题目

208. 实现 Trie (前缀树)

720. 词典中最长的单词

思路和解答

📌208. 实现 Trie (前缀树)

class Trie {
    class TrieNode{
        boolean end;  //标识是否是接收字符的结束
        TrieNode[] next =new TrieNode[26];  //初始化都为null
    }
    TrieNode root;
    //构造函数
    public Trie() {
        root = new TrieNode();
    }
    
    public void insert(String word) {
        TrieNode p = root;
        for(int i=0;i<word.length();i++){
            int t = word.charAt(i)-'a';
            if(p.next[t]==null){
                p.next[t] = new TrieNode();
            }
            p = p.next[t];
        }
        p.end = true;  //标记结束
    }
    
    public boolean search(String word) {
        TrieNode p = root;
        for(int i=0;i<word.length();i++){
            int t = word.charAt(i)-'a';
            if(p.next[t]==null){
                return false;
            }
            p = p.next[t];
        }
        return p.end;
    }
    
    public boolean startsWith(String prefix) {
        TrieNode p = root;
        for(int i=0;i<prefix.length();i++){
            int t = prefix.charAt(i)-'a';
            if(p.next[t]==null){
                return false;
            }
            p = p.next[t];
        }
        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);
 */

📌

题解：力扣

方法一  时空复杂度粗略写成O(n^2)

class Solution {
    public String longestWord(String[] words) {
        //用时15分钟
        //按字典序从小到大 长度从大到小排序 
        Arrays.sort(words,(a,b)->{
            if(a.length()==b.length()) return a.compareTo(b);
            else return b.length()-a.length();
        });
        Set<String> st = new HashSet<>();
        for(String word:words){
            st.add(word);
        }
        for(String word:words){
           boolean flag = true;
           for(int i=1;i<=word.length();i++){
               if(!st.contains(word.substring(0,i))){  //如果一个字符串的所有前缀都被包含则直接返回
                   flag = false;
                   break;
               }
           }
           if(flag) return word;
        }   
        return "";
    }
}

方法二：前缀树 dfs