C++ Program For Sorting Linked List Which Is Already Sorted On Absolute Values

Given a linked list that is sorted based on absolute values. Sort the list based on actual values.
Examples:

Input:  1 -> -10 
Output: -10 -> 1

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

Input: -5 -> -10 
Output: -10 -> -5

Input: 5 -> 10 
Output: 5 -> 10

Source : Amazon Interview

A simple solution is to traverse the linked list from beginning to end. For every visited node, check if it is out of order. If it is, remove it from its current position and insert it at the correct position. This is the implementation of insertion sort for linked list and the time complexity of this solution is O(n*n).
A better solution is to sort the linked list using merge sort. Time complexity of this solution is O(n Log n).
An efficient solution can work in O(n) time. An important observation is, all negative elements are present in reverse order. So we traverse the list, whenever we find an element that is out of order, we move it to the front of the linked list. 
Below is the implementation of the above idea.

ever we find an element that is out of order, we move it to the front of the linked list. 
Below is the implementation of the above idea. 

C++

// C++ program to sort a linked list,
// already sorted by absolute values
#include <bits/stdc++.h>
using namespace std;

// Linked List Node
struct Node
{
    Node* next;
    int data;
};

// Utility function to insert a
// node at the beginning
void push(Node** head, int data)
{
    Node* newNode = new Node;
    newNode->next = (*head);
    newNode->data = data;
    (*head) = newNode;
}

// Utility function to print
// a linked list
void printList(Node* head)
{
    while (head != NULL)
    {
        cout << head->data;
        if (head->next != NULL)
            cout << " -> ";
        head = head->next;
    }
    cout<<endl;
}

// To sort a linked list by actual values.
// The list is assumed to be sorted by
// absolute values.
void sortList(Node** head)
{
    // Initialize previous and current
    // nodes
    Node* prev = (*head);
    Node* curr = (*head)->next;

    // Traverse list
    while (curr != NULL)
    {
        // If curr is smaller than prev,
        // then it must be moved to head
        if (curr->data < prev->data)
        {
            // Detach curr from linked list
            prev->next = curr->next;

            // Move current node to beginning
            curr->next = (*head);
            (*head) = curr;

            // Update current
            curr = prev;
        }

        // Nothing to do if current
        // element is at right place
        else
            prev = curr;

        // Move current
        curr = curr->next;
    }
}

// Driver code
int main()
{
    Node* head = NULL;
    push(&head, -5);
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, -2);
    push(&head, 1);
    push(&head, 0);

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

    sortList(&head);

    cout << "Sorted list :";
    printList(head);

    return 0;
}

Output:

Original list :
0 -> 1 -> -2 -> 3 -> 4 -> 5 -> -5
Sorted list :
-5 -> -2 -> 0 -> 1 -> 3 -> 4 -> 5

Time Complexity: O(N)

Auxiliary Space: O(1)

Please refer complete article on Sort linked list which is already sorted on absolute values for more details!