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C++ Program For Rearranging A Given Linked List In-Place.

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  • Last Updated : 15 Jun, 2022

Given a singly linked list L0 -> L1 -> … -> Ln-1 -> Ln. Rearrange the nodes in the list so that the new formed list is : L0 -> Ln -> L1 -> Ln-1 -> L2 -> Ln-2
You are required to do this in place without altering the nodes’ values. 

Examples: 

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

Input: 1 -> 2 -> 3 -> 4 -> 5
Output: 1 -> 5 -> 2 -> 4 -> 3

Simple Solution:

1) Initialize current node as head.
2) While next of current node is not null, do following
    a) Find the last node, remove it from the end and insert it as next
       of the current node.
    b) Move current to next to next of current

The time complexity of the above simple solution is O(n2) where n is the number of nodes in the linked list.

Better Solution: 
1) Copy contents of the given linked list to a vector. 
2) Rearrange the given vector by swapping nodes from both ends. 
3) Copy the modified vector back to the linked list. 
Implementation of this approach: https://ide.geeksforgeeks.org/1eGSEy 
Thanks to Arushi Dhamija for suggesting this approach.

Efficient Solution:

1) Find the middle point using tortoise and hare method.
2) Split the linked list into two halves using found middle point in step 1.
3) Reverse the second half.
4) Do alternate merge of first and second halves.

The Time Complexity of this solution is O(n). 

Below is the implementation of this method.

C++




// C++ program to rearrange a
// linked list in-place
#include <bits/stdc++.h>
using namespace std;
 
// Linkedlist Node structure
struct Node
{
    int data;
    struct Node* next;
};
 
// 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 reverse the
// linked list
void reverselist(Node** head)
{
    // Initialize prev and current
    // pointers
    Node *prev = NULL,
         *curr = *head, *next;
 
    while (curr)
    {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
 
    *head = prev;
}
 
// Function to print the
// linked list
void printlist(Node* head)
{
    while (head != NULL)
    {
        cout << head->data << " ";
        if (head->next)
            cout << "-> ";
        head = head->next;
    }
    cout << endl;
}
 
// Function to rearrange a
// linked list
void rearrange(Node** head)
{
    // 1) Find the middle point using
    // tortoise and hare method
    Node *slow = *head,
         *fast = slow->next;
    while (fast && fast->next)
    {
        slow = slow->next;
        fast = fast->next->next;
    }
 
    // 2) Split the linked list in
    // two halves
    // head1, head of first half- 1 -> 2
    // head2, head of second half- 3 -> 4
    Node* head1 = *head;
    Node* head2 = slow->next;
    slow->next = NULL;
 
    // 3) Reverse the second half,
    // i.e.,  4 -> 3
    reverselist(&head2);
 
    // 4) Merge alternate nodes
    // Assign dummy Node
    *head = newNode(0);
 
    // curr is the pointer to this
    // dummy Node, which will be
    // used to form the new list
    Node* curr = *head;
    while (head1 || head2)
    {
        // First add the element
        // from list
        if (head1)
        {
            curr->next = head1;
            curr = curr->next;
            head1 = head1->next;
        }
 
        // Then add the element from
        // the second list
        if (head2)
        {
            curr->next = head2;
            curr = curr->next;
            head2 = head2->next;
        }
    }
 
    // Assign the head of the new
    // list to head pointer
    *head = (*head)->next;
}
 
// Driver program
int main()
{
    Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(5);
 
    printlist(head); // Print original list
    rearrange(&head); // Modify the list
    printlist(head); // Print modified list
    return 0;
}


Output:

1 -> 2 -> 3 -> 4 -> 5 
1 -> 5 -> 2 -> 4 -> 3

Time Complexity: O(n) 
Auxiliary Space: O(1)
Thanks to Gaurav Ahirwar for suggesting the above approach.

Another approach: 
1. Take two pointers prev and curr, which hold the addresses of head and head-> next. 
2. Compare their data and swap. 
After that, a new linked list is formed. 

Below is the implementation: 

C++




// C++ code to rearrange linked list
// in place
#include <bits/stdc++.h>
using namespace std;
 
struct node
{
    int data;
    struct node* next;
};
typedef struct node Node;
 
// Function for rearranging a
// linked list with high and low
// value.
void rearrange(Node* head)
{
    // Base case.
    if (head == NULL)
        return;
 
    // Two pointer variable.
    Node *prev = head,
         *curr = head->next;
 
    while (curr)
    {
        // Swap function for swapping data.
        if (prev->data > curr->data)
            swap(prev->data, curr->data);
 
        // Swap function for swapping data.
        if (curr->next &&
            curr->next->data > curr->data)
            swap(curr->next->data, curr->data);
 
        prev = curr->next;
 
        if (!curr->next)
            break;
        curr = curr->next->next;
    }
}
 
// Function to insert a node in the
// linked list at the beginning.
void push(Node** head, int k)
{
    Node* tem = (Node*)malloc(sizeof(Node));
    tem->data = k;
    tem->next = *head;
    *head = tem;
}
 
// Function to display node of
// linked list.
void display(Node* head)
{
    Node* curr = head;
    while (curr != NULL)
    {
        printf("%d ", curr->data);
        curr = curr->next;
    }
}
 
// Driver code
int main()
{
    Node* head = NULL;
 
    // Let create a linked list.
    // 9 -> 6 -> 8 -> 3 -> 7
    push(&head, 7);
    push(&head, 3);
    push(&head, 8);
    push(&head, 6);
    push(&head, 9);
 
    rearrange(head);
    display(head);
 
    return 0;
}


Output:

6 9 3 8 7

Time Complexity : O(n) 
Auxiliary Space : O(1) 
Thanks to Aditya for suggesting this approach.

Another Approach: (Using recursion)

  1. Hold a pointer to the head node and go till the last node using recursion
  2. Once the last node is reached, start swapping the last node to the next of head node
  3. Move the head pointer to the next node
  4. Repeat this until the head and the last node meet or come adjacent to each other
  5. Once the Stop condition met, we need to discard the left nodes to fix the loop created in the list while swapping nodes.

C++




// C/C++ program to implement
// the above approach
#include <stdio.h>
#include <stdlib.h>
 
// Creating the structure
// for node
struct Node
{
    int data;
    struct Node* next;
};
 
// Function to create newNode
// in a linkedlist
struct Node* newNode(int key)
{
    struct Node* temp =
           malloc(sizeof(struct Node));
    temp->data = key;
    temp->next = NULL;
    return temp;
}
 
// Function to print the list
void printlist(struct Node* head)
{
    while (head)
    {
        printf("%d ", head->data);
        if (head->next)
            printf("->");
        head = head->next;
    }
    printf("");
}
 
// Function to rearrange
void rearrange(struct Node** head,
               struct Node* last)
{
    if (!last)
        return;
 
    // Recursive call
    rearrange(head, last->next);
 
    // (*head)->next will be set to NULL
    // after rearrangement.
    // Need not do any operation further
    // Just return here to come out of recursion
    if (!(*head)->next)
        return;
 
    // Rearrange the list until both head
    // and last meet or next to each other.
    if ((*head) != last && (*head)->next != last)
    {
        struct Node* tmp = (*head)->next;
        (*head)->next = last;
        last->next = tmp;
        *head = tmp;
    }
    else
    {
        if ((*head) != last)
            *head = (*head)->next;
        (*head)->next = NULL;
    }
}
 
// Drivers Code
int main()
{
    struct Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(5);
 
    // Print original list
    printlist(head);
 
    struct Node* tmp = head;
 
    // Modify the list
    rearrange(&tmp, head);
 
    // Print modified list
    printlist(head);
    return 0;
}


Output:

1 ->2 ->3 ->4 ->5 
1 ->5 ->2 ->4 ->3

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), for recursive stack where n represents the length of the given linked list.

Please refer complete article on Rearrange a given linked list in-place. for more details!


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