PĀRBAUDIET, VAI DIVKĀRŠI SAISTĪTS RAKSTZĪMJU SARAKSTS IR PALINDROMS

Ņemot vērā a divkārši saistīts saraksts no rakstzīmes uzdevums ir pārbaudīt, vai divkārši saistīts saraksts ir a palindroms vai ne.

Piemēri:

Ievade:

Izvade: Taisnība
Paskaidrojums: Saraksts atbilst 'LEVEL', kas ir palindroms.
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Ievade:

Izvade: Nepatiesi
Paskaidrojums: Saraksts atbilst “LEVES”, kas nav palindroms.

Pieeja:

Ideja ir inicializēt divus rādītājus: pa kreisi (sākotnēji iestatīts uz galvu) un pareizi (sākotnēji iestatīts uz asti). Salīdziniet abu rādītāju vērtības, kamēr pa kreisi nav vienāds ar null vai pa kreisi ir pārcēlies uz nākamo no pareizi. Ja abu rādītāju vērtības ir vienāds kustēties pa kreisi uz nākamo rādītāju un pareizi uz iepriekšējo rādītāju. Pretējā gadījumā atgrieziet nepatiesu.

Tālāk ir aprakstīta iepriekš minētās pieejas īstenošana.

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');

Izvade

True

Laika sarežģītība: O(n) kur n ir mezglu skaits divkārši saistītajā sarakstā.
Palīgtelpa: O(1)

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Saistītie raksti:

Funkcija, lai pārbaudītu, vai atsevišķi saistītais saraksts ir palindroms
Pārbaudiet, vai saistītais virkņu saraksts veido palindromu

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Pārbaudiet, vai divkārši saistīts rakstzīmju saraksts ir palindroms