zoukankan      html  css  js  c++  java
  • 1032. Stream of Characters

    package LeetCode_1032
    
    /**
     * 1032. Stream of Characters
     * https://leetcode.com/problems/stream-of-characters/
     * 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. 1 <= words.length <= 2000
    2. 1 <= words[i].length <= 2000
    3. Words will only consist of lowercase English letters.
    4. Queries will only consist of lowercase English letters.
    5. The number of queries is at most 40000.
     * */

    /* * Solution: Trie Tree, * Time complexity: O(M) M is the longest length of the word, * Space complexity: O(M*N), N is the number of word * */ class StreamChecker(words: Array<String>) { var root: Trie? = null //save letter to help to check val sb = StringBuilder() init { root = Trie() for (word in words) { addWord(word) } } fun query(letter: Char): Boolean { sb.append(letter) var node = root //here also check form last for (i in sb.length - 1 downTo 0) { if (node == null) { continue } val c = sb[i] node = node.children[c - 'a'] if (node != null && node.end) { return true } } return false } private fun addWord(word: String) { var node = root //create by reverse order, because can check from last letter //for (ch in word) { for (i in word.length - 1 downTo 0) { val pos = word[i] - 'a' if (node!!.children[pos] == null) { node.children[pos] = Trie() } //set the children for current node node = node.children[pos] } node!!.end = true } class Trie { var end = false var children = arrayOfNulls<Trie>(26) } } /** * Your StreamChecker object will be instantiated and called as such: * var obj = StreamChecker(words) * var param_1 = obj.query(letter) */
  • 相关阅读:
    软件项目管理
    asterisk channel driver dev ref
    标 题: 有什么办法快速把pc上的网址发送到手机上
    dongle0
    ubuntu
    Huawei E1750 Asterisk
    Jquery重新学习之七[Ajax运用总结A]
    Jquery重新学习之六[操作XML数据]
    Jquery重新学习之五[操作JSON数据]
    Jquery重新学习之四[核心属性与文档处理属性]
  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/14075275.html
Copyright © 2011-2022 走看看