查询是否在Trie中有单词以prefix为前缀
- prefix在Trie中都找到的情况下,不用再判断prefix.charAt(prefix.length - 1)所对应的节点是否是单词,直接返回true即可;
// 查询是否在Trie中有单词以prefix为前缀
public boolean isPrefix(String prefix){
Node cur = root;
for(int i = 0 ; i < prefix.length() ; i ++){
char c = prefix.charAt(i);
if(cur.next.get(c) == null)
return false;
cur = cur.next.get(c);
}
return true;
}
A Problem In Leetcode
public class Trie {
private class Node{
public boolean isWord;
public TreeMap<Character, Node> next;
public Node(boolean isWord){
this.isWord = isWord;
next = new TreeMap<>();
}
public Node(){
this(false);
}
}
private Node root;
public Trie(){
root = new Node();
}
// 向Trie中添加一个新的单词word
public void insert(String word){
Node cur = root;
for(int i = 0 ; i < word.length() ; i ++){
char c = word.charAt(i);
if(cur.next.get(c) == null)
cur.next.put(c, new Node());
cur = cur.next.get(c);
}
cur.isWord = true;
}
// 查询单词word是否在Trie中
public boolean search(String word){
Node cur = root;
for(int i = 0 ; i < word.length() ; i ++){
char c = word.charAt(i);
if(cur.next.get(c) == null)
return false;
cur = cur.next.get(c);
}
return cur.isWord;
}
// 查询是否在Trie中有单词以prefix为前缀
public boolean startsWith(String isPrefix){
Node cur = root;
for(int i = 0 ; i < isPrefix.length() ; i ++){
char c = isPrefix.charAt(i);
if(cur.next.get(c) == null)
return false;
cur = cur.next.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);
*/