#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)