trie

问题 I'm looking to use the following code to not check whether there is a word matching in the Trie but to return a list all words beginning with the prefix inputted by the user. Can someone point me in the right direction? I can't get it working at all..... public boolean search(String s) { Node current = root; System.out.println("\nSearching for string: "+s); while(current != null) { for(int i=0;i<s.length();i++) { if(current.child[(int)(s.charAt(i)-'a')] == null) { System.out.println("Cannot

问题 Given a set of strings (large set), and an input string, you need to find all the anagrams of the input string efficiently. What data structure will you use. And using that, how will you find the anagrams? Things that I have thought of are these: Using maps a) eliminate all words with more/less letters than the input. b) put the input characters in map c) Traverse the map for each string and see if all letters are present with their count. Using Tries a) Put all strings which have the right

问题 I'm stuck at one problem i.e. to find a subarray whose xor is 0. I read somewhere that this can be done using TRIE data structure but I want the starting and ending indices of the array. For example, consider an array a = [3, 6, 13, 8 15] The subarray from 0 to 3 i.e. [3, 6, 13, 8] has xor equal to 0. (3 xor 6 xor 13 xor 8 = 0) I'm in search for an algorithm than can find those indices ([0, 3] in this case). Detailed answer would be very helpful. Update I tried the brute Force approach find

问题 i just construct a trie for a vocabulary, and then I found that there are many branches shared the same struct. i want to combine them together result to be a DAWG. What algorithm would I use to convert a trie to a DAWG? 回答1: The standard algorithm for converting a trie into a DAWG works by treating the trie as a deterministic finite automaton, then converting the trie into a minimum-state DFA. There are many algorithms for performing this conversion. The algorithm I'm most familiar with is

问题 I have to implement the T9 Dictionary . Essentially, when I am pressing any of the 9 keys, it should show me the top 5 words that can be started with that combination of keys. If I type '46', it can give 'hotel' or 'good' depending on whether I intended 'g' or 'h' when I pressed 4. The priority is based on which words are relatively popular - you can use, say, the first 5000 words from the top 100 000 words. The code I am doing is: Import import java.io.BufferedReader; import java.io.File;

问题 I'm not too well-versed about the actual algorithms used in string matching with tries. I'm wondering why there seems to be more focus on suffix tries for string matching rather than prefix tries. Can we not use prefix tries for substring matching also? Put in another way, what are the advantages of suffix tries over prefix tries? 回答1: .retteb era seirt xiferp ,drawkcab daer uoy fI Seriously. Suffix tries allow you to traverse from the beginning of a string. 来源： https://stackoverflow.com

问题 So, I just ported the Trie from OCaml. Unfortunately, it runs slower than the standard Map in terms of tryFind. I don't understand this - the trie seems like it should be faster. Is F#'s code libraries built in some special way as to make them faster than the code that the user typically deploys? Here's the code - [<RequireQualifiedAccess>] module Trie type Node<'k, 'v when 'k : comparison> = { TrieMap : Map<'k, Node<'k, 'v>> TrieKvp : ('k list * 'v) option } member inline x.IsEmpty = x

问题 Over the holidays, my family loves to play Boggle. Problem is, I'm terrible at Boggle. So I did what any good programmer would do: wrote a program to play for me. At the core of the algorithm is a simple prefix trie, where each node is a dict of references to the next letters. This is the trie:add implementation: add([], Trie) -> dict:store(stop, true, Trie); add([Ch|Rest], Trie) -> % setdefault(Key, Default, Dict) -> % case dict:find(Key, Dict) of % { ok, Val } -> { Dict, Val } % error -> {

问题 I have the following C++ code: #include <vector> #include <string> using namespace std; struct Trie { bool eow; //end of word char val; vector<Trie> chd; //children void push_word(const string& word){ Trie& trie = *this; for (char c: word){ if (trie.chd.empty() || trie.chd.back().val != c) { trie.chd.push_back(Trie{false, c, vector<Trie>{}}); } trie = trie.chd.back(); } trie.eow = true; } }; It's a trie for strings. push_word is supposed to only accept strings that are lexicographically

问题 I have a simple Trie that I'm using to store about 80k words of length 2 - 15. It works great for checking to see if a string is a word; However, now I need a way of getting a random word of a given length. In other words, I need "getRandomWord(5)" to return a 5 letter word, with all 5 letter words having an equal chance of being returned. The only way I can think of is to pick a random number and traverse the tree breadth-first until I've passed that many words of the desired length. Is