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### 题目描述

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.

### 例子

`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.

### 解题思路

#### 方法一

• 时间复杂度：Solution: O(n)， Query: O(N) 最坏情况是所有节点都在候选队列中（类似于 `aaaaa...` 的情况）
• 空间复杂度：O(N)

#### 方法二

• 时间复杂度：Solution: O(n)， Query: O(k) k 为单词最大长度
• 空间复杂度：O(N)