208. Implement Trie (Prefix Tree)
Last updated
Last updated
time: O(n), n is word len
space: O(26*n), each node have 26 chars, in worst case, tire will have n level , n is word len
class Trie {
class TrieNode {
Map<Character, TrieNode> sons;
boolean isWord;
String word;
TrieNode() {
this.sons = new HashMap<>();
this.isWord = false;
this.word = null;
}
}
/** Initialize your data structure here. */
private TrieNode root;
public Trie() {
root = new TrieNode();
}
/** Inserts a word into the trie. */
public void insert(String word) {
TrieNode node = root;
for (char c : word.toCharArray()) {
node.sons.computeIfAbsent(c, k -> new TrieNode());
node = node.sons.get(c);
}
node.isWord = true;
node.word = word;
}
private TrieNode getNode(String word) {
TrieNode node = root;
for (char c : word.toCharArray()) {
TrieNode next = node.sons.get(c);
if (next == null) {
return null;
}
node = next;
}
return node;
}
/** Returns if the word is in the trie. */
public boolean search(String word) {
TrieNode node = getNode(word);
return node != null && node.isWord;
}
/** Returns if there is any word in the trie that starts with the given prefix. */
// ex: build: apple
// input prefix: app (prefix 比較短
// so getNode will return a node is going to app's last p, so next not null!
public boolean startsWith(String prefix) {
TrieNode node = getNode(prefix);
return node != null;
}
}
/**
* Your Trie object will be instantiated and called as such:
* Trie obj = new Trie();
* obj.insert(word);
* boolean param_2 = obj.search(word);
* boolean param_3 = obj.startsWith(prefix);
*/