Problem
https://leetcode.com/explore/learn/card/trie/147/basic-operations/1047/
Solution
/**
1. Initialize: cur = root
2. for each char c in target string S:
3. if cur does not have a child c:
4. cur.children[c] = new Trie node
5. cur = cur.children[c]
6. cur is the node which represents the string S
**/
class Trie {
class TrieNode {
private Map<Character, TrieNode> childrenMap = new HashMap<>();
private boolean isWord = false;
public Map<Character, TrieNode> getChildrenMap() {
return this.childrenMap;
}
public boolean isWord() {
return this.isWord;
}
public void setWord() {
this.isWord = true;
}
}
TrieNode root;
public Trie() {
root = new TrieNode();
}
public void insert(String word) {
char[] charArr = word.toCharArray();
TrieNode current = root;
for (int i=0;i<charArr.length;i++) {
char c = charArr[i];
if (!current.getChildrenMap().containsKey(c)) {
TrieNode next = new TrieNode();
current.getChildrenMap().put(c, next);
}
current = current.getChildrenMap().get(c);
}
current.setWord();
}
public boolean search(String word) {
char[] charArr = word.toCharArray();
TrieNode current = root;
for (int i=0;i<charArr.length;i++) {
char c = charArr[i];
if (!current.getChildrenMap().containsKey(c)) {
return false;
}
current = current.getChildrenMap().get(c);
}
if (!current.isWord()) {
return false;
}
return true;
}
public boolean startsWith(String prefix) {
char[] charArr = prefix.toCharArray();
TrieNode current = root;
for (int i=0;i<charArr.length;i++) {
char c = charArr[i];
if (!current.getChildrenMap().containsKey(c)) {
return false;
}
current = current.getChildrenMap().get(c);
}
return true;
}
}
/**
* 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);
*/
- Trie 문제의 대표적인 코드
- TrieNode 라는 클래스를 선언하는 것이 포인트!
- 시작할때 root 를 만들고, current 라는 포인터를 기준으로 순회하는 것이 중요.

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