UDRULLET LINKET LISTE | SÆT 1 (INTRODUKTION)

Ligesom array og linket liste er den udrullede linkede liste også en lineær datastruktur og er en variant af en linket liste.

Hvorfor har vi brug for en udrullet linket liste?

En af de største fordele ved sammenkædede lister i forhold til arrays er, at indsættelse af et element på et hvilket som helst sted kun tager O(1). Men fangsten her er, at det kræver O(n) at søge efter et element i en sammenkædet liste. Så for at løse problemet med at søge, dvs. at reducere tiden til at søge i elementet, blev konceptet med udrullede linkede lister fremsat. Den udrullede sammenkædede liste dækker fordelene ved både array og sammenkædet liste, da den reducerer hukommelsesomkostningerne sammenlignet med simple sammenkædede lister ved at gemme flere elementer ved hver node, og den har også fordelen ved hurtig indsættelse og sletning som en sammenkædet liste.

Udrullet linket liste | Sæt 1 (introduktion) unrolledlinked list' title=

Fordele:

På grund af Cache-adfærden er lineær søgning meget hurtigere i udrullede sammenkædede lister.
Sammenlignet med den almindelige linkede liste kræver det mindre lagerplads til pointere/referencer.
Den udfører operationer som indsættelsessletning og gennemgang hurtigere end almindelige linkede lister (fordi søgningen er hurtigere).

Ulemper:

python bytes til streng

Overhead pr. node er forholdsvis høj end enkelt-linkede lister. Se en eksempelknude i koden nedenfor

Eksempel: Lad os sige, at vi har 8 elementer, så sqrt(8)=2,82, som rundes af til 3. Så hver blok vil gemme 3 elementer. Derfor vil der for at gemme 8 elementer blive oprettet 3 blokke, hvoraf de første to blokke vil gemme 3 elementer, og den sidste blok vil gemme 2 elementer.

Hvordan søgning bliver bedre i udrullede linkede lister?

Så tager vi ovenstående eksempel, hvis vi vil søge efter det 7. element i listen, krydser vi listen af blokke til den, der indeholder det 7. element. Det kræver kun O(sqrt(n)), da vi fandt det ved ikke at gå mere end sqrt(n) blokke.

Simpel implementering:

Nedenstående program opretter en simpel udrullet sammenkædet liste med 3 noder, der indeholder et variabelt antal elementer i hver. Den krydser også den oprettede liste.

C++

// C++ program to implement unrolled linked list  // and traversing it.  #include    using namespace std; #define maxElements 4  // Unrolled Linked List Node  class Node  {   public:  int numElements;   int array[maxElements];   Node *next;  };  /* Function to traverse an unrolled linked list  and print all the elements*/ void printUnrolledList(Node *n)  {   while (n != NULL)   {   // Print elements in current node   for (int i=0; i<n->numElements; i++)   cout<<n->array[i]<<' ';   // Move to next node   n = n->next;   }  }  // Program to create an unrolled linked list  // with 3 Nodes  int main()  {   Node* head = NULL;   Node* second = NULL;   Node* third = NULL;   // allocate 3 Nodes   head = new Node();  second = new Node();  third = new Node();  // Let us put some values in second node (Number   // of values must be less than or equal to   // maxElement)   head->numElements = 3;   head->array[0] = 1;   head->array[1] = 2;   head->array[2] = 3;   // Link first Node with the second Node   head->next = second;   // Let us put some values in second node (Number   // of values must be less than or equal to   // maxElement)   second->numElements = 3;   second->array[0] = 4;   second->array[1] = 5;   second->array[2] = 6;   // Link second Node with the third Node   second->next = third;   // Let us put some values in third node (Number   // of values must be less than or equal to   // maxElement)   third->numElements = 3;   third->array[0] = 7;   third->array[1] = 8;   third->array[2] = 9;   third->next = NULL;   printUnrolledList(head);   return 0;  }  // This is code is contributed by rathbhupendra

// C program to implement unrolled linked list // and traversing it. #include #include #define maxElements 4 // Unrolled Linked List Node struct Node {  int numElements;  int array[maxElements];  struct Node *next; }; /* Function to traverse an unrolled linked list  and print all the elements*/ void printUnrolledList(struct Node *n) {  while (n != NULL)  {  // Print elements in current node  for (int i=0; i<n->numElements; i++)  printf('%d ' n->array[i]);  // Move to next node   n = n->next;  } } // Program to create an unrolled linked list // with 3 Nodes int main() {  struct Node* head = NULL;  struct Node* second = NULL;  struct Node* third = NULL;  // allocate 3 Nodes  head = (struct Node*)malloc(sizeof(struct Node));  second = (struct Node*)malloc(sizeof(struct Node));  third = (struct Node*)malloc(sizeof(struct Node));  // Let us put some values in second node (Number  // of values must be less than or equal to  // maxElement)  head->numElements = 3;  head->array[0] = 1;  head->array[1] = 2;  head->array[2] = 3;  // Link first Node with the second Node  head->next = second;  // Let us put some values in second node (Number  // of values must be less than or equal to  // maxElement)  second->numElements = 3;  second->array[0] = 4;  second->array[1] = 5;  second->array[2] = 6;  // Link second Node with the third Node  second->next = third;  // Let us put some values in third node (Number  // of values must be less than or equal to  // maxElement)  third->numElements = 3;  third->array[0] = 7;  third->array[1] = 8;  third->array[2] = 9;  third->next = NULL;  printUnrolledList(head);  return 0; }

Java

// Java program to implement unrolled // linked list and traversing it.  import java.util.*; class GFG{   static final int maxElements = 4; // Unrolled Linked List Node  static class Node  {   int numElements;   int []array = new int[maxElements];   Node next;  };  // Function to traverse an unrolled  // linked list and print all the elements static void printUnrolledList(Node n)  {   while (n != null)   {     // Print elements in current node   for(int i = 0; i < n.numElements; i++)   System.out.print(n.array[i] + ' ');   // Move to next node   n = n.next;   }  }  // Program to create an unrolled linked list  // with 3 Nodes  public static void main(String[] args)  {   Node head = null;   Node second = null;   Node third = null;   // Allocate 3 Nodes   head = new Node();  second = new Node();  third = new Node();  // Let us put some values in second   // node (Number of values must be   // less than or equal to maxElement)   head.numElements = 3;   head.array[0] = 1;   head.array[1] = 2;   head.array[2] = 3;   // Link first Node with the   // second Node   head.next = second;   // Let us put some values in   // second node (Number of values  // must be less than or equal to   // maxElement)   second.numElements = 3;   second.array[0] = 4;   second.array[1] = 5;   second.array[2] = 6;   // Link second Node with the third Node   second.next = third;   // Let us put some values in third   // node (Number of values must be  // less than or equal to maxElement)   third.numElements = 3;   third.array[0] = 7;   third.array[1] = 8;   third.array[2] = 9;   third.next = null;   printUnrolledList(head);  }  }  // This code is contributed by amal kumar choubey

Python3

# Python3 program to implement unrolled # linked list and traversing it.  maxElements = 4 # Unrolled Linked List Node  class Node: def __init__(self): self.numElements = 0 self.array = [0 for i in range(maxElements)] self.next = None # Function to traverse an unrolled linked list  # and print all the elements def printUnrolledList(n): while (n != None): # Print elements in current node  for i in range(n.numElements): print(n.array[i] end = ' ') # Move to next node  n = n.next # Driver Code if __name__=='__main__': head = None second = None third = None # Allocate 3 Nodes  head = Node() second = Node() third = Node() # Let us put some values in second # node (Number of values must be  # less than or equal to  # maxElement)  head.numElements = 3 head.array[0] = 1 head.array[1] = 2 head.array[2] = 3 # Link first Node with the second Node  head.next = second # Let us put some values in second node # (Number of values must be less than # or equal to maxElement)  second.numElements = 3 second.array[0] = 4 second.array[1] = 5 second.array[2] = 6 # Link second Node with the third Node  second.next = third # Let us put some values in third node # (Number of values must be less than  # or equal to maxElement)  third.numElements = 3 third.array[0] = 7 third.array[1] = 8 third.array[2] = 9 third.next = None printUnrolledList(head) # This code is contributed by rutvik_56

// C# program to implement unrolled // linked list and traversing it.  using System; class GFG{   static readonly int maxElements = 4; // Unrolled Linked List Node  class Node  {   public int numElements;   public int []array = new int[maxElements];   public Node next;  };  // Function to traverse an unrolled  // linked list and print all the elements static void printUnrolledList(Node n)  {   while (n != null)   {   // Print elements in current node   for(int i = 0; i < n.numElements; i++)   Console.Write(n.array[i] + ' ');   // Move to next node   n = n.next;   }  }  // Program to create an unrolled linked list  // with 3 Nodes  public static void Main(String[] args)  {   Node head = null;   Node second = null;   Node third = null;   // Allocate 3 Nodes   head = new Node();  second = new Node();  third = new Node();  // Let us put some values in second   // node (Number of values must be   // less than or equal to maxElement)   head.numElements = 3;   head.array[0] = 1;   head.array[1] = 2;   head.array[2] = 3;   // Link first Node with the   // second Node   head.next = second;   // Let us put some values in   // second node (Number of values  // must be less than or equal to   // maxElement)   second.numElements = 3;   second.array[0] = 4;   second.array[1] = 5;   second.array[2] = 6;   // Link second Node with the third Node   second.next = third;   // Let us put some values in third   // node (Number of values must be  // less than or equal to maxElement)   third.numElements = 3;   third.array[0] = 7;   third.array[1] = 8;   third.array[2] = 9;   third.next = null;   printUnrolledList(head);  }  }  // This code is contributed by Rajput-Ji

JavaScript

<script>  // JavaScript program to implement unrolled  // linked list and traversing it.  const maxElements = 4;  // Unrolled Linked List Node  class Node {  constructor() {  this.numElements = 0;  this.array = new Array(maxElements);  this.next = null;  }  }  // Function to traverse an unrolled  // linked list and print all the elements  function printUnrolledList(n) {  while (n != null) {  // Print elements in current node  for (var i = 0; i < n.numElements; i++)  document.write(n.array[i] + ' ');  // Move to next node  n = n.next;  }  }  // Program to create an unrolled linked list  // with 3 Nodes  var head = null;  var second = null;  var third = null;  // Allocate 3 Nodes  head = new Node();  second = new Node();  third = new Node();  // Let us put some values in second  // node (Number of values must be  // less than or equal to maxElement)  head.numElements = 3;  head.array[0] = 1;  head.array[1] = 2;  head.array[2] = 3;  // Link first Node with the  // second Node  head.next = second;  // Let us put some values in  // second node (Number of values  // must be less than or equal to  // maxElement)  second.numElements = 3;  second.array[0] = 4;  second.array[1] = 5;  second.array[2] = 6;  // Link second Node with the third Node  second.next = third;  // Let us put some values in third  // node (Number of values must be  // less than or equal to maxElement)  third.numElements = 3;  third.array[0] = 7;  third.array[1] = 8;  third.array[2] = 9;  third.next = null;  printUnrolledList(head);   </script>

Produktion

1 2 3 4 5 6 7 8 9

Kompleksitetsanalyse:

Tidskompleksitet: O(n). Rumkompleksitet: O(n).

I denne artikel har vi introduceret en udrullet liste og fordelene ved den. Vi har også vist, hvordan man krydser listen. I den næste artikel vil vi diskutere indsættelsessletningen og værdierne af maxElements/numElements i detaljer.

Indsættelse i Unrolled Linked List