Replace even nodes of a doubly linked list with the elements of array

Given a doubly linked list and an array with only odd values. Both are of equal size N. The task is replace all node which have even value with the Array elements from left to right.

Examples:

Input : List = 6 9 8 7 4
Arr[] = {3, 5, 23, 17, 1}
Output : List = 3 9 5 7 23

Input : List = 9 14 7 12 8 13
Arr[] = {5, 1, 17, 21, 11, 7}
Output : List = 9 5 7 1 17 13

Approach: The idea is to traverse the nodes of the doubly linked list one by one and get the pointer of the nodes having even data then replace by the value of the array and increment the index of the array and move to the next node in the linked list.

Below is the implementation of above approach:

C++

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// C++ implementation to create
// odd doubly linked list
#include <bits/stdc++.h>
using namespace std;
  
// Node of the doubly linked list
struct Node {
    int data;
    Node *prev, *next;
};
  
// function to insert a node at the beginning
// of the Doubly Linked List
void push(Node** head_ref, int new_data)
{
    // allocate node
    Node* new_node = (Node*)malloc(sizeof(struct Node));
  
    // put in the data
    new_node->data = new_data;
  
    // since we are adding at the begining,
    // prev is always NULL
    new_node->prev = NULL;
  
    // link the old list off the new node
    new_node->next = (*head_ref);
  
    // change prev of head node to new node
    if ((*head_ref) != NULL)
        (*head_ref)->prev = new_node;
  
    // move the head to point to the new node
    (*head_ref) = new_node;
}
  
// function to make all node is odd
void makeOddNode(Node** head_ref, int A[], int n)
{
    Node* ptr = *head_ref;
    Node* next;
    int i = 0;
    // traves list till last node
    while (ptr != NULL) {
        next = ptr->next;
        // check if node is even then
        // replace it and incriment in i
        if (ptr->data % 2 == 0) {
            ptr->data = A[i];
            i++;
        }
        ptr = next;
    }
    // return sum of nodes which is divided by K
}
// function to print nodes in a
// given doubly linked list
void printList(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
}
  
// Driver program to test above
int main()
{
    // start with the empty list
    Node* head = NULL;
  
    // create the doubly linked list
    // 6 <=> 9 <=> 8 <=> 7 <=> 4
    int Arr[] = { 3, 5, 23, 17, 1 };
    push(&head, 4);
    push(&head, 7);
    push(&head, 8);
    push(&head, 9);
    push(&head, 6);
    int n = sizeof(Arr) / sizeof(Arr[0]);
    cout << "Original List: ";
    printList(head);
    cout << endl;
    makeOddNode(&head, Arr, n);
    cout << "New odd List: ";
    printList(head);
}

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Java

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// Java implementation to create 
// odd doubly linked list 
class GFG
{
  
// Node of the doubly linked list 
static class Node
    int data; 
    Node prev, next; 
}; 
  
// function to insert a node at the beginning 
// of the Doubly Linked List 
static Node push(Node head_ref, int new_data) 
    // allocate node 
    Node new_node = new Node(); 
  
    // put in the data 
    new_node.data = new_data; 
  
    // since we are adding at the begining, 
    // prev is always null 
    new_node.prev = null
  
    // link the old list off the new node 
    new_node.next = (head_ref); 
  
    // change prev of head node to new node 
    if ((head_ref) != null
        (head_ref).prev = new_node; 
  
    // move the head to point to the new node 
    (head_ref) = new_node; 
    return head_ref;
  
// function to make all node is odd 
static Node makeOddNode(Node head_ref, int A[], int n) 
    Node ptr = head_ref; 
    Node next; 
    int i = 0
    // traves list till last node 
    while (ptr != null
    
        next = ptr.next; 
          
        // check if node is even then 
        // replace it and incriment in i 
        if (ptr.data % 2 == 0)
        
  
            ptr.data = A[i]; 
            i++; 
        
        ptr = next; 
    
      
    // return sum of nodes which is divided by K 
    return head_ref;
  
// function to print nodes in a 
// given doubly linked list 
static void printList(Node head) 
    while (head != null)
    
        System.out.print( head.data + " "); 
        head = head.next; 
    
  
// Driver code 
public static void main(String args[])
    // start with the empty list 
    Node head = null
  
    // create the doubly linked list 
    // 6 <=> 9 <=> 8 <=> 7 <=> 4 
    int Arr[] = { 3, 5, 23, 17, 1 }; 
    head = push(head, 4); 
    head = push(head, 7); 
    head = push(head, 8); 
    head = push(head, 9); 
    head = push(head, 6); 
    int n = Arr.length; 
    System.out.print( "Original List: "); 
    printList(head); 
    System.out.println(); 
    head = makeOddNode(head, Arr, n); 
    System.out.print("New odd List: "); 
    printList(head); 
}
  
// This code is contributed by Arnab Kundu

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Python3

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# Python3 implementation to 
# create odd doubly linked list 
  
# Node of the doubly linked list 
class Node:
      
    def __init__(self, data):
        self.data = data
        self.prev = None
        self.next = None
  
# Function to insert a node at the 
# beginning of the Doubly Linked List 
def push(head_ref, new_data): 
  
    # allocate node 
    new_node = Node(new_data) 
  
    # link the old list off the new node 
    new_node.next = head_ref 
  
    # change prev of head node to new node 
    if head_ref != None
        head_ref.prev = new_node 
  
    # move the head to point to the new node 
    head_ref = new_node
    return head_ref
  
# Function to make all node is odd 
def makeOddNode(head_ref, A, n): 
  
    ptr = head_ref 
    i = 0
      
    # traves list till last node 
    while ptr != None:
          
        next = ptr.next
          
        # check if node is even then 
        # replace it and incriment in i 
        if ptr.data % 2 == 0
            ptr.data = A[i] 
            i += 1
          
        ptr = next
      
    # return sum of nodes which is divided by K 
  
# Function to print nodes in a 
# given doubly linked list 
def printList(head): 
  
    while head != None:
        print(head.data, end = " "
        head = head.next
  
# Driver Code
if __name__ == "__main__"
  
    # start with the empty list 
    head = None
  
    # create the doubly linked list 
    # 6 <=> 9 <=> 8 <=> 7 <=> 4 
    Arr = [3, 5, 23, 17, 1
    head = push(head, 4
    head = push(head, 7
    head = push(head, 8
    head = push(head, 9
    head = push(head, 6
    n = len(Arr) 
      
    print("Original List:", end = " "
    printList(head) 
    print()
      
    makeOddNode(head, Arr, n) 
    print("New odd List:", end = " "
    printList(head) 
  
# This code is contributed by Rituraj Jain

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C#

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// C# implementation to create 
// odd doubly linked list
using System;
      
class GFG
{
  
// Node of the doubly linked list 
public class Node
    public int data; 
    public Node prev, next; 
}; 
  
// function to insert a node at the beginning 
// of the Doubly Linked List 
static Node push(Node head_ref, int new_data) 
    // allocate node 
    Node new_node = new Node(); 
  
    // put in the data 
    new_node.data = new_data; 
  
    // since we are adding at the begining, 
    // prev is always null 
    new_node.prev = null
  
    // link the old list off the new node 
    new_node.next = (head_ref); 
  
    // change prev of head node to new node 
    if ((head_ref) != null
        (head_ref).prev = new_node; 
  
    // move the head to point to the new node 
    (head_ref) = new_node; 
    return head_ref;
  
// function to make all node is odd 
static Node makeOddNode(Node head_ref, int []A, int n) 
    Node ptr = head_ref; 
    Node next; 
    int i = 0; 
      
    // traves list till last node 
    while (ptr != null
    
        next = ptr.next; 
          
        // check if node is even then 
        // replace it and incriment in i 
        if (ptr.data % 2 == 0)
        
  
            ptr.data = A[i]; 
            i++; 
        
        ptr = next; 
    
      
    // return sum of nodes which is divided by K 
    return head_ref;
  
// function to print nodes in a 
// given doubly linked list 
static void printList(Node head) 
    while (head != null)
    
        Console.Write( head.data + " "); 
        head = head.next; 
    
  
// Driver code 
public static void Main(String []args)
    // start with the empty list 
    Node head = null
  
    // create the doubly linked list 
    // 6 <=> 9 <=> 8 <=> 7 <=> 4 
    int []Arr = { 3, 5, 23, 17, 1 }; 
    head = push(head, 4); 
    head = push(head, 7); 
    head = push(head, 8); 
    head = push(head, 9); 
    head = push(head, 6); 
    int n = Arr.Length; 
    Console.WriteLine( "Original List: "); 
    printList(head); 
    Console.WriteLine(); 
    head = makeOddNode(head, Arr, n); 
    Console.WriteLine("New odd List: "); 
    printList(head); 
}
  
// This code contributed by Rajput-Ji

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Output:

Original List: 6 9 8 7 4 
New odd List: 3 9 5 7 23

Time Complexity: O(N), where N is the total number of nodes.



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