Segregate even and odd nodes in a Linked List using Deque

Given a Linked List of integers. The task is to write a program to modify the linked list such that all even numbers appear before all the odd numbers in the modified linked list. It is not needed to keep the order of even and odd nodes same as that of the original list, the task is just to rearrange the nodes such that all even valued nodes appear before the odd valued nodes.

See Also: Segregate even and odd nodes in a Linked List

Examples:

Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 -> NULL
Output: 10 -> 8 -> 6 -> 4 -> 2 -> 1 -> 3 -> 5 -> 7 -> 9 -> NULL

Input: 4 -> 3 -> 2 -> 1 -> NULL
Output: 2 -> 4 -> 3 -> 1 -> NULL



The idea is to iteratively push all the elements of the linked list to deque as per the below conditions:

  • Start traversing the linked list and if an element is even then push it to the front of the Deque and,
  • If the element is odd then push it to the back of the Deque.

Finally, replace all elements of the linked list with the elements of Deque starting from the first element.

Below is the implementation of the above approach:

C++

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// CPP program to segregate even and
// odd noeds in a linked list using deque
#include <bits/stdc++.h>
using namespace std;
  
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
  
/*UTILITY FUNCTIONS*/
/* Push a node to linked list. Note that this function
changes the head */
void push(struct Node** head_ref, char new_data)
{
    /* allocate node */
    struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
  
    /* put in the data */
    new_node->data = new_data;
  
    /* link the old list off the new node */
    new_node->next = (*head_ref);
  
    /* move the head to pochar to the new node */
    (*head_ref) = new_node;
}
  
// printing the linked list
void printList(struct Node* head)
{
    struct Node* temp = head;
    while (temp != NULL) {
                printf("%d ", temp->data);
        temp = temp->next;
    }
}
  
// Function to rearrange even and odd
// elements in a linked list using deque
void evenOdd(struct Node* head)
{
    struct Node* temp = head;
      
    // Declaring a Deque
    deque<int> d; 
  
    // Push all the elements of
    // linked list in to deque
    while (temp != NULL) {
  
        // if element is even push it
        // to front of the deque
        if (temp->data % 2 == 0) 
            d.push_front(temp->data);
  
        else // else push at the back of the deque
            d.push_back(temp->data);
        temp = temp->next; // increase temp
    }
      
    temp = head;
      
    // Replace all elements of the linked list
    // with the elements of Deque starting from 
    // the first element
    while (!d.empty()) {
        temp->data = d.front();
        d.pop_front();
        temp = temp->next; 
    }
}
  
// Driver code
int main()
{
    struct Node* head = NULL;
    push(&head, 10);
    push(&head, 9);
    push(&head, 8);
    push(&head, 7);
    push(&head, 6);
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
      
    cout << "Given linked list: ";
        printList(head);
          
    evenOdd(head);
      
    cout << "\nAfter rearrangement: ";
        printList(head);
          
    return 0;
}

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Python

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# Python program to segregate even and
# odd noeds in a linked list using deque
import collections
  
# Node class 
class Node: 
      
    # Function to initialise the node object 
    def __init__(self, data): 
        self.data = data # Assign data 
        self.next = None
  
# UTILITY FUNCTIONS
# Push a node to linked list. Note that this function
# changes the head 
def push( head_ref, new_data):
  
    # allocate node 
    new_node = Node(0)
  
    # put in the data 
    new_node.data = new_data
  
    # link the old list off the new node 
    new_node.next = (head_ref)
  
    # move the head to pochar to the new node 
    (head_ref) = new_node
      
    return head_ref
  
# printing the linked list
def printList( head):
  
    temp = head
    while (temp != None): 
        print( temp.data, end = " ")
        temp = temp.next
      
# Function to rearrange even and odd
# elements in a linked list using deque
def evenOdd( head):
  
    temp = head
      
    # Declaring a Deque
    d = collections.deque([]) 
  
    # Push all the elements of
    # linked list in to deque
    while (temp != None) :
  
        # if element is even push it
        # to front of the deque
        if (temp.data % 2 == 0): 
            d.appendleft(temp.data)
  
        else: # else push at the back of the deque
            d.append(temp.data)
        temp = temp.next # increase temp
      
    temp = head
      
    # Replace all elements of the linked list
    # with the elements of Deque starting from 
    # the first element
    while (len(d) > 0) :
        temp.data = d[0]
        d.popleft()
        temp = temp.next
      
# Driver code
  
head = None
head = push(head, 10)
head = push(head, 9)
head = push(head, 8)
head = push(head, 7)
head = push(head, 6)
head = push(head, 5)
head = push(head, 4)
head = push(head, 3)
head = push(head, 2)
head = push(head, 1)
      
print( "Given linked list: ", end = "")
printList(head)
          
evenOdd(head)
      
print("\nAfter rearrangement: ", end = "")
printList(head)
  
# This code is contributed by Arnab Kundu

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Output:

Given linked list: 1 2 3 4 5 6 7 8 9 10 
After rearrangement: 10 8 6 4 2 1 3 5 7 9

Time complexity: O(N)
Auxiliary Space: O(N), where N is the total number of nodes in the linked list.

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