C++ Program For Segregating Even And Odd Nodes In A Linked List

Given a Linked List of integers, write a function to modify the linked list such that all even numbers appear before all the odd numbers in the modified linked list. Also, keep the order of even and odd numbers the same.

Examples:

Input: 17->15->8->12->10->5->4->1->7->6->NULL
Output: 8->12->10->4->6->17->15->5->1->7->NULL

Input: 8->12->10->5->4->1->6->NULL
Output: 8->12->10->4->6->5->1->NULL

// If all numbers are even then do not change the list
Input: 8->12->10->NULL
Output: 8->12->10->NULL

// If all numbers are odd then do not change the list
Input: 1->3->5->7->NULL
Output: 1->3->5->7->NULL

Method 1:
The idea is to get pointer to the last node of list. And then traverse the list starting from the head node and move the odd valued nodes from their current position to end of the list.
Thanks to blunderboy for suggesting this method. 
Algorithm: 

• Get pointer to the last node.
• Move all the odd nodes to the end.
  • Consider all odd nodes before the first even node and move them to end.
  • Change the head pointer to point to the first even node.
  • Consider all odd nodes after the first even node and move them to the end.

Below is the implementation of the above approach:

C++

// C++ program to segregate even and
// odd nodes in a Linked List
#include <bits/stdc++.h>
using namespace std;

// A node of the singly linked list
class Node
{
    public:
    int data;
    Node *next;
};

void segregateEvenOdd(Node **head_ref)
{
    Node *end = *head_ref;
    Node *prev = NULL;
    Node *curr = *head_ref;

    // Get pointer to the last node
    while (end->next != NULL)
        end = end->next;

    Node *new_end = end;

    /* Consider all odd nodes before
       the first even node and move
       then after end */
    while (curr->data % 2 != 0 &&
           curr != end)
    {
        new_end->next = curr;
        curr = curr->next;
        new_end->next->next = NULL;
        new_end = new_end->next;
    }

    // 10->8->17->17->15
    /* Do following steps only if
       there is any even node */
    if (curr->data%2 == 0)
    {
        /* Change the head pointer to
           point to first even node */
        *head_ref = curr;

        /* Now current points to
           the first even node */
        while (curr != end)
        {
            if ( (curr->data) % 2 == 0 )
            {
                prev = curr;
                curr = curr->next;
            }
            else
            {
                /* Break the link between
                   prev and current */
                prev->next = curr->next;

                // Make next of curr as NULL
                curr->next = NULL;

                // Move curr to end
                new_end->next = curr;

                // Make curr as new end of list
                new_end = curr;

                /* Update current pointer to
                   next of the moved node */
                curr = prev->next;
            }
        }
    }

    /* We must have prev set before
       executing lines following this
       statement */
    else prev = curr;

    /* If there are more than 1 odd nodes
       and end of original list is odd then
       move this node to end to maintain
       same order of odd numbers in modified
       list */
    if (new_end != end &&
       (end->data) % 2 != 0)
    {
        prev->next = end->next;
        end->next = NULL;
        new_end->next = end;
    }
    return;
}

// UTILITY FUNCTIONS
/* Function to insert a node at
   the beginning */
void push(Node** head_ref,
          int new_data)
{
    // Allocate node
    Node* new_node = new 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 point to
    // the new node
    (*head_ref) = new_node;
}

// Function to print nodes in a
// given linked list
void printList(Node *node)
{
    while (node != NULL)
    {
        cout << node->data <<" ";
        node = node->next;
    }
}

// Driver code
int main()
{
    // Start with the empty list
    Node* head = NULL;

    /* Let us create a sample
       linked list as following
       0->2->4->6->8->10->11 */
    push(&head, 11);
    push(&head, 10);
    push(&head, 8);
    push(&head, 6);
    push(&head, 4);
    push(&head, 2);
    push(&head, 0);

    cout << "Original Linked list ";
    printList(head);

    segregateEvenOdd(&head);

    cout << "Modified Linked list ";
    printList(head);

    return 0;
}
// This code is contributed by rathbhupendra

Output

Original Linked list 0 2 4 6 8 10 11 Modified Linked list 0 2 4 6 8 10 11 

Time Complexity: O(n)
Auxiliary Space: O(1)

Method 2:
The idea is to split the linked list into two: one containing all even nodes and other containing all odd nodes. And finally, attach the odd node linked list after the even node linked list. 
To split the Linked List, traverse the original Linked List and move all odd nodes to a separate Linked List of all odd nodes. At the end of the loop, the original list will have all the even nodes and the odd node list will have all the odd nodes. To keep the ordering of all nodes the same, we must insert all the odd nodes at the end of the odd node list. And to do that in constant time, we must keep track of the last pointer in the odd node list.

C++

// CPP program to segregate even and
// odd nodes in a Linked List
#include <stdio.h>
#include <stdlib.h>

// A node of the singly linked list
struct Node
{
    int data;
    struct Node *next;
};

// Function to segregate even and
// odd nodes.
void segregateEvenOdd(struct Node **head_ref)
{
    // Starting node of list having
    // even values.
    Node *evenStart = NULL;

    // Ending node of even values list.
    Node *evenEnd = NULL;

    // Starting node of odd values list.
    Node *oddStart = NULL;

    // Ending node of odd values list.
    Node *oddEnd = NULL;

    // Node to traverse the list.
    Node *currNode = *head_ref;

    while(currNode != NULL)
    {
        int val = currNode -> data;

        // If current value is even, add
        // it to even values list.
        if(val % 2 == 0)
        {
            if(evenStart == NULL)
            {
                evenStart = currNode;
                evenEnd = evenStart;
            }

            else
            {
                evenEnd -> next = currNode;
                evenEnd = evenEnd -> next;
            }
        }

        // If current value is odd, add
        // it to odd values list.
        else
        {
            if(oddStart == NULL)
            {
                oddStart = currNode;
                oddEnd = oddStart;
            }
            else
            {
                oddEnd -> next = currNode;
                oddEnd = oddEnd -> next;
            }
        }

        // Move head pointer one step in
        // forward direction
        currNode = currNode -> next;
    }

    // If either odd list or even list
    // is empty, no change is required
    // as all elements are either even
    // or odd.
    if(oddStart == NULL ||
       evenStart == NULL)
    {
        return;
    }

    // Add odd list after even list.
    evenEnd -> next = oddStart;
    oddEnd -> next = NULL;

    // Modify head pointer to
    // starting of even list.
    *head_ref = evenStart;
}

// UTILITY FUNCTIONS
/* Function to insert a node at
   the beginning */
void push(struct Node** head_ref,
          int 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 point to the
    // new node
    (*head_ref) = new_node;
}

/* Function to print nodes in a
   given linked list */
void printList(struct Node *node)
{
    while (node!=NULL)
    {
        printf("%d ", node->data);
        node = node->next;
    }
}

// Driver code
int main()
{
    // Start with the empty list
    struct Node* head = NULL;

    /* Let us create a sample
       linked list as following
       0->1->4->6->9->10->11 */
    push(&head, 11);
    push(&head, 10);
    push(&head, 9);
    push(&head, 6);
    push(&head, 4);
    push(&head, 1);
    push(&head, 0);

    printf("Original Linked list ");
    printList(head);

    segregateEvenOdd(&head);

    printf("Modified Linked list ");
    printList(head);

    return 0;
}
// This code is contributed by NIKHIL JINDAL.

Output

Original Linked list 0 1 4 6 9 10 11 Modified Linked list 0 4 6 10 1 9 11 

Time complexity: O(n)
Auxiliary space: O(1) because using constant space

Please refer complete article on Segregate even and odd nodes in a Linked List for more details!