frequently mispronounced words

. Follow the middle sequence to output all 10,000 numbers in the current binary sort tree.Basically the time complexity of the algorithm is O (N) times the comparisonThe spatial complexity of the algorithm is 10000 (constant)Based on the idea above, it can be implemented with the smallest heap, so that the complexity is log10000 when the smallest number in the 10,000 tree is not added.Related Similar issues:1, a text file, about 10,000 lines, one word per line, asked to count the most frequent o

possible that the machine has multiple cores, so that multithreading can be used to process the entire dataset. This article introduces solutions suitable for different application scenarios. 3. Solution 3.1 single-host + single-core + large enough memory If each query word occupies an average of 8 bytes, the memory required for the 1 billion query words is about 10 ^ 9*8 = 8 GB. If you have such a large memory, you can sort the query

Import java. Io .*; Import java. util .*; Public class maxlength {// find the word with the longest Length Static void max (int A [], string mystr []) {String [] sigstr;Int max = A [0];For (INT I = 0; I If (A [I]> MAX) max = A [I];}// System. Out. println (max );For (INT I = 0; I If (A [I] = max) system. Out. println ("the longest word is:" + mystr [I]);}}Static void Nums (string mystr []) {// find the most frequently occurring word and OutputInt []

The top 10 most frequently used words in the statistics file (C # TPL DataFlow ), Recently, the company held a program writing competition, requesting that the top 10 words frequently appear in 2G files. The original idea was to use the dictionary tree. Later, it was found that the dictionary tree is more suitable for

Linux classes on the assignment Wednesday, if there is a poor consideration to the place, please do not hesitate to enlighten.Shell Processing Text See this blog for related frequently used commands##假设输入两个參数 is the first to count the number of words. The second article to be counted#假设输入一个參数 the default count of words is 10#Detailed ideas:Replacing various symbo

Python practice every day (1): Calculate the most frequently used words in each article in a folder, # Coding: utf-8import OS, repath = 'test' files = OS. listdir (path) def count_word (words): dic ={} max = 0 marked_key = ''# calculate the number of times each word appears for word in words: if dic. has_key (word) is

The following function is JS to find an English article in the most frequently occurring word (composed of 26 letters of the letter), the output of the word and the number of occurrences, not case-sensitive, mainly the use of regular:functioncounts (article) {article=Article.trim (). toUpperCase (); varArray = Article.match (/[a-z]+/g); Article= "" +array.join ("") + ""; varmax = 0,word,num = 0,maxword= ""; for(vari = 0; i ) {Word=NewREGEXP ("+array[

#!/usr/bin/python#filename:readlinepy.pyimport sys,reurldir=r "C:\python27\a.txt" distone={}numten=[] #先文档变成一个字典f = Open (Urldir, ' R ') for line in F.readlines (): #去掉非字符的符号 line = re.sub (' \w ', "", line) Lineone=line.split () For Keyone in Lineone: if not Distone.get (keyone): distone[keyone]=1 Else: distone[keyone]+= 1f.close () #整理前10出现的单词的次数for v in Distone.values (): if v not in Numten: numten.append (v) numten.sort (

();Findenglishnum (str);}public static void Findenglishnum (String text) {Scanner scan=new Scanner (system.in);int i=0;String[] Array = {".", ",", "?", "!"};for (int i1 = 0; i1 Text = Text.replace (Array[i1], "");}string[] Textarray = Text.split ("");for (int i=0;i// {System.out.println (Textarray[i]);// }mapfor (int i1 = 0; i1 String key = Textarray[i1];Convert to lowercaseString key_l = Key.tolowercase ();if (! "". Equals (key_l)) {Integer num = map.get (key_l);if (num = = NULL | | num = = 0)