Create linked list from a given array

Given an array arr[] of size N. The task is to create linked list from the given array.

Examples:

Input : arr[]={1, 2, 3, 4, 5}
Output : 1->2->3->4->5

Input :arr[]={10, 11, 12, 13, 14}
Output : 10->11->12->13->14

Simple Approach: For each element of an array arr[] we create a node in a linked list and insert it at the end.

C++



filter_none

edit
close

play_arrow

link
brightness_4
code

#include <iostream>
using namespace std;
  
// Representation of a node
struct Node {
    int data;
    Node* next;
};
  
// Function to insert node
void insert(Node** root, int item)
{
    Node* temp = new Node;
    Node* ptr;
    temp->data = item;
    temp->next = NULL;
  
    if (*root == NULL)
        *root = temp;
    else {
        ptr = *root;
        while (ptr->next != NULL)
            ptr = ptr->next;
        ptr->next = temp;
    }
}
  
void display(Node* root)
{
    while (root != NULL) {
        cout << root->data << " ";
        root = root->next;
    }
}
  
Node *arrayToList(int arr[], int n)
{
    Node *root = NULL;
    for (int i = 0; i < n; i++)
        insert(&root, arr[i]);
   return root;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    Node* root = arrayToList(arr, n);
    display(root);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
class GFG
      
// Representation of a node
static class Node 
{
    int data;
    Node next;
};
  
// Function to insert node
static Node insert(Node root, int item)
{
    Node temp = new Node();
    Node ptr;
    temp.data = item;
    temp.next = null;
  
    if (root == null)
        root = temp;
    else 
    {
        ptr = root;
        while (ptr.next != null)
            ptr = ptr.next;
        ptr.next = temp;
    }
    return root;
}
  
static void display(Node root)
{
    while (root != null
    {
        System.out.print( root.data + " ");
        root = root.next;
    }
}
  
static Node arrayToList(int arr[], int n)
{
    Node root = null;
    for (int i = 0; i < n; i++)
        root = insert(root, arr[i]);
    return root;
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = arr.length;
    Node root = arrayToList(arr, n);
    display(root);
}
}
  
// This code is contributed by Arnab Kundu

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the above approach 
using System; 
      
class GFG
      
// Representation of a node
public class Node 
{
    public int data;
    public Node next;
};
  
// Function to insert node
static Node insert(Node root, int item)
{
    Node temp = new Node();
    Node ptr;
    temp.data = item;
    temp.next = null;
  
    if (root == null)
        root = temp;
    else
    {
        ptr = root;
        while (ptr.next != null)
            ptr = ptr.next;
        ptr.next = temp;
    }
    return root;
}
  
static void display(Node root)
{
    while (root != null
    {
        Console.Write(root.data + " ");
        root = root.next;
    }
}
  
static Node arrayToList(int []arr, int n)
{
    Node root = null;
    for (int i = 0; i < n; i++)
        root = insert(root, arr[i]);
    return root;
}
  
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;
    Node root = arrayToList(arr, n);
    display(root);
}
}
  
// This code is contributed by PrinciRaj1992 

chevron_right


Output:

1 2 3 4 5

Time Complexity : O(n*n)

Efficient Approach: We traverse array from end and insert every element at the beginning of the list.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

#include <iostream>
using namespace std;
  
// Representation of a node
struct Node {
    int data;
    Node* next;
};
  
// Function to insert node
void insert(Node** root, int item)
{
    Node* temp = new Node;
    temp->data = item;
    temp->next = *root;
    *root = temp;
}
  
void display(Node* root)
{
    while (root != NULL) {
        cout << root->data << " ";
        root = root->next;
    }
}
  
Node *arrayToList(int arr[], int n)
{
    Node *root = NULL;
    for (int i = n-1; i >= 0 ; i--)
        insert(&root, arr[i]);
    return root;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    Node* root = arrayToList(arr, n);
    display(root);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to print level order traversal
// in spiral form using one queue and one stack.
import java.util.*;
class GFG 
  
// Representation of a node
static class Node 
{
    int data;
    Node next;
};
static Node root; 
  
// Function to insert node
static Node insert(Node root, int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.next = root;
    root = temp;
    return root;
}
  
static void display(Node root)
{
    while (root != null
    {
        System.out.print(root.data + " ");
        root = root.next;
    }
}
  
static Node arrayToList(int arr[], int n)
{
    root = null;
    for (int i = n - 1; i >= 0 ; i--)
        root = insert(root, arr[i]);
    return root;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = arr.length;
    Node root = arrayToList(arr, n);
    display(root);
}
}
  
// This code is contributed by Princi Singh

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to print level order traversal
// in spiral form using one queue and one stack.
using System;
      
class GFG 
  
// Representation of a node
public class Node 
{
    public int data;
    public Node next;
};
static Node root; 
  
// Function to insert node
static Node insert(Node root, int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.next = root;
    root = temp;
    return root;
}
  
static void display(Node root)
{
    while (root != null
    {
        Console.Write(root.data + " ");
        root = root.next;
    }
}
  
static Node arrayToList(int []arr, int n)
{
    root = null;
    for (int i = n - 1; i >= 0 ; i--)
        root = insert(root, arr[i]);
    return root;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;
    Node root = arrayToList(arr, n);
    display(root);
}
}
  
// This code is contributed by Rajput-Ji

chevron_right


Output:

1 2 3 4 5

Time Complexity : O(n)

Alternate Efficient Solution is maintain tail pointer, traverse array elements from left to right, insert at tail and update tail after insertion.



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.