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C++ Program For Arranging Single Linked List In Alternate Odd and Even Nodes Order

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  • Last Updated : 18 Jan, 2022

Given a singly linked list, rearrange the list so that even and odd nodes are alternate in the list.
There are two possible forms of this rearrangement. If the first data is odd, then the second node must be even. The third node must be odd and so on. Notice that another arrangement is possible where the first node is even, second odd, third even and so on. 
Examples:

Input: 11 -> 20 -> 40 -> 55 -> 77 -> 80 -> NULL
Output: 11 -> 20 -> 55 -> 40 -> 77 -> 80 -> NULL
20, 40, 80 occur in even positions and 11, 55, 77
occur in odd positions.

Input: 10 -> 1 -> 2 -> 3 -> 5 -> 6 -> 7 -> 8 -> NULL
Output: 1 -> 10 -> 3 -> 2 -> 5 -> 6 -> 7 -> 8 -> NULL
1, 3, 5, 7 occur in odd positions and 10, 2, 6, 8 occur
at even positions in the list

Method 1 (Simple): 
In this method, we create two stacks-Odd and Even. We traverse the list and when we encounter an even node in an odd position we push this node’s address onto Even Stack. If we encounter an odd node in an even position we push this node’s address onto Odd Stack. 
After traversing the list, we simply pop the nodes at the top of the two stacks and exchange their data. We keep repeating this step until the stacks become empty.

Step 1: Create two stacks Odd and Even.  
        These stacks will store the pointers to the nodes in the list 
Step 2: Traverse list from start to end, using the variable current. 
        Repeat following steps 3-4.
Step 3: If the current node is even and it occurs at an odd position, 
        push this node's address to stack Even.
Step 4: If the current node is odd and it occurs at an even position, 
        push this node's address to stack Odd. 
[END OF TRAVERSAL]. Step 5: The size of both the stacks will be the same. While both the stacks are not empty exchange the nodes at the top of the two stacks and pop both nodes from their respective stacks.  Step 6: The List is now rearranged. STOP

C++




// CPP program to rearrange nodes
// as alternate odd even nodes in
// a Singly Linked List
#include <bits/stdc++.h>
using namespace std;
  
// Structure node
struct Node 
{
    int data;
    struct Node* next;
};
  
// A utility function to print
// linked list
void printList(struct Node* node)
{
    while (node != NULL) 
    {
        cout << node->data << " ";
        node = node->next;
    }
    cout << endl;
}
  
// Function to create newNode
// in a linkedlist
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->data = key;
    temp->next = NULL;
    return temp;
}
  
// Function to insert at beginning
Node* insertBeg(Node* head, 
                int val)
{
    Node* temp = newNode(val);
    temp->next = head;
    head = temp;
    return head;
}
  
// Function to rearrange the
// odd and even nodes
void rearrangeOddEven(Node* head)
{
    stack<Node*> odd;
    stack<Node*> even;
    int i = 1;
  
    while (head != nullptr) 
    {
        if (head->data % 2 != 0 && 
            i % 2 == 0) 
        {
            // Odd Value in Even Position
            // Add pointer to current node
            // in odd stack
            odd.push(head);
        }
  
        else if (head->data % 2 == 0 && 
                 i % 2 != 0) 
        {
            // Even Value in Odd Position
            // Add pointer to current node
            // in even stack
            even.push(head);
        }
  
        head = head->next;
        i++;
    }
  
    while (!odd.empty() && 
           !even.empty()) 
    {
        // Swap Data at the top of 
        // two stacks
        swap(odd.top()->data, 
             even.top()->data);
        odd.pop();
        even.pop();
    }
}
  
// Driver code
int main()
{
    Node* head = newNode(8);
    head = insertBeg(head, 7);
    head = insertBeg(head, 6);
    head = insertBeg(head, 5);
    head = insertBeg(head, 3);
    head = insertBeg(head, 2);
    head = insertBeg(head, 1);
  
    cout << "Linked List:" << endl;
    printList(head);
    rearrangeOddEven(head);
  
    cout << "Linked List after " << 
            "Rearranging:" << endl;
    printList(head);
  
    return 0;
}

Output: 

Linked List:
1 2 3 5 6 7 8 
Linked List after Rearranging:
1 2 3 6 5 8 7

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

Method 2 (Efficient):

  1. Segregate odd and even values in the list. After this, all odd values will occur together followed by all even values. 
  2. Split the list into two lists odd and even.
  3. Merge the even list into odd list
REARRANGE (HEAD)
Step 1: Traverse the list using NODE TEMP. 
           If TEMP is odd
               Add TEMP to the beginning of the List
           [END OF IF]
        [END OF TRAVERSAL]
Step 2: Set TEMP to 2nd element of LIST.
Step 3: Set PREV_TEMP to 1st element of List
Step 4: Traverse using node TEMP as long as an even
        node is not encountered.
            PREV_TEMP = TEMP, TEMP = TEMP->NEXT
        [END OF TRAVERSAL]
Step 5: Set EVEN to TEMP. Set PREV_TEMP->NEXT to NULL
Step 6: I = HEAD, J = EVEN
Step 7: Repeat while I != NULL and J != NULL
            Store next nodes of I and J in K and L
            K = I->NEXT, L = J->NEXT
            I->NEXT = J, J->NEXT = K, PTR = J
            I = K and J = L 
       [END OF LOOP]
Step 8: if I == NULL 
            PTR->NEXT = J
        [END of IF]
Step 8: Return HEAD.
Step 9: End

C++




// Cpp program to rearrange nodes
// as alternate odd even nodes in
// a Singly Linked List
#include <bits/stdc++.h>
using namespace std;
  
// Structure node
struct Node 
{
    int data;
    struct Node* next;
};
  
// A utility function to print
// linked list
void printList(struct Node* node)
{
    while (node != NULL) 
    {
        cout << node->data << " ";
        node = node->next;
    }
    cout << endl;
}
  
// Function to create newNode
// in a linkedlist
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->data = key;
    temp->next = NULL;
    return temp;
}
  
// Function to insert at beginning
Node* insertBeg(Node* head, 
                int val)
{
    Node* temp = newNode(val);
    temp->next = head;
    head = temp;
    return head;
}
  
// Function to rearrange the
// odd and even nodes
void rearrange(Node** head)
{
    // Step 1: Segregate even and odd nodes
    // Step 2: Split odd and even lists
    // Step 3: Merge even list into odd list
    Node* even;
    Node *temp, *prev_temp;
    Node *i, *j, *k, *l, *ptr;
  
    // Step 1: Segregate Odd and 
    // Even Nodes
    temp = (*head)->next;
    prev_temp = *head;
  
    while (temp != nullptr) 
    {
        // Backup next pointer of temp
        Node* x = temp->next;
  
        // If temp is odd move the node
        // to beginning of list
        if (temp->data % 2 != 0)  
        {
            prev_temp->next = x;
            temp->next = (*head);
            (*head) = temp;
        }
        else 
        {
            prev_temp = temp;
        }
  
        // Advance Temp Pointer
        temp = x;
    }
  
    // Step 2
    // Split the List into Odd and 
    // even
    temp = (*head)->next;
    prev_temp = (*head);
  
    while (temp != nullptr && 
           temp->data % 2 != 0) 
    {
        prev_temp = temp;
        temp = temp->next;
    }
  
    even = temp;
  
    // End the odd List (Make 
    // last node null)
    prev_temp->next = nullptr;
  
    // Step 3:
    // Merge Even List into odd
    i = *head;
    j = even;
  
    while (j != nullptr && 
           i != nullptr) 
    {
        // While both lists are not
        // exhausted Backup next
        // pointers of i and j
        k = i->next;
        l = j->next;
  
        i->next = j;
        j->next = k;
  
        // ptr points to the latest 
        // node added
        ptr = j;
  
        // Advance i and j pointers
        i = k;
        j = l;
    }
  
    if (i == nullptr) 
    {
        // Odd list exhausts before even,
        // append remainder of even list 
        // to odd.
        ptr->next = j;
    }
  
    // The case where even list exhausts 
    // before odd list is automatically 
    // handled since we merge the even 
    // list into the odd list
}
  
// Driver Code
int main()
{
    Node* head = newNode(8);
    head = insertBeg(head, 7);
    head = insertBeg(head, 6);
    head = insertBeg(head, 3);
    head = insertBeg(head, 5);
    head = insertBeg(head, 1);
    head = insertBeg(head, 2);
    head = insertBeg(head, 10);
  
    cout << "Linked List:" << endl;
    printList(head);
    cout << "Rearranged List" << endl;
    rearrange(&head);
    printList(head);
}

Output:

Linked List:
10 2 1 5 3 6 7 8 
Rearranged List
7 10 3 2 5 6 1 8

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

Please refer complete article on Alternate Odd and Even Nodes in a Singly Linked List for more details!


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