#practiceLinkDiv { display: none !important; }Gegeben ist eine Liste von Kontakten, die in einem Telefonverzeichnis vorhanden sind. Die Aufgabe besteht darin, eine Suchanfrage für das Telefonbuch umzusetzen. Die Suchanfrage nach einer Zeichenfolge „ str ' zeigt alle Kontakte an, deren Präfixe ' str '. Eine besondere Eigenschaft der Suchfunktion besteht darin, dass, wenn ein Benutzer nach einem Kontakt aus der Kontaktliste sucht, Vorschläge (Kontakte mit Präfix als eingegebene Zeichenfolge) angezeigt werden, nachdem der Benutzer jedes Zeichen eingegeben hat.
Notiz: Die Kontakte in der Liste bestehen nur aus Kleinbuchstaben. Beispiel:
Sara Ali Khan Alter
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' Empfohlen: Bitte lösen Sie es auf ÜBEN bevor Sie mit der Lösung fortfahren.
Das Telefonverzeichnis kann effizient implementiert werden mit Versuchen Sie es Datenstruktur. Wir fügen alle Kontakte in Trie ein. Im Allgemeinen dient eine Suchabfrage bei einem Trie dazu, festzustellen, ob die Zeichenfolge im Trie vorhanden ist oder nicht. In diesem Fall werden wir jedoch gebeten, alle Zeichenfolgen mit jedem Präfix „str“ zu finden. Dies entspricht der Ausführung von a DFS-Durchlauf in einem Diagramm . Besuchen Sie von einem Trie-Knoten aus benachbarte Trie-Knoten und führen Sie dies rekursiv aus, bis keine benachbarten Knoten mehr vorhanden sind. Diese rekursive Funktion benötigt zwei Argumente: eines als Trie-Knoten, der auf den aktuell besuchten Trie-Knoten verweist, und das andere als Zeichenfolge, die die bisher gefundene Zeichenfolge mit dem Präfix „str“ speichert. Jeder Trie-Knoten speichert eine boolesche Variable „isLast“, die wahr ist, wenn der Knoten das Ende eines Kontakts (Worts) darstellt.
// 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)
Der Benutzer gibt die Zeichenfolge Zeichen für Zeichen ein und wir müssen Vorschläge mit dem Präfix anzeigen, das nach jedem eingegebenen Zeichen gebildet wird. Ein Ansatz, um das Präfix zu finden, das mit der gebildeten Zeichenfolge beginnt, besteht darin, zu prüfen, ob das Präfix im Trie vorhanden ist. Wenn ja, rufen Sie dann die Funktion displayContacts() auf. Bei diesem Ansatz prüfen wir nach jedem eingegebenen Zeichen, ob die Zeichenfolge im Trie vorhanden ist. Anstatt immer wieder zu überprüfen, können wir einen Zeiger beibehalten prevNode ', das auf den TrieNode verweist, der dem zuletzt vom Benutzer eingegebenen Zeichen entspricht. Jetzt müssen wir den untergeordneten Knoten auf den 'prevNode' überprüfen, wenn der Benutzer ein anderes Zeichen eingibt, um zu überprüfen, ob es im Trie vorhanden ist. Wenn das neue Präfix nicht im Trie enthalten ist, können alle Zeichenfolgen, die durch die Eingabe von Zeichen nach „Präfix“ gebildet werden, auch nicht im Trie gefunden werden. Deshalb unterbrechen wir die Schleife, die zum Generieren von Präfixen nacheinander verwendet wird, und geben für alle verbleibenden Zeichen „Kein Ergebnis gefunden“ aus.
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>
Ausgabe
Suggestions based on gare geeksquiz gforgeeks Suggestions based on geare geeksquiz No Results Found for gek No Results Found for gekk
Zeitkomplexität: O(n*m) wobei n die Anzahl der Kontakte und m die maximale Länge einer Kontaktzeichenfolge ist.
Hilfsraum: O(n*m)