Tuesday, April 5, 2016

LeetCode Q211: Add and Search Word - Data structure design

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:
Design and implement a prefix tree (Trie)

class WordDictionary {
public:
class TrieNode {
public:
// Initialize your data structure here.
TrieNode() {
wordsEndHere=false;
memset(subTries, NULL, sizeof(subTries));
}
TrieNode(char _c): c(_c){
wordsEndHere=false;
memset(subTries, NULL, sizeof(subTries));
}
char c;
bool wordsEndHere;
TrieNode* subTries[26];
};
WordDictionary(){
root = new TrieNode();
}
// Adds a word into the data structure.
void addWord(string word) {
TrieNode* p = root;
for(int i=0; i<word.length(); i++){
int c=word[i]-'a';
if(p->subTries[c]==NULL){
TrieNode * node = new TrieNode(c);
p->subTries[c]=node;
}
p=p->subTries[c];
}
p->wordsEndHere=true;
}
// Returns if the word is in the data structure. A word could
// contain the dot character '.' to represent any one letter.
bool helper(TrieNode* root, string word){
bool found=false;
if(word.length()==0)
return root->wordsEndHere;
if(word[0]=='.'){
for(int i=0; i<26; i++)
if(root->subTries[i]!=NULL)
found = found | helper(root->subTries[i], word.substr(1, word.length()-1));
}else{
int c = word[0]-'a';
if(root->subTries[c]!=NULL)
found = helper(root->subTries[c], word.substr(1, word.length()-1));
}
return found;
}
bool search(string word) {
bool found=false;
TrieNode* p =root;
if(word.length()==0)
return false;
found = helper(p, word);
return found;
}
TrieNode* root;
};
// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary;
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");

No comments:

Post a Comment