92 lines
3.0 KiB
C++
92 lines
3.0 KiB
C++
class TrieNode {
|
|
public:
|
|
// initializing one node
|
|
TrieNode(){
|
|
count = 0;
|
|
// each node can have at most 26 children: a to z.
|
|
// and when accessing child, always remember do that -'a'
|
|
for(int i=0;i<26;i++){
|
|
children.push_back(nullptr);
|
|
}
|
|
}
|
|
// so that we can access private member variables.
|
|
friend class Trie;
|
|
// private:
|
|
// count how many words use this node
|
|
int count;
|
|
std::vector<TrieNode*> children;
|
|
};
|
|
|
|
class Trie {
|
|
public:
|
|
// initializing the prefix tree
|
|
Trie() {
|
|
// which basically is initializing the root
|
|
// and we guarantee that the root always exists
|
|
root = new TrieNode;
|
|
}
|
|
|
|
void insert(string word) {
|
|
int len = word.length();
|
|
TrieNode* current = root;
|
|
for(int i=0; i<len; i++){
|
|
// inserting a node at the branch of [word[i]-'a'], as the children of current
|
|
// if no one has inserted here yet
|
|
if(current->children[word[i]-'a'] == nullptr){
|
|
// create a node
|
|
current->children[word[i]-'a'] = new TrieNode;
|
|
current->children[word[i]-'a']->count++;
|
|
// after the insertion, update current
|
|
current = current->children[word[i]-'a'];
|
|
}else{
|
|
// else, it means this node is already there, just update current
|
|
// we make the code duplicated here, but this way it's easier to understand.
|
|
current->children[word[i]-'a']->count++;
|
|
current = current->children[word[i]-'a'];
|
|
}
|
|
}
|
|
}
|
|
|
|
// private:
|
|
// just like every tree, we should have a root.
|
|
TrieNode* root;
|
|
};
|
|
|
|
/**
|
|
* Your Trie object will be instantiated and called as such:
|
|
* Trie* obj = new Trie();
|
|
* obj->insert(word);
|
|
* bool param_2 = obj->search(word);
|
|
* bool param_3 = obj->startsWith(prefix);
|
|
*/
|
|
|
|
class Solution {
|
|
public:
|
|
string longestCommonPrefix(vector<string>& strs) {
|
|
int size = strs.size();
|
|
std::string result;
|
|
Trie trie;
|
|
TrieNode* current = trie.root;
|
|
// step 1, insert every word in trie
|
|
for(int i=0;i<size;i++){
|
|
trie.insert(strs[i]);
|
|
}
|
|
// step 2, search prefix
|
|
int len = strs[0].length();
|
|
// the common prefix can never be longer than the length of strs[0].
|
|
for(int i=0;i<len;i++){
|
|
if(current && (current->children[strs[0][i]-'a']!=nullptr)){
|
|
// std::cout << current->children[strs[0][i]-'a']->count << ":" << size << std::endl;
|
|
}
|
|
if(current && (current->children[strs[0][i]-'a']!=nullptr) && current->children[strs[0][i]-'a']->count == size){
|
|
result.push_back(strs[0][i]);
|
|
// once it's pushed back, update current so as to move on to the next level.
|
|
current = current->children[strs[0][i]-'a'];
|
|
}else{
|
|
break;
|
|
}
|
|
}
|
|
return result;
|
|
}
|
|
};
|