#practiceLinkDiv { görüntü: yok !önemli; }Bir telefon rehberinde bulunan kişilerin listesi verilmiştir. Görev, telefon rehberi için bir arama sorgusu uygulamaktır. Bir dizedeki arama sorgusu ' cadde ' öneki' olan tüm kişileri görüntüler cadde '. Arama fonksiyonunun özel bir özelliği, bir kullanıcı kişi listesinden bir kişiyi aradığında, kullanıcı her karakteri girdikten sonra önerilerin (bunun için girilen dize olarak öneki olan kişiler) gösterilmesidir.
Not: Örnek:
bilgisayar nasıl çalışır
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' Önerilen: Lütfen sorunu çözün PRATİK Çözüme geçmeden önce ilk olarak.
Üçlü Veri Yapısı. Tüm kişileri Trie'ye ekliyoruz. Genellikle bir Trie'deki arama sorgusu, dizenin trie'de mevcut olup olmadığını belirlemek içindir ancak bu durumda bizden her 'str' önekine sahip tüm dizeleri bulmamız istenir. Bu, bir şey yapmaya eşdeğerdir Grafikte DFS geçişi . Bir Trie düğümünden bitişik Trie düğümlerini ziyaret edin ve bunu, artık bitişik düğüm kalmayana kadar yinelemeli olarak yapın. Bu özyinelemeli işlev, biri ziyaret edilen geçerli Trie Düğümünü gösteren Trie Düğümü, diğeri ise şu ana kadar bulunan dizeyi 'str' önekiyle saklayan dize olarak 2 argüman alacaktır. Her Trie Düğümü, düğüm bir kişinin (kelime) sonunu temsil ediyorsa doğru olan 'isLast' boolean değişkenini saklar.
// 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)
Kullanıcı stringi karakter karakter girecek ve girilen her karakterden sonra oluşan önekle önerileri görüntülememiz gerekiyor. Bu nedenle, oluşturulan dizeyle başlayan öneki bulmaya yönelik bir yaklaşım, önekin Trie'de var olup olmadığını kontrol etmek, eğer evetse, ardından displayContacts() işlevini çağırmaktır. Bu yaklaşımda girilen her karakterden sonra dizenin Trie'de var olup olmadığını kontrol ederiz. Tekrar tekrar kontrol etmek yerine bir işaretçiyi koruyabiliriz öncekiDüğüm Kullanıcı tarafından girilen son karaktere karşılık gelen TrieNode'a işaret eden 'prevNode'un alt düğümünü, kullanıcı başka bir karakter girdiğinde Trie'de var olup olmadığını kontrol etmek için kontrol etmemiz gerekiyor. Yeni önek Trie'de değilse 'önek'ten sonra karakterlerin girilmesiyle oluşturulan dizelerin tümü Trie'de de bulunamaz. Böylece önekleri tek tek oluşturmak için kullanılan döngüyü kırıyoruz ve kalan tüm karakterlere 'Sonuç Bulunamadı' yazdırıyoruz.
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>
Çıkış
Suggestions based on gare geeksquiz gforgeeks Suggestions based on geare geeksquiz No Results Found for gek No Results Found for gekk
Zaman Karmaşıklığı: O(n*m) burada n, kontakların sayısı ve m, bir kontak dizisinin maksimum uzunluğudur.
Yardımcı Alan: O(n*m)