#practiceLinkDiv { display : aucun !important; }Étant donné une liste de contacts qui existent dans un annuaire téléphonique. La tâche consiste à implémenter une requête de recherche pour l'annuaire téléphonique. La requête de recherche sur une chaîne ' str ' affiche tous les contacts dont le préfixe est ' str '. Une propriété spéciale de la fonction de recherche est que lorsqu'un utilisateur recherche un contact dans la liste de contacts, des suggestions (Contacts avec préfixe comme chaîne saisie ainsi) sont affichées après que l'utilisateur a saisi chaque caractère.
Note: Les contacts de la liste sont uniquement constitués d’alphabets minuscules. Exemple:
qu'est-ce que jquery
Input : contacts [] = {gforgeeks geeksquiz } Query String = gekk Output : Suggestions based on 'g' are geeksquiz gforgeeks Suggestions based on 'ge' are geeksquiz No Results Found for 'gek' No Results Found for 'gekk' Recommandé : veuillez le résoudre sur PRATIQUE d'abord avant de passer à la solution.
L'annuaire téléphonique peut être mis en œuvre efficacement en utilisant Essayer Structure des données. Nous insérons tous les contacts dans Trie. Généralement, la requête de recherche sur un Trie consiste à déterminer si la chaîne est présente ou non dans le trie, mais dans ce cas, il nous est demandé de trouver toutes les chaînes avec chaque préfixe « str ». Cela équivaut à faire un Parcours DFS sur un graphique . À partir d'un nœud Trie, visitez les nœuds Trie adjacents et faites-le de manière récursive jusqu'à ce qu'il n'y en ait plus. Cette fonction récursive prendra 2 arguments, l'un comme Trie Node qui pointe vers le Trie Node actuel visité et l'autre comme chaîne qui stocke la chaîne trouvée jusqu'à présent avec le préfixe « str ». Chaque nœud Trie stocke une variable booléenne « isLast » qui est vraie si le nœud représente la fin d'un contact (mot).
// This function displays all words with given // prefix. 'node' represents last node when // path from root follows characters of 'prefix'. displayContacts (TreiNode node string prefix) If (node.isLast is true) display prefix // finding adjacent nodes for each character ‘i’ in lower case Alphabets if (node.child[i] != NULL) displayContacts(node.child[i] prefix+i)
L'utilisateur entrera la chaîne caractère par caractère et nous devons afficher des suggestions avec le préfixe formé après chaque caractère saisi. Ainsi, une approche pour trouver le préfixe commençant par la chaîne formée consiste à vérifier si le préfixe existe dans le Trie. Si oui, puis à appeler la fonction displayContacts(). Dans cette approche, après chaque caractère saisi, nous vérifions si la chaîne existe dans le Trie. Au lieu de vérifier encore et encore, nous pouvons maintenir un pointeur précédentNœud ' qui pointe vers le TrieNode qui correspond au dernier caractère saisi par l'utilisateur. Nous devons maintenant vérifier le nœud enfant pour le 'prevNode' lorsque l'utilisateur entre un autre caractère pour vérifier s'il existe dans le Trie. Si le nouveau préfixe n'est pas dans Trie, alors toutes les chaînes formées en saisissant des caractères après « préfixe » ne peuvent pas non plus être trouvées dans Trie. Nous cassons donc la boucle utilisée pour générer les préfixes un par un et imprimons « Aucun résultat trouvé » pour tous les caractères restants.
C++
// C++ Program to Implement a Phone // Directory Using Trie Data Structure #include
// Java Program to Implement a Phone // Directory Using Trie Data Structure import java.util.*; class TrieNode { // Each Trie Node contains a Map 'child' // where each alphabet points to a Trie // Node. HashMap<CharacterTrieNode> child; // 'isLast' is true if the node represents // end of a contact boolean isLast; // Default Constructor public TrieNode() { child = new HashMap<CharacterTrieNode>(); // Initialize all the Trie nodes with NULL for (char i = 'a'; i <= 'z'; i++) child.put(inull); isLast = false; } } class Trie { TrieNode root; // Insert all the Contacts into the Trie public void insertIntoTrie(String contacts[]) { root = new TrieNode(); int n = contacts.length; for (int i = 0; i < n; i++) { insert(contacts[i]); } } // Insert a Contact into the Trie public void insert(String s) { int len = s.length(); // 'itr' is used to iterate the Trie Nodes TrieNode itr = root; for (int i = 0; i < len; i++) { // Check if the s[i] is already present in // Trie TrieNode nextNode = itr.child.get(s.charAt(i)); if (nextNode == null) { // If not found then create a new TrieNode nextNode = new TrieNode(); // Insert into the HashMap itr.child.put(s.charAt(i)nextNode); } // Move the iterator('itr') to point to next // Trie Node itr = nextNode; // If its the last character of the string 's' // then mark 'isLast' as true if (i == len - 1) itr.isLast = true; } } // This function simply displays all dictionary words // going through current node. String 'prefix' // represents string corresponding to the path from // root to curNode. public void displayContactsUtil(TrieNode curNode String prefix) { // Check if the string 'prefix' ends at this Node // If yes then display the string found so far if (curNode.isLast) System.out.println(prefix); // Find all the adjacent Nodes to the current // Node and then call the function recursively // This is similar to performing DFS on a graph for (char i = 'a'; i <= 'z'; i++) { TrieNode nextNode = curNode.child.get(i); if (nextNode != null) { displayContactsUtil(nextNode prefix + i); } } } // Display suggestions after every character enter by // the user for a given string 'str' void displayContacts(String str) { TrieNode prevNode = root; // 'flag' denotes whether the string entered // so far is present in the Contact List String prefix = ''; int len = str.length(); // Display the contact List for string formed // after entering every character int i; for (i = 0; i < len; i++) { // 'str' stores the string entered so far prefix += str.charAt(i); // Get the last character entered char lastChar = prefix.charAt(i); // Find the Node corresponding to the last // character of 'str' which is pointed by // prevNode of the Trie TrieNode curNode = prevNode.child.get(lastChar); // If nothing found then break the loop as // no more prefixes are going to be present. if (curNode == null) { System.out.println('nNo Results Found for ' + prefix); i++; break; } // If present in trie then display all // the contacts with given prefix. System.out.println('nSuggestions based on ' + prefix + ' are '); displayContactsUtil(curNode prefix); // Change prevNode for next prefix prevNode = curNode; } for ( ; i < len; i++) { prefix += str.charAt(i); System.out.println('nNo Results Found for ' + prefix); } } } // Driver code class Main { public static void main(String args[]) { Trie trie = new Trie(); String contacts [] = {'gforgeeks' 'geeksquiz'}; trie.insertIntoTrie(contacts); String query = 'gekk'; // Note that the user will enter 'g' then 'e' so // first display all the strings with prefix as 'g' // and then all the strings with prefix as 'ge' trie.displayContacts(query); } }
# Python Program to Implement a Phone # Directory Using Trie Data Structure class TrieNode: def __init__(self): # Each Trie Node contains a Map 'child' # where each alphabet points to a Trie # Node. self.child = {} self.is_last = False # Making root NULL for ease so that it doesn't # have to be passed to all functions. root = TrieNode() # Insert a Contact into the Trie def insert(string): # 'itr' is used to iterate the Trie Nodes itr = root for char in string: # Check if the s[i] is already present in # Trie if char not in itr.child: # If not found then create a new TrieNode itr.child[char] = TrieNode() # Move the iterator('itr') to point to next # Trie Node itr = itr.child[char] # If its the last character of the string 's' # then mark 'isLast' as true itr.is_last = True # This function simply displays all dictionary words # going through current node. String 'prefix' # represents string corresponding to the path from # root to curNode. def display_contacts_util(cur_node prefix): # Check if the string 'prefix' ends at this Node # If yes then display the string found so far if cur_node.is_last: print(prefix) # Find all the adjacent Nodes to the current # Node and then call the function recursively # This is similar to performing DFS on a graph for i in range(ord('a') ord('z') + 1): char = chr(i) next_node = cur_node.child.get(char) if next_node: display_contacts_util(next_node prefix + char) # Display suggestions after every character enter by # the user for a given query string 'str' def displayContacts(string): prev_node = root prefix = '' # Display the contact List for string formed # after entering every character for i char in enumerate(string): # 'prefix' stores the string formed so far prefix += char # Find the Node corresponding to the last # character of 'prefix' which is pointed by # prevNode of the Trie cur_node = prev_node.child.get(char) # If nothing found then break the loop as # no more prefixes are going to be present. if not cur_node: print(f'No Results Found for {prefix}n') break # If present in trie then display all # the contacts with given prefix. print(f'Suggestions based on {prefix} are 'end=' ') display_contacts_util(cur_node prefix) print() # Change prevNode for next prefix prev_node = cur_node # Once search fails for a prefix we print # 'Not Results Found' for all remaining # characters of current query string 'str'. for char in string[i+1:]: prefix += char print(f'No Results Found for {prefix}n') # Insert all the Contacts into the Trie def insertIntoTrie(contacts): # Insert each contact into the trie for contact in contacts: insert(contact) # Driver program to test above functions # Contact list of the User contacts=['gforgeeks''geeksquiz'] # Size of the Contact List n=len(contacts) # Insert all the Contacts into Trie insertIntoTrie(contacts) query = 'gekk' # Note that the user will enter 'g' then 'e' so # first display all the strings with prefix as 'g' # and then all the strings with prefix as 'ge' displayContacts(query) # This code is contributed by Aman Kumar
// C# Program to Implement a Phone // Directory Using Trie Data Structure using System; using System.Collections.Generic; class TrieNode { // Each Trie Node contains a Map 'child' // where each alphabet points to a Trie // Node. public Dictionary<char TrieNode> child; // 'isLast' is true if the node represents // end of a contact public bool isLast; // Default Constructor public TrieNode() { child = new Dictionary<char TrieNode>(); // Initialize all the Trie nodes with NULL for (char i = 'a'; i <= 'z'; i++) child.Add(i null); isLast = false; } } class Trie { public TrieNode root; // Insert all the Contacts into the Trie public void insertIntoTrie(String []contacts) { root = new TrieNode(); int n = contacts.Length; for (int i = 0; i < n; i++) { insert(contacts[i]); } } // Insert a Contact into the Trie public void insert(String s) { int len = s.Length; // 'itr' is used to iterate the Trie Nodes TrieNode itr = root; for (int i = 0; i < len; i++) { // Check if the s[i] is already present in // Trie TrieNode nextNode = itr.child[s[i]]; if (nextNode == null) { // If not found then create a new TrieNode nextNode = new TrieNode(); // Insert into the Dictionary if(itr.child.ContainsKey(s[i])) itr.child[s[i]] = nextNode; else itr.child.Add(s[i] nextNode); } // Move the iterator('itr') to point to next // Trie Node itr = nextNode; // If its the last character of the string 's' // then mark 'isLast' as true if (i == len - 1) itr.isLast = true; } } // This function simply displays all dictionary words // going through current node. String 'prefix' // represents string corresponding to the path from // root to curNode. public void displayContactsUtil(TrieNode curNode String prefix) { // Check if the string 'prefix' ends at this Node // If yes then display the string found so far if (curNode.isLast) Console.WriteLine(prefix); // Find all the adjacent Nodes to the current // Node and then call the function recursively // This is similar to performing DFS on a graph for (char i = 'a'; i <= 'z'; i++) { TrieNode nextNode = curNode.child[i]; if (nextNode != null) { displayContactsUtil(nextNode prefix + i); } } } // Display suggestions after every character enter by // the user for a given string 'str' public void displayContacts(String str) { TrieNode prevNode = root; // 'flag' denotes whether the string entered // so far is present in the Contact List String prefix = ''; int len = str.Length; // Display the contact List for string formed // after entering every character int i; for (i = 0; i < len; i++) { // 'str' stores the string entered so far prefix += str[i]; // Get the last character entered char lastChar = prefix[i]; // Find the Node corresponding to the last // character of 'str' which is pointed by // prevNode of the Trie TrieNode curNode = prevNode.child[lastChar]; // If nothing found then break the loop as // no more prefixes are going to be present. if (curNode == null) { Console.WriteLine('nNo Results Found for ' + prefix); i++; break; } // If present in trie then display all // the contacts with given prefix. Console.WriteLine('nSuggestions based on ' + prefix + ' are '); displayContactsUtil(curNode prefix); // Change prevNode for next prefix prevNode = curNode; } for ( ; i < len; i++) { prefix += str[i]; Console.WriteLine('nNo Results Found for ' + prefix); } } } // Driver code public class GFG { public static void Main(String []args) { Trie trie = new Trie(); String []contacts = {'gforgeeks' 'geeksquiz'}; trie.insertIntoTrie(contacts); String query = 'gekk'; // Note that the user will enter 'g' then 'e' so // first display all the strings with prefix as 'g' // and then all the strings with prefix as 'ge' trie.displayContacts(query); } } // This code is contributed by PrinciRaj1992
<script> // Javascript Program to Implement a Phone // Directory Using Trie Data Structure class TrieNode { constructor() { // Each Trie Node contains a Map 'child' // where each alphabet points to a Trie // Node. // We can also use a fixed size array of // size 256. this.child = {}; // 'isLast' is true if the node represents // end of a contact this.isLast = false; } } // Making root NULL for ease so that it doesn't // have to be passed to all functions. let root = null; // Insert a Contact into the Trie function insert(s) { const len = s.length; // 'itr' is used to iterate the Trie Nodes let itr = root; for (let i = 0; i < len; i++) { // Check if the s[i] is already present in // Trie const char = s[i]; let nextNode = itr.child[char]; if (nextNode === undefined) { // If not found then create a new TrieNode nextNode = new TrieNode(); // Insert into the Map itr.child[char] = nextNode; } // Move the iterator('itr') to point to next // Trie Node itr = nextNode; // If its the last character of the string 's' // then mark 'isLast' as true if (i === len - 1) { itr.isLast = true; } } } // This function simply displays all dictionary words // going through current node. String 'prefix' // represents string corresponding to the path from // root to curNode. function displayContactsUtil(curNode prefix) { // Check if the string 'prefix' ends at this Node // If yes then display the string found so far if (curNode.isLast) { document.write(prefix+'
'); } // Find all the adjacent Nodes to the current // Node and then call the function recursively // This is similar to performing DFS on a graph for (let i = 97; i <= 122; i++) { const char = String.fromCharCode(i); const nextNode = curNode.child[char]; if (nextNode !== undefined) { displayContactsUtil(nextNode prefix + char); } } } // Display suggestions after every character enter by // the user for a given query string 'str' function displayContacts(str) { let prevNode = root; let prefix = ''; const len = str.length; // Display the contact List for string formed // after entering every character let i; for (i = 0; i < len; i++) { // 'prefix' stores the string formed so far prefix += str[i]; // Get the last character entered const lastChar = prefix[i]; // Find the Node corresponding to the last // character of 'prefix' which is pointed by // prevNode of the Trie const curNode = prevNode.child[lastChar]; // If nothing found then break the loop as // no more prefixes are going to be present. if (curNode === undefined) { document.write(`No Results Found for ${prefix}`+'
'); i++; break; } // If present in trie then display all // the contacts with given prefix. document.write(`Suggestions based on ${prefix} are `); displayContactsUtil(curNode prefix); document.write('
'); // Change prevNode for next prefix prevNode = curNode; } document.write('
'); // Once search fails for a prefix we print // 'Not Results Found' for all remaining // characters of current query string 'str'. for (; i < len; i++) { prefix += str[i]; document.write('No Results Found for ' + prefix + '
'); } } // Insert all the Contacts into the Trie function insertIntoTrie(contacts) { // Initialize root Node root = new TrieNode(); const n = contacts.length; // Insert each contact into the trie for (let i = 0; i < n; i++) { insert(contacts[i]); } } // Driver program to test above functions // Contact list of the User const contacts = ['gforgeeks' 'geeksquiz']; //Insert all the Contacts into Trie insertIntoTrie(contacts); const query = 'gekk'; // Note that the user will enter 'g' then 'e' so // first display all the strings with prefix as 'g' // and then all the strings with prefix as 'ge' displayContacts(query); // This code is contributed by Utkarsh Kumar. </script>
Sortir
Suggestions based on gare geeksquiz gforgeeks Suggestions based on geare geeksquiz No Results Found for gek No Results Found for gekk
Complexité temporelle : O(n*m) où n est le nombre de contacts et m est la longueur maximale d'une chaîne de contacts.
Espace auxiliaire : O(n*m)