logo

TechCodeview

Обришите средину повезане листе

С обзиром на појединачну листу задатак је да избрише средњи чвор листе.

  • Ако листа садржи парни број чворова, биће два средња чворова. У овом случају избришите други средњи чвор.
  • Ако је повезана листа састоји од само једног чвора, а затим се врати нулл.

Пример:



Улаз: Лингедлист: 1-> 2-> 3-> 4-> 5
Излаз: 1-> 2-> 4-> 5
Објашњење:

Обришите средину повезане листе

Улаз: Лингедлист: 2-> 4-> 6-> 7-> 5-> 1
Излаз: 2-> 4-> 6-> 5-> 1
Објашњење:

Обришите средину повезане листе

Улаз: Лингедлист: 7
Излаз:



Табела садржаја

[Наивни приступ] Коришћење двотрашњег прелазног прелаза - О (Н) Време и О (1) Простор

Основна идеја овог приступа је прва да пређе целу повезану листу да бисте пребројали укупан број чворова. Једном када знамо укупан број чворова, можемо израчунати положај средњег чвора који је у индексу Н / 2 (где је н укупан број чворова). Затим поново прођите кроз повезану листу, али овај пут заустављамо непосредно пре средњег чвора. Једном када тамо модификујемо следећи показивач чвора пре средњег чвора тако да прескочи преко средње чвора и бодове директно на чвор након њега

Испод је примена горе наведеног приступа:



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);

Излаз 
Original Linked List: 1 -> 2 -> 3 -> 4 -> 5 -> nullptr Linked List after deleting the middle node: 1 -> 2 -> 4 -> 5 -> nullptr

Сложеност времена: О (н). Потребне су два преласка повезане листе
Помоћни простор: О (1). Није потребан додатни простор.

[Очекивани приступ] Једноставни прелаз са спорим и брзим показивачима - о (н) време и О (1) простора

Горње решење захтева два преласка повезане листе. Средњи чвор може се избрисати помоћу једног преласка. Идеја је да се користи два показивача спор_птр и фаст_птр . Брзи показивач помера два чворова у исто време док спор показивач помера један по један по један. Када брзи показивач дође до краја листе, спор показивач ће бити постављен на средњем чвору. Затим морате да повежете чвор који долази пре средњег чвора ( прев ) На чвор који долази после средње чвора. Ово ефикасно прескаче преко средњег чвора уклањајући га са листе.

Испод је примена горе наведеног приступа

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);

Излаз 
Original Linked List: 1 -> 2 -> 3 -> 4 -> 5 -> NULL Linked List after deleting the middle node: 1 -> 2 -> 4 -> 5 -> NULL

Сложеност времена: О (н). Потребан је само један прелазак повезане листе
Помоћни простор: О (1). Јер није потребан додатни простор.

Сродни чланак:

  • Пронађите средину повезане листе