IŠVYNIOTAS SUSIETŲ SĄRAŠAS | 1 RINKINYS (ĮVADAS)

Kaip ir masyvas ir susietasis sąrašas, išvyniotas susietasis sąrašas taip pat yra linijinė duomenų struktūra ir susieto sąrašo variantas.

Kodėl mums reikia išvynioto susieto sąrašo?

Vienas didžiausių susietų sąrašų pranašumų, palyginti su masyvais, yra tas, kad elemento įterpimas į bet kurią vietą užtrunka tik O(1). Tačiau svarbiausia yra tai, kad norint ieškoti elemento susietame sąraše, reikia O(n). Taigi, siekiant išspręsti paieškos problemą, ty sutrumpinti elemento paieškos laiką, buvo pasiūlyta išvyniotų susietų sąrašų koncepcija. Išvyniotas susietas sąrašas apima ir masyvo, ir susieto sąrašo pranašumus, nes sumažina atminties sąnaudas, palyginti su paprastais susietais sąrašais, nes kiekviename mazge saugo kelis elementus, taip pat turi greito įterpimo ir ištrynimo pranašumą, kaip ir susieto sąrašo.

Java nuorodų tipai

Išvyniotas susietų sąrašas | 1 rinkinys (įvadas) išvyniotas nuorodų sąrašas' title=

Privalumai:

Dėl talpyklos veikimo linijinė paieška yra daug greitesnė išvyniotuose susietuose sąrašuose.
Palyginti su įprastu susietu sąrašu, jam reikia mažiau vietos nuorodoms / nuorodoms laikyti.
Jis atlieka tokias operacijas kaip įterpimo ištrynimas ir perėjimas greičiau nei įprasti susieti sąrašai (nes paieška yra greitesnė).

Trūkumai:

10 iš 50

Vieno mazgo pridėtinės išlaidos yra palyginti didesnės nei atskirai susietų sąrašų. Žr. pavyzdinį mazgą toliau pateiktame kode

Pavyzdys: Tarkime, kad turime 8 elementus, taigi sqrt(8)=2.82, kuris suapvalinamas iki 3. Taigi kiekviename bloke bus saugomi 3 elementai. Taigi 8 elementams saugoti bus sukurti 3 blokai, iš kurių pirmuose dviejuose blokuose bus saugomi 3 elementai, o paskutiniame bloke – 2 elementai.

Kaip paieška tampa geresnė išvyniotuose susietuose sąrašuose?

Taigi, atsižvelgiant į aukščiau pateiktą pavyzdį, jei norime ieškoti 7-ojo sąrašo elemento, perkeliame blokų sąrašą į tą, kuriame yra 7-asis elementas. Tai užtrunka tik O(sqrt(n)), nes radome ne daugiau nei sqrt(n) blokus.

Paprastas įgyvendinimas:

Toliau pateikta programa sukuria paprastą išvyniotą susietą sąrašą su 3 mazgais, kurių kiekviename yra kintamas elementų skaičius. Jis taip pat kerta sukurtą sąrašą.

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>

Išvestis

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Sudėtingumo analizė:

Laiko sudėtingumas: O(n). Erdvės sudėtingumas: O(n).

Šiame straipsnyje mes pristatėme išvyniotą sąrašą ir jo pranašumus. Taip pat parodėme, kaip pereiti sąrašą. Kitame straipsnyje mes išsamiai aptarsime įterpimo ištrynimą ir maxElements / numElements reikšmes.

Įterpimas į išvyniotą susietų sąrašą

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