CONTROLEER OF EEN DUBBEL GEKOPPELDE LIJST MET KARAKTERS EEN PALINDROOM IS OF NIET

Gegeven een dubbel gekoppelde lijst van karakters de taak is om te controleren of de dubbel gekoppelde lijst a is palindroom of niet.

Voorbeelden:

Invoer:

Uitgang: WAAR
Uitleg: De lijst komt overeen met 'LEVEL', wat een palindroom is.
omkerende tekenreeks java

Invoer:

Uitgang: Vals
Uitleg: De lijst komt overeen met 'LEVES', wat geen palindroom is.

Benadering:

Het idee is om twee pointers te initialiseren: links (aanvankelijk ingesteld op hoofd) en rechts (aanvankelijk ingesteld op staart). Vergelijk de waarden van de twee wijzers while links is niet gelijk aan nul of links is verhuisd naar de volgende rechts. Als de waarden van de twee wijzers zijn gelijkwaardig beweging links naar de volgende aanwijzer en rechts naar de vorige aanwijzer. Anders retourneert false.

Hieronder vindt u de implementatie van de bovenstaande aanpak:

C++

// C++ program to check if a doubly  // linked list is palindrome. #include    using namespace std; class Node { public:  char data;  Node* prev *next;  Node (char x) {  data = x;  prev = nullptr;  next = nullptr;  } }; // Function that returns true if the // doubly linked list is a palindrome bool isPalindrome(Node* head) {  if (head == nullptr) return true;    // Find the tail ptr.  Node *left=head *right=head;  while (right->next != nullptr) {  right = right->next;  }    // Check if the doubly linked list is   // a palindrome.  while (left!=right && left->prev!=right) {    // If char mismatch return  // false.  if (left->data != right->data)  return false;    // Move the pointers  left = left->next;  right = right->prev;  }    return true; } int main() {    // Doubly Linked list:   // L <-> E <-> V <-> E <-> L  Node *head = new Node('L');  head->next = new Node('E');  head->next->prev = head;  head->next->next = new Node('V');  head->next->next->prev = head->next;  head->next->next->next = new Node('E');  head->next->next->next->prev = head->next->next;  head->next->next->next->next = new Node('L');  head->next->next->next->next->prev = head->next->next->next;  if (isPalindrome(head))  cout << 'True';  else  cout << 'False';  return 0; }

// C program to check if a doubly  // linked list is palindrome. #include  #include  struct Node {  char data;  struct Node* prev;  struct Node* next; }; // Function that returns true if the // doubly linked list is a palindrome int isPalindrome(struct Node* head) {  if (head == NULL) return 1;    // Find the tail ptr.  struct Node *left = head *right = head;  while (right->next != NULL) {  right = right->next;  }    // Check if the doubly linked list is   // a palindrome.  while (left != right && left->prev != right) {    // If char mismatch return  // false.  if (left->data != right->data)  return 0;    // Move the pointers  left = left->next;  right = right->prev;  }    return 1; } struct Node* createNode(char x) {  struct Node* newNode =   (struct Node*)malloc(sizeof(struct Node));  newNode->data = x;  newNode->prev = NULL;  newNode->next = NULL;  return newNode; } int main() {    // Doubly Linked list:   // L <-> E <-> V <-> E <-> L  struct Node *head = createNode('L');  head->next = createNode('E');  head->next->prev = head;  head->next->next = createNode('V');  head->next->next->prev = head->next;  head->next->next->next = createNode('E');  head->next->next->next->prev = head->next->next;  head->next->next->next->next = createNode('L');  head->next->next->next->next->prev = head->next->next->next;  if (isPalindrome(head))  printf('Truen');  else  printf('Falsen');  return 0; }

Java

// Java program to check if a doubly  // linked list is palindrome. class Node {  char data;  Node prev next;  Node(char x) {  data = x;  prev = null;  next = null;  } } class GfG {  // Function that returns true if the  // doubly linked list is a palindrome  static boolean isPalindrome(Node head) {  if (head == null) return true;    // Find the tail ptr.  Node left = head right = head;  while (right.next != null) {  right = right.next;  }    // Check if the doubly linked list is   // a palindrome.  while (left != right && left.prev != right) {    // If char mismatch return  // false.  if (left.data != right.data)  return false;    // Move the pointers  left = left.next;  right = right.prev;  }    return true;  }  public static void main(String[] args) {  // Doubly Linked list:   // L <-> E <-> V <-> E <-> L  Node head = new Node('L');  head.next = new Node('E');  head.next.prev = head;  head.next.next = new Node('V');  head.next.next.prev = head.next;  head.next.next.next = new Node('E');  head.next.next.next.prev = head.next.next;  head.next.next.next.next = new Node('L');  head.next.next.next.next.prev = head.next.next.next;  if (isPalindrome(head))  System.out.println('True');  else  System.out.println('False');  } }

Python

# Python program to check if a doubly  # linked list is palindrome. class Node: def __init__(self x): self.data = x self.prev = None self.next = None # Function that returns true if the # doubly linked list is a palindrome def isPalindrome(head): if head is None: return True # Find the tail ptr. left = head right = head while right.next is not None: right = right.next # Check if the doubly linked list is  # a palindrome. while left != right and left.prev != right: # If char mismatch return # false. if left.data != right.data: return False # Move the pointers left = left.next right = right.prev return True if __name__ == '__main__': # Doubly Linked list:  # L <-> E <-> V <-> E <-> L head = Node('L') head.next = Node('E') head.next.prev = head head.next.next = Node('V') head.next.next.prev = head.next head.next.next.next = Node('E') head.next.next.next.prev = head.next.next head.next.next.next.next = Node('L') head.next.next.next.next.prev = head.next.next.next if isPalindrome(head): print('True') else: print('False')

// C# program to check if a doubly  // linked list is palindrome. using System; class Node {  public char data;  public Node prev next;  public Node(char x) {  data = x;  prev = null;  next = null;  } } class GfG {  // Function that returns true if the  // doubly linked list is a palindrome  static bool isPalindrome(Node head) {  if (head == null) return true;    // Find the tail ptr.  Node left = head right = head;  while (right.next != null) {  right = right.next;  }    // Check if the doubly linked list is   // a palindrome.  while (left != right && left.prev != right) {    // If char mismatch return  // false.  if (left.data != right.data)  return false;    // Move the pointers  left = left.next;  right = right.prev;  }    return true;  }  static void Main(string[] args) {  // Doubly Linked list:   // L <-> E <-> V <-> E <-> L  Node head = new Node('L');  head.next = new Node('E');  head.next.prev = head;  head.next.next = new Node('V');  head.next.next.prev = head.next;  head.next.next.next = new Node('E');  head.next.next.next.prev = head.next.next;  head.next.next.next.next = new Node('L');  head.next.next.next.next.prev = head.next.next.next;    if (isPalindrome(head))  Console.WriteLine('True');  else  Console.WriteLine('False');  } }

JavaScript

// JavaScript program to check if a doubly  // linked list is palindrome. class Node {  constructor(x) {  this.data = x;  this.prev = null;  this.next = null;  } } // Function that returns true if the // doubly linked list is a palindrome function isPalindrome(head) {  if (head === null) return true;    // Find the tail ptr.  let left = head right = head;  while (right.next !== null) {  right = right.next;  }    // Check if the doubly linked list is   // a palindrome.  while (left !== right && left.prev !== right) {    // If char mismatch return  // false.  if (left.data !== right.data)  return false;    // Move the pointers  left = left.next;  right = right.prev;  }    return true; } // Doubly Linked list:  // L <-> E <-> V <-> E <-> L let head = new Node('L'); head.next = new Node('E'); head.next.prev = head; head.next.next = new Node('V'); head.next.next.prev = head.next; head.next.next.next = new Node('E'); head.next.next.next.prev = head.next.next; head.next.next.next.next = new Node('L'); head.next.next.next.next.prev = head.next.next.next; if (isPalindrome(head))  console.log('True'); else  console.log('False');

Uitvoer

True

Tijdcomplexiteit: O(n) waarbij n het aantal knooppunten in de dubbel gekoppelde lijst is.
Hulpruimte: O(1)

in.volgende java

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