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전화번호부 구현

GfG Practice에서 사용해 보세요. ' title= #practiceLinkDiv { 표시: 없음 !중요; }

전화번호부에 존재하는 연락처 목록이 제공됩니다. 작업은 전화번호부에 대한 검색 쿼리를 구현하는 것입니다. 문자열 '에 대한 검색어 str ’는 접두사가 ‘로 붙은 모든 연락처를 표시합니다. str '. 검색 기능의 특별한 속성 중 하나는 사용자가 연락처 목록에서 연락처를 검색할 때 사용자가 각 문자를 입력한 후 제안 사항(입력한 문자열로 접두사가 있는 연락처)이 표시된다는 것입니다.
메모: 목록의 연락처는 소문자 알파벳으로만 구성됩니다. 예:

자바의 char + int
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' 

권장 사항: 해결해 보세요. 관행 먼저 솔루션으로 넘어가기 전에

전화번호부는 다음을 사용하여 효율적으로 구현할 수 있습니다. 트라이 데이터 구조. 모든 연락처를 Trie에 삽입합니다. 일반적으로 Trie에 대한 검색 쿼리는 문자열이 trie에 존재하는지 여부를 확인하는 것이지만 이 경우 각 접두사 'str'이 있는 모든 문자열을 찾으라는 요청을 받습니다. 이는 다음을 수행하는 것과 동일합니다. 그래프의 DFS 순회 . Trie 노드에서 인접한 Trie 노드를 방문하고 더 이상 인접한 노드가 없을 때까지 이 작업을 반복적으로 수행합니다. 이 재귀 함수는 두 개의 인수를 사용합니다. 하나는 방문 중인 현재 Trie 노드를 가리키는 Trie 노드이고 다른 하나는 'str'이라는 접두사를 사용하여 지금까지 발견된 문자열을 저장하는 문자열입니다. 각 Trie 노드는 노드가 연락처(단어)의 끝을 나타내는 경우 true인 부울 변수 'isLast'를 저장합니다.

// 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)

사용자는 문자열을 문자별로 입력하며 입력한 모든 문자 뒤에 접두사가 형성된 제안을 표시해야 합니다. 따라서 형성된 문자열로 시작하는 접두어를 찾는 한 가지 방법은 접두어가 Trie에 존재하는지 확인한 다음, 그렇다면 displayContacts() 함수를 호출하는 것입니다. 이 접근 방식에서는 입력된 모든 문자 후에 해당 문자열이 Trie에 존재하는지 확인합니다. 계속해서 확인하는 대신 포인터를 유지할 수 있습니다. 이전노드 '는 사용자가 마지막으로 입력한 문자에 해당하는 TrieNode를 가리키는데 이제 사용자가 다른 문자를 입력하면 Trie에 존재하는지 확인하기 위해 'prevNode'에 대한 하위 노드를 확인해야 합니다. 새 접두사가 Trie에 없으면 '접두사' 뒤에 문자를 입력하여 형성된 모든 문자열을 Trie에서도 찾을 수 없습니다. 따라서 접두사를 하나씩 생성하는 데 사용되는 루프를 중단하고 나머지 모든 문자에 대해 '결과를 찾을 수 없음'을 인쇄합니다. 



C++
// C++ Program to Implement a Phone // Directory Using Trie Data Structure #include    using namespace std; struct TrieNode {  // 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.  unordered_map<char TrieNode*> child;  // 'isLast' is true if the node represents  // end of a contact  bool isLast;  // Default Constructor  TrieNode()  {  // Initialize all the Trie nodes with NULL  for (char i = 'a'; i <= 'z'; i++)  child[i] = NULL;  isLast = false;  } }; // Making root NULL for ease so that it doesn't // have to be passed to all functions. TrieNode* root = NULL; // Insert a Contact into the Trie 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 Map  itr->child[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. 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)  cout << prefix << endl;  // 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 + (char)i);  } } // Display suggestions after every character enter by // the user for a given query string 'str' void displayContacts(string str) {  TrieNode* prevNode = root;  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++) {  // 'prefix' stores the string formed so far  prefix += (char)str[i];  // Get the last character entered  char lastChar = prefix[i];  // Find the Node corresponding to the last  // character of 'prefix' 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) {  cout << 'nNo Results Found for ' << prefix  << 'n';  i++;  break;  }  // If present in trie then display all  // the contacts with given prefix.  cout << 'nSuggestions based on ' << prefix  << 'are ';  displayContactsUtil(curNode prefix);  // Change prevNode for next prefix  prevNode = curNode;  }  // 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 += (char)str[i];  cout << 'nNo Results Found for ' << prefix << 'n';  } } // Insert all the Contacts into the Trie void insertIntoTrie(string contacts[] int n) {  // Initialize root Node  root = new TrieNode();  // Insert each contact into the trie  for (int i = 0; i < n; i++)  insert(contacts[i]); } // Driver program to test above functions int main() {  // Contact list of the User  string contacts[] = { 'gforgeeks' 'geeksquiz' };  // Size of the Contact List  int n = sizeof(contacts) / sizeof(string);  // Insert all the Contacts into Trie  insertIntoTrie(contacts n);  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'  displayContacts(query);  return 0; } 
Java
// 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);  } } 
Python3
# 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#
// 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 
JavaScript
<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>

산출
Suggestions based on gare geeksquiz gforgeeks Suggestions based on geare geeksquiz No Results Found for gek No Results Found for gekk

시간 복잡도: O(n*m) 여기서 n은 접촉 수이고 m은 접촉 문자열의 최대 길이입니다.
보조 공간: O(n*m)