Iterative Merge Sort for Linked List

Given a singly linked list of integers, the task is to sort it using iterative merge sort.

Merge Sort is often preferred for sorting a linked list. It is discussed here. However, the method discussed above uses Stack for storing recursion calls. This may consume a lot of memory if the linked list to be sorted is too large. Hence, a purely iterative method for Merge Sort with no recursive calls is discussed in this post.

We use bottom-up approach of Merge Sort in this post. We know that Merge Sort first merges two items, then 4 items and so on. The idea is to use an integer variable to store the gap to find the midpoint around which the linked list needs to be sorted. So the problem reduces to merging two sorted Linked List which is discussed here. However, we do not use an additional list to keep the merged list. Instead we merge the lists within itself. The gap is incremented exponentially by 2 in each iteration and the process is repeated.





// Iterative C++ program to do merge sort on 
// linked list
#include <iostream>
using namespace std;
/* Structure of the Node */
struct Node {
    int data;
    struct Node* next;
/* Function to calculate length of linked list */
int length(struct Node* current)
    int count = 0;
    while (current != NULL) {
        current = current->next;
    return count;
/* Merge function of Merge Sort to Merge the two sorted parts
   of the Linked List. We compare the next value of start1 and 
   current value of start2 and insert start2 after start1 if 
   it's smaller than next value of start1. We do this until
   start1 or start2 end. If start1 ends, then we assign next 
   of start1 to start2 because start2 may have some elements
   left out which are greater than the last value of start1. 
   If start2 ends then we assign end2 to end1. This is necessary
   because we use end2 in another function (mergeSort function) 
   to determine the next start1 (i.e) start1 for next
   iteration = end2->next */
void merge(struct Node** start1, struct Node** end1, 
          struct Node** start2, struct Node** end2)
    // Making sure that first node of second
    // list is higher.
    struct Node* temp = NULL;
    if ((*start1)->data > (*start2)->data) {
        swap(*start1, *start2);
        swap(*end1, *end2);
    // Merging remaining nodes
    struct Node* astart = *start1, *aend = *end1;
    struct Node* bstart = *start2, *bend = *end2;
    struct Node* bendnext = (*end2)->next;
    while (astart != aend && bstart != bendnext) {
        if (astart->next->data > bstart->data) {
            temp = bstart->next;
            bstart->next = astart->next;
            astart->next = bstart;
            bstart = temp;
        astart = astart->next;
    if (astart == aend)
        astart->next = bstart;
        *end2 = *end1;
/* MergeSort of Linked List
   The gap is initially 1. It is incremented as 
   2, 4, 8, .. until it reaches the length of the 
   linked list. For each gap, the linked list is 
   sorted around the gap. 
   The prevend stores the address of the last node after
   sorting a part of linked list so that it's next node
   can be assigned after sorting the succeeding list. 
   temp is used to store the next start1 because after 
   sorting, the last node will be different. So it 
   is necessary to store the address of start1 before 
   sorting. We select the start1, end1, start2, end2 for 
   sorting. start1 - end1 may be considered as a list 
   and start2 - end2 may be considered as another list 
   and we are merging these two sorted list in merge 
   function and assigning the starting address to the 
   previous end address. */
void mergeSort(struct Node** head)
    if (*head == NULL)
    struct Node* start1 = NULL, *end1 = NULL;
    struct Node* start2 = NULL, *end2 = NULL;
    struct Node* prevend = NULL;
    int len = length(*head);
    for (int gap = 1; gap < len; gap = gap*2) {
        start1 = *head;
        while (start1) {
            // If this is first iteration
            bool isFirstIter = 0;
            if (start1 == *head)
                isFirstIter = 1;
            // First part for merging
            int counter = gap;
            end1 = start1;
            while (--counter && end1->next)
                end1 = end1->next;
            // Second part for merging
            start2 = end1->next;
            if (!start2)
            counter = gap;
            end2 = start2;
            while (--counter && end2->next)
                end2 = end2->next;
            // To store for next iteration.
            Node *temp = end2->next;
            // Merging two parts.
            merge(&start1, &end1, &start2, &end2);
            // Update head for first iteration, else
            // append after previous list
            if (isFirstIter)
                *head = start1;
                prevend->next = start1;
            prevend = end2;
            start1 = temp;
        prevend->next = start1;
/* Function to print the Linked List */
void print(struct Node** head)
    if ((*head) == NULL)
    struct Node* temp = *head;
    while (temp != NULL) {
        printf("%d ", temp->data);
        temp = temp->next;
/* Given a reference (pointer to   
   pointer) to the head of a list  
   and an int, push a new node on  
   the front of the list. */
void push(struct Node** head_ref, 
          int new_data) 
    struct Node* new_node = new Node; 
    new_node->data = new_data; 
    new_node->next = (*head_ref); 
    (*head_ref) = new_node; 
int main()
    // start with empty list 
    struct Node* head = NULL; 
    // create linked list 
    // 1->2->3->4->5->6->7 
    push(&head, 7); 
    push(&head, 6); 
    push(&head, 5); 
    push(&head, 4); 
    push(&head, 3); 
    push(&head, 2); 
    push(&head, 1); 



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

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