leetcode [208]Implement Trie (Prefix Tree)

佚名 7年前 (2019-04-26) 随笔 930人围观抢沙发百度已收录

Implement a trie with insert, search, and startsWith methods.

Example:

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Trie trie = new Trie();

trie.insert("apple");
trie.search("apple");   // returns true
trie.search("app");     // returns false
trie.startsWith("app"); // returns true
trie.insert("app");   
trie.search("app");     // returns true

Note:

You may assume that all inputs are consist of lowercase letters a-z.
All inputs are guaranteed to be non-empty strings.

题目大意:

前缀树的具体实现。

解法：

前缀树的每一个节点保存着当前节点的字符，子节点，从根节点到当前节点是否构成一个词。这道题目里面所有词都是由小写字母组成的，所以子节点的数组大小为24。

java:

public class TrieNode{
    public char val;
    public boolean isWord;
    public TrieNode[] children=new TrieNode[26];
    TrieNode(){}
    TrieNode(char val){
        this.val=val;
    }
}

class Trie {
    public TrieNode root;
    /** Initialize your data structure here. */
    public Trie() {
        root=new TrieNode(' ');
    }

    /** Inserts a word into the trie. */
    public void insert(String word) {
        TrieNode t=root;
        for(int i=0;i<word.length();i++){
            if(t.children[word.charAt(i)-'a']==null){
                t.children[word.charAt(i)-'a']=new TrieNode(word.charAt(i));
            }
            t=t.children[word.charAt(i)-'a'];
            if(i==word.length()-1) t.isWord=true;
        }
    }

    /** Returns if the word is in the trie. */
    public boolean search(String word) {
        TrieNode t=root;
        for(int i=0;i<word.length();i++){
            char c=word.charAt(i);
            if (t.children[c-'a']==null) return false;
            t=t.children[c-'a'];
            if(i==word.length()-1 && t.isWord) return true;
        }

        return false;
    }

    /** Returns if there is any word in the trie that starts with the given prefix. */
    public boolean startsWith(String prefix) {
        TrieNode t=root;
            for(int i=0;i<prefix.length();i++){
                char c=prefix.charAt(i);
                if (t.children[c-'a']==null) return false;
                t=t.children[c-'a'];
            }
            return true;
    }
}