Implement Trie (Prefix Tree)
Link
Description
A trie (pronounced as "try") or prefix tree is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.
Implement the Trie class:
Trie()
Initializes the trie object.
void insert(String word)
Inserts the string word
into the trie.
boolean search(String word)
Returns true
if the string word
is in the trie (i.e., was inserted before), and false otherwise.
boolean startsWith(String prefix)
Returns true
if there is a previously inserted string word
that has the prefix prefix
, and false
otherwise.
Example 1:
| ["Trie", "insert", "search", "search", "startsWith", "insert", "search"]
[[], ["apple"], ["apple"], ["app"], ["app"], ["app"], ["app"]]
|
| [null, null, true, false, true, null, true]
|
| Trie trie = new Trie();
trie.insert("apple");
trie.search("apple"); // return True
trie.search("app"); // return False
trie.startsWith("app"); // return True
trie.insert("app");
trie.search("app"); // return True
|
Constraints:
1 <= word.length, prefix.length <= 2000
word
and prefix
consist only of lowercase English letters.
- At most
3 * 10^4
calls in total will be made to insert
, search
, and startsWith
.
Solution
Way 1
| class Trie {
private:
const static int N = 31010;
int son[N][26] {0};
int cnt[N] {0};
int idx = 0;
public:
Trie() {
}
void insert(string word) {
int p = 0;
for (char ch : word) {
int u = ch - 'a';
if (!son[p][u]) son[p][u] = ++idx;
p = son[p][u];
}
cnt[p]++;
}
bool search(string word) {
int p = 0;
for (char ch : word) {
int u = ch - 'a';
if (!son[p][u]) return false;
p = son[p][u];
}
return cnt[p] > 0;
}
bool startsWith(string prefix) {
int p = 0;
for (char ch : prefix) {
int u = ch - 'a';
if (!son[p][u]) return false;
p = son[p][u];
}
return true;
}
};
/**
* 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);
*/
|
Solution with better variable names:
| class Trie {
private:
const static int N = 31010;
int son[N][26] {0};
int ending[N] {0};
int cnt = 0;
public:
Trie() {
}
void insert(string word) {
int current = 0;
for (char& ch : word) {
int index = ch - 'a';
if (!son[current][index]) {
son[current][index] = ++cnt;
}
current = son[current][index];
}
ending[current]++;
}
bool search(string word) {
int current = 0;
for (char& ch : word) {
int index = ch - 'a';
if (!son[current][index]) {
return false;
}
current = son[current][index];
}
return ending[current] > 0;
}
bool startsWith(string prefix) {
int current = 0;
for (char& ch : prefix) {
int index = ch - 'a';
if (!son[current][index]) {
return false;
}
current = son[current][index];
}
return true;
}
};
/**
* 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);
*/
|
Way 2
| class Trie {
private:
struct Node {
bool is_tail;
Node *son[26];
Node() {
is_tail = false;
for (int i = 0; i < 26; i++) son[i] = nullptr;
}
~Node() {
for (int i = 0; i < 26; i++) delete son[i];
}
} *root;
public:
Trie() {
root = new Node();
}
~Trie() {
delete root;
}
void insert(string word) {
auto p = root;
for (auto ch : word) {
int u = ch - 'a';
if (!p->son[u]) p->son[u] = new Node();
p = p->son[u];
}
p->is_tail = true;
}
bool search(string word) {
auto p = root;
for (auto ch : word) {
int u = ch - 'a';
if (!p->son[u]) return false;
p = p->son[u];
}
return p->is_tail;
}
bool startsWith(string prefix) {
auto p = root;
for (auto ch : prefix) {
int u = ch - 'a';
if (!p->son[u]) return false;
p = p->son[u];
}
return true;
}
};
/**
* 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);
*/
|