LeetCode 1032. Stream of Characters

Problem Description: 

Implement the StreamChecker class as follows:

• StreamChecker(words): Constructor, init the data structure with the given words.
• query(letter): returns true if and only if for some k >= 1, the last k characters queried (in order from oldest to newest, including this letter just queried) spell one of the words in the given list.

Example:

StreamChecker streamChecker = new StreamChecker(["cd","f","kl"]); // init the dictionary.
streamChecker.query('a');          // return false
streamChecker.query('b');          // return false
streamChecker.query('c');          // return false
streamChecker.query('d');          // return true, because 'cd' is in the wordlist
streamChecker.query('e');          // return false
streamChecker.query('f');          // return true, because 'f' is in the wordlist
streamChecker.query('g');          // return false
streamChecker.query('h');          // return false
streamChecker.query('i');          // return false
streamChecker.query('j');          // return false
streamChecker.query('k');          // return false
streamChecker.query('l');          // return true, because 'kl' is in the wordlist
 

Note:

1 <= words.length <= 2000
1 <= words[i].length <= 2000
Words will only consist of lowercase English letters.
Queries will only consist of lowercase English letters.
The number of queries is at most 40000.

题解：

看到这个题很快想到了该用前缀树trie来存下单词表，但是此题因为是要判断以query传入的当前字母结尾的单词是否出现在单词表中，要倒序存。

而此题的关键也是倒序存和倒序的查找，在trie的基础上稍微修改代码即可完成。

时间复杂度：O(L * Q)

其中L是最大单词长度，Q是query的次数。

代码如下

class StreamChecker {
    class Node{
        Node[] children;
        boolean endOfWord;
        public Node() {
            children = new Node[26];
            endOfWord = false;
        }
    }
    
    Node root;
    StringBuilder sb;
    
    private void insert(String word) {
        Node cur = root;
        for(int i = word.length() - 1; i >= 0; i--) {
            char ch = word.charAt(i);
            Node next = cur.children[ch - 'a'];
            if(next == null) {
                cur.children[ch - 'a'] = next = new Node();
            }
            cur = next;
        }
        cur.endOfWord = true;
    }
    
    public StreamChecker(String[] words) {
        root = new Node();
        sb = new StringBuilder();
        for(String word : words) {
            insert(word);
        }
    }
    
    public boolean query(char letter) {
        boolean res = false;
        
        sb.append(letter);
        Node cur = root;
        for(int i = sb.length() - 1; i >= 0; i--) {
            char ch = sb.charAt(i);
            if(cur.children[ch - 'a'] == null) {
                return false;
            }
            cur = cur.children[ch - 'a'];
            if(cur.endOfWord) return true;
        }

        return false;
    }
}