     }
32 
33 private:
34     TrieNode* root;
35 
36     bool query(const char* word, TrieNode* node) {
37         TrieNode* run = node;
38         for (int i = 0; word[i]; i++) {
39             if (run && word[i] != '.')
40                 run = run -> children[word[i] - 'a'];
41             else if (run && word[i] == '.') { 
42                 TrieNode* tmp = run;
43                 for (int j = 0; j < 26; j++) {
44                     run = tmp -> children[j];
45                     if (query(word + i + 1, run))
46                         return true;
47                 }
48             }
49             else break;
50         }
51         return run && run -> isKey; 
52     }
53 };
54 
55 // Your WordDictionary object will be instantiated and called as such:
56 // WordDictionary wordDictionary;
57 // wordDictionary.addWord("word");
58 // wordDictionary.search("pattern");