بالنظر إلى قائمة مرتبطة بشكل فردي، تكون المهمة هي حذف العقدة الوسطى من القائمة.
- إذا كانت القائمة تحتوي على عدد زوجي من العقد، فسيكون هناك عقدتان وسطيان. في هذه الحالة، قم بحذف العقدة الوسطى الثانية.
- إذا كانت القائمة المرتبطة تتكون من عقدة واحدة فقط، فسيتم إرجاع NULL.
مثال:
مدخل: القائمة المرتبطة: 1-> 2-> 3-> 4-> 5
الإخراج: 1->2->4->5
توضيح:![]()
مدخل: القائمة المرتبطة: 2-> 4-> 6-> 7-> 5-> 1
الإخراج: 2->4->6->5->1
الشرح:![]()
مدخل: القائمة المرتبطة: 7
الإخراج:
جدول المحتويات
- [نهج ساذج] استخدام اجتياز ثنائي المسار - وقت O(n) ومساحة O(1).
- [النهج المتوقع] اجتياز مسار واحد بمؤشرات بطيئة وسريعة - وقت O(n) ومسافة O(1)
[نهج ساذج] استخدام اجتياز ثنائي المسار - وقت O(n) ومساحة O(1).
الفكرة الأساسية وراء هذا الأسلوب هي اجتياز القائمة المرتبطة بالكامل أولاً لحساب العدد الإجمالي للعقد. بمجرد أن نعرف العدد الإجمالي للعقد يمكننا حساب موضع العقدة الوسطى الموجودة في الفهرس ن/2 (حيث n هو العدد الإجمالي للعقد). ثم انتقل إلى القائمة المرتبطة مرة أخرى ولكن هذه المرة نتوقف مباشرة قبل العقدة الوسطى. وبمجرد الوصول إلى هناك، نقوم بتعديل المؤشر التالي للعقدة قبل العقدة الوسطى بحيث يتخطى العقدة الوسطى ويشير مباشرة إلى العقدة التي تليها
وفيما يلي تنفيذ النهج المذكور أعلاه:
C++
// C++ program to delete middle of a linked list #include using namespace std; struct Node { int data; Node* next; Node(int x){ data = x; next = nullptr; } }; // Function to delete middle node from linked list. Node* deleteMid(Node* head) { // Edge case: return nullptr if there is only // one node. if (head->next == nullptr) return nullptr; int count = 0; Node *p1 = head *p2 = head; // First pass count the number of nodes // in the linked list using 'p1'. while (p1 != nullptr) { count++; p1 = p1->next; } // Get the index of the node to be deleted. int middleIndex = count / 2; // Second pass let 'p2' move toward the predecessor // of the middle node. for (int i = 0; i < middleIndex - 1; ++i) p2 = p2->next; // Delete the middle node and return 'head'. p2->next = p2->next->next; return head; } void printList(Node* head) { Node* temp = head; while (temp != nullptr) { cout << temp->data << ' -> '; temp = temp->next; } cout << 'nullptr' << endl; } int main() { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5. Node* head = new Node(1); head->next = new Node(2); head->next->next = new Node(3); head->next->next->next = new Node(4); head->next->next->next->next = new Node(5); cout << 'Original Linked List: '; printList(head); // Delete the middle node. head = deleteMid(head); cout << 'Linked List after deleting the middle node: '; printList(head); return 0; }
C // C program to delete middle of a linked list #include #include struct Node { int data; struct Node* next; }; // Function to delete middle node from linked list. struct Node* deleteMid(struct Node* head) { // Edge case: return NULL if there is only // one node. if (head->next == NULL) return NULL; int count = 0; struct Node *p1 = head *p2 = head; // First pass count the number of nodes // in the linked list using 'p1'. while (p1 != NULL) { count++; p1 = p1->next; } // Get the index of the node to be deleted. int middleIndex = count / 2; // Second pass let 'p2' move toward the predecessor // of the middle node. for (int i = 0; i < middleIndex - 1; ++i) p2 = p2->next; // Delete the middle node and return 'head'. p2->next = p2->next->next; return head; } void printList(struct Node* head) { struct Node* temp = head; while (temp != NULL) { printf('%d -> ' temp->data); temp = temp->next; } printf('NULLn'); } struct Node* newNode(int x) { struct Node* temp = (struct Node*)malloc(sizeof(struct Node)); temp->data = x; temp->next = NULL; return temp; } int main() { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5. struct Node* head = newNode(1); head->next = newNode(2); head->next->next = newNode(3); head->next->next->next = newNode(4); head->next->next->next->next = newNode(5); printf('Original Linked List: '); printList(head); // Delete the middle node. head = deleteMid(head); printf('Linked List after deleting the middle node: '); printList(head); return 0; }
Java // Java program to delete middle of a linked list class Node { int data; Node next; Node(int x) { data = x; next = null; } } public class GfG { // Function to delete middle node from linked list. public static Node deleteMid(Node head) { // Edge case: return null if there is only // one node. if (head.next == null) return null; int count = 0; Node p1 = head p2 = head; // First pass count the number of nodes // in the linked list using 'p1'. while (p1 != null) { count++; p1 = p1.next; } // Get the index of the node to be deleted. int middleIndex = count / 2; // Second pass let 'p2' move toward predecessor // of the middle node. for (int i = 0; i < middleIndex - 1; ++i) p2 = p2.next; // Delete the middle node and return 'head'. p2.next = p2.next.next; return head; } public static void printList(Node head) { Node temp = head; while (temp != null) { System.out.print(temp.data + ' -> '); temp = temp.next; } System.out.println('null'); } public static void main(String[] args) { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5. Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); System.out.print('Original Linked List: '); printList(head); // Delete the middle node. head = deleteMid(head); System.out.print ('Linked List after deleting the middle node: '); printList(head); } }
Python # Python3 program to delete middle of a linked list class Node: def __init__(self data): self.data = data self.next = None # Function to delete middle node from linked list. def deleteMid(head): # Edge case: return None if there is only # one node. if head.next is None: return None count = 0 p1 = head p2 = head # First pass count the number of nodes # in the linked list using 'p1'. while p1 is not None: count += 1 p1 = p1.next # Get the index of the node to be deleted. middleIndex = count // 2 # Second pass let 'p2' move toward the predecessor # of the middle node. for i in range(middleIndex - 1): p2 = p2.next # Delete the middle node and return 'head'. p2.next = p2.next.next return head def printList(head): temp = head while temp is not None: print(temp.data end=' -> ') temp = temp.next print('None') if __name__ == '__main__': # Create a static hardcoded linked list: # 1 -> 2 -> 3 -> 4 -> 5. head = Node(1) head.next = Node(2) head.next.next = Node(3) head.next.next.next = Node(4) head.next.next.next.next = Node(5) print('Original Linked List:' end=' ') printList(head) # Delete the middle node. head = deleteMid(head) print('Linked List after deleting the middle node:' end=' ') printList(head)
C# // C# program to delete middle of a linked list using System; class Node { public int data; public Node next; public Node(int x) { data = x; next = null; } } class GfG { // Function to delete middle node from linked list. static Node deleteMid(Node head) { // Edge case: return null if there is only // one node. if (head.next == null) return null; int count = 0; Node p1 = head p2 = head; // First pass count the number of nodes // in the linked list using 'p1'. while (p1 != null) { count++; p1 = p1.next; } // Get the index of the node to be deleted. int middleIndex = count / 2; // Second pass let 'p2' move toward the predecessor // of the middle node. for (int i = 0; i < middleIndex - 1; ++i) p2 = p2.next; // Delete the middle node and return 'head'. p2.next = p2.next.next; return head; } static void printList(Node head) { Node temp = head; while (temp != null) { Console.Write(temp.data + ' -> '); temp = temp.next; } Console.WriteLine('null'); } static void Main(string[] args) { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5. Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); Console.Write('Original Linked List: '); printList(head); // Delete the middle node. head = deleteMid(head); Console.Write ('Linked List after deleting the middle node: '); printList(head); } }
JavaScript class Node { constructor(data) { this.data = data; this.next = null; } } // Function to delete middle node from linked list. function deleteMid(head) { // Edge case: return null if there is only // one node. if (head.next === null) return null; let count = 0; let p1 = head p2 = head; // First pass count the number of nodes // in the linked list using 'p1'. while (p1 !== null) { count++; p1 = p1.next; } // Get the index of the node to be deleted. let middleIndex = Math.floor(count / 2); // Second pass let 'p2' move toward the predecessor // of the middle node. for (let i = 0; i < middleIndex - 1; ++i) p2 = p2.next; // Delete the middle node and return 'head'. p2.next = p2.next.next; return head; } function printList(head) { let temp = head; while (temp !== null) { console.log(temp.data + ' -> '); temp = temp.next; } console.log('null'); } // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5. let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); console.log('Original Linked List: '); printList(head); // Delete the middle node. head = deleteMid(head); console.log('Linked List after deleting the middle node: '); printList(head);
الإخراج
Original Linked List: 1 -> 2 -> 3 -> 4 -> 5 -> nullptr Linked List after deleting the middle node: 1 -> 2 -> 4 -> 5 -> nullptr
تعقيد الوقت: على). هناك حاجة إلى اجتيازين للقائمة المرتبطة
المساحة المساعدة: يا(1). ليس هناك حاجة إلى مساحة إضافية.
[النهج المتوقع] اجتياز مسار واحد بمؤشرات بطيئة وسريعة - وقت O(n) ومسافة O(1)
يتطلب الحل أعلاه اجتياز القائمة المرتبطة مرتين. يمكن حذف العقدة الوسطى باستخدام اجتياز واحد. والفكرة هي استخدام مؤشرين Slow_ptr و fast_ptr . يحرك المؤشر السريع عقدتين في كل مرة بينما يتحرك المؤشر البطيء عقدة واحدة في كل مرة. عندما يصل المؤشر السريع إلى نهاية القائمة، سيتم وضع المؤشر البطيء في العقدة الوسطى. بعد ذلك، تحتاج إلى توصيل العقدة التي تأتي قبل العقدة الوسطى ( السابق ) إلى العقدة التي تأتي بعد العقدة الوسطى. يؤدي هذا إلى تخطي العقدة الوسطى بشكل فعال وإزالتها من القائمة.
وفيما يلي تنفيذ النهج المذكور أعلاه
C++// C++ program to delete middle of a linked list #include using namespace std; struct Node { int data; Node* next; Node(int x){ data = x; next = nullptr; } }; // Function to delete middle node from linked list struct Node* deleteMid(struct Node* head) { // If the list is empty return NULL if (head == NULL) return NULL; // If the list has only one node // delete it and return NULL if (head->next == NULL) { delete head; return NULL; } struct Node* prev = NULL; struct Node* slow_ptr = head; struct Node* fast_ptr = head; // Move the fast pointer 2 nodes ahead // and the slow pointer 1 node ahead // until fast pointer reaches end of the list while (fast_ptr != NULL && fast_ptr->next != NULL) { fast_ptr = fast_ptr->next->next; // Update prev to hold the previous // slow pointer value prev = slow_ptr; slow_ptr = slow_ptr->next; } // At this point slow_ptr points to middle node // Bypass the middle node prev->next = slow_ptr->next; // Delete the middle node delete slow_ptr; // Return the head of the modified list return head; } void printList(struct Node* head) { struct Node* temp = head; while (temp != NULL) { cout << temp->data << ' -> '; temp = temp->next; } cout << 'NULL' << endl; } int main() { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node* head = new Node(1); head->next = new Node(2); head->next->next = new Node(3); head->next->next->next = new Node(4); head->next->next->next->next = new Node(5); cout << 'Original Linked List: '; printList(head); // Delete the middle node head = deleteMid(head); cout << 'Linked List after deleting the middle node: '; printList(head); return 0; }
C // C program to delete middle of a linked list #include #include struct Node { int data; struct Node* next; }; // Function to delete middle node from linked list struct Node* deleteMid(struct Node* head) { // If the list is empty return NULL if (head == NULL) return NULL; // If the list has only one node // delete it and return NULL if (head->next == NULL) { free(head); return NULL; } struct Node* prev = NULL; struct Node* slow_ptr = head; struct Node* fast_ptr = head; // Move the fast pointer 2 nodes ahead // and the slow pointer 1 node ahead // until fast pointer reaches end of the list while (fast_ptr != NULL && fast_ptr->next != NULL) { fast_ptr = fast_ptr->next->next; // Update prev to hold the previous // slow pointer value prev = slow_ptr; slow_ptr = slow_ptr->next; } // At this point slow_ptr points to middle node // Bypass the middle node prev->next = slow_ptr->next; // Delete the middle node free(slow_ptr); // Return the head of the modified list return head; } void printList(struct Node* head) { struct Node* temp = head; while (temp != NULL) { printf('%d -> ' temp->data); temp = temp->next; } printf('NULLn'); } struct Node* newNode(int x) { struct Node* temp = (struct Node*)malloc(sizeof(struct Node)); temp->data = x; temp->next = NULL; return temp; } int main() { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5. struct Node* head = newNode(1); head->next = newNode(2); head->next->next = newNode(3); head->next->next->next = newNode(4); head->next->next->next->next = newNode(5); printf('Original Linked List: '); printList(head); // Delete the middle node. head = deleteMid(head); printf('Linked List after deleting the middle node: '); printList(head); return 0; }
Java // Java program to delete the middle of a linked list class Node { int data; Node next; Node(int x) { data = x; next = null; } } class GfG { // Function to delete middle node from linked list static Node deleteMid(Node head) { // If the list is empty return null if (head == null) return null; // If the list has only one node // delete it and return null if (head.next == null) { return null; } Node prev = null; Node slow_ptr = head; Node fast_ptr = head; // Move the fast pointer 2 nodes ahead // and the slow pointer 1 node ahead // until fast pointer reaches end of list while (fast_ptr != null && fast_ptr.next != null) { fast_ptr = fast_ptr.next.next; // Update prev to hold the previous // slow pointer value prev = slow_ptr; slow_ptr = slow_ptr.next; } // At this pointslow_ptr points to middle node // Bypass the middle node prev.next = slow_ptr.next; // Return the head of the modified list return head; } static void printList(Node head) { Node temp = head; while (temp != null) { System.out.print(temp.data + ' -> '); temp = temp.next; } System.out.println('NULL'); } public static void main(String[] args) { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); System.out.print('Original Linked List: '); printList(head); // Delete the middle node head = deleteMid(head); System.out.print ('Linked List after deleting the middle node: '); printList(head); } }
Python # Python program to delete the middle of a linked list class Node: def __init__(self data): self.data = data self.next = None # Function to delete middle node from linked list def deleteMid(head): # If the list is empty return None if head is None: return None # If the list has only one node # delete it and return None if head.next is None: return None prev = None slow_ptr = head fast_ptr = head # Move the fast pointer 2 nodes ahead # and the slow pointer 1 node ahead # until fast pointer reaches end of the list while fast_ptr is not None and fast_ptr.next is not None: fast_ptr = fast_ptr.next.next # Update prev to hold the previous # slow pointer value prev = slow_ptr slow_ptr = slow_ptr.next # At this point slow_ptr points to middle node # Bypass the middle node prev.next = slow_ptr.next # Return the head of the modified list return head def printList(head): temp = head while temp: print(temp.data end=' -> ') temp = temp.next print('NULL') if __name__ == '__main__': # Create a static hardcoded linked list: # 1 -> 2 -> 3 -> 4 -> 5 head = Node(1) head.next = Node(2) head.next.next = Node(3) head.next.next.next = Node(4) head.next.next.next.next = Node(5) print('Original Linked List: ' end='') printList(head) # Delete the middle node head = deleteMid(head) print('Linked List after deleting the middle node: ' end='') printList(head)
C# // C# program to delete middle of a linked list using System; class Node { public int data; public Node next; public Node(int x) { data = x; next = null; } } class GfG { // Function to delete middle node from linked list public static Node deleteMid(Node head) { // If the list is empty return null if (head == null) return null; // If the list has only one node // delete it and return null if (head.next == null) { return null; } Node prev = null; Node slow_ptr = head; Node fast_ptr = head; // Move the fast pointer 2 nodes ahead // and the slow pointer 1 node ahead // until fast pointer reaches end of the list while (fast_ptr != null && fast_ptr.next != null) { fast_ptr = fast_ptr.next.next; // Update prev to hold the previous // slow pointer value prev = slow_ptr; slow_ptr = slow_ptr.next; } // At this point slow_ptr points to middle node // Bypass the middle node prev.next = slow_ptr.next; // Return the head of the modified list return head; } // Function to print the linked list public static void printList(Node head) { Node temp = head; while (temp != null) { Console.Write(temp.data + ' -> '); temp = temp.next; } Console.WriteLine('NULL'); } public static void Main(string[] args) { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); Console.Write('Original Linked List: '); printList(head); // Delete the middle node head = deleteMid(head); Console.Write ('Linked List after deleting the middle node: '); printList(head); } }
JavaScript // javascript program to delete middle of a linked list class Node { constructor(data) { this.data = data; this.next = null; } } // Function to delete the middle node from the linked list function deleteMid(head) { // If the list is empty return null if (head === null) { return null; } // If the list has only one node delete it and return // null if (head.next === null) { return null; } let prev = null; let slow_ptr = head; let fast_ptr = head; // Move the fast pointer 2 nodes ahead // and the slow pointer 1 node ahead // until the fast pointer reaches the end of the list while (fast_ptr !== null && fast_ptr.next !== null) { fast_ptr = fast_ptr.next.next; // Update prev to hold the previous slow pointer // value prev = slow_ptr; slow_ptr = slow_ptr.next; } // At this point slow_ptr points to the middle node // Bypass the middle node prev.next = slow_ptr.next; // Return the head of the modified list return head; } function printList(head) { let temp = head; while (temp !== null) { process.stdout.write(temp.data + ' -> '); temp = temp.next; } console.log('null'); } // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5 let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); process.stdout.write('Original Linked List: '); printList(head); // Delete the middle node head = deleteMid(head); process.stdout.write( 'Linked List after deleting the middle node: '); printList(head);
الإخراج
Original Linked List: 1 -> 2 -> 3 -> 4 -> 5 -> NULL Linked List after deleting the middle node: 1 -> 2 -> 4 -> 5 -> NULL
تعقيد الوقت: على). هناك حاجة إلى اجتياز واحد فقط للقائمة المرتبطة
المساحة المساعدة: يا(1). حيث لا توجد حاجة إلى مساحة إضافية.
مقالة ذات صلة:
- ابحث عن منتصف القائمة المرتبطة