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 23Input : 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++
// 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 beginning, // 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
// 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 beginning, // 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
# 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#
// 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 beginning, // 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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