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Merge two sorted lists (in-place)
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Iterative Merge Sort for Linked List

Last Updated : 23 Oct, 2024
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Given a singly linked list of integers, the task is to sort it using iterative merge sort.

Examples:

Input: 40 -> 20 -> 60 -> 10 -> 50 -> 30 -> NULL
Output: 10 -> 20 -> 30 -> 40 -> 50 -> 60 -> NULL

Input: 9 -> 5 -> 2 -> 8 -> NULL
Output: 2 -> 5 -> 8 -> 9 -> NULL

Merge Sort is often preferred for sorting a linked list. It is discussed Merge Sort for Linked Lists. However, the recursive 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.

Approach:

The idea is to use bottom-up approach of Merge Sort. 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 Merge two sorted linked lists. 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.

Below is the implementation of the above approach:

C++ 
// Iterative C++ program to sort a linked  // list using merge sort #include <iostream> using namespace std;  class Node { public:     int data;     Node* next;      Node(int x) {         data = x;         next = nullptr;     } };  // Function to merge two sorted sublists Node* merge(Node* left, Node* right) {        // Handle the base case when either      // left or right is null     if (!left) return right;     if (!right) return left;      Node* result = nullptr;     Node* curr = nullptr;      // Initialize result to the smaller node      // and set the current pointer     if (left->data <= right->data) {         result = left;         left = left->next;     } else {         result = right;         right = right->next;     }     curr = result;      // Merge the nodes in sorted order     while (left && right) {         if (left->data <= right->data) {             curr->next = left;             left = left->next;         } else {             curr->next = right;             right = right->next;         }         curr = curr->next;     }      // Append the remaining nodes   	// from either list     if (left) curr->next = left;     else curr->next = right;      return result; }  // Function to split the list into // sublists of given size Node* split(Node* head, int size) {     for (int i = 1; head && i < size; i++) {         head = head->next;     }     if (!head) return nullptr;      Node* second = head->next;     head->next = nullptr;      return second; }  // Iterative merge sort function for linked list Node* mergeSort(Node* head) {     if (!head || !head->next) return head;      // Calculate the total length of the list     int len = 0;     for (Node* temp = head; temp; temp = temp->next) len++;      Node* left, *right;     Node* sortedHead = nullptr;     Node* sortedTail = nullptr;      // Bottom-up merge sort, iterating over sublist sizes     for (int size = 1; size < len; size *= 2) {         Node* curr = head;         sortedHead = nullptr;         sortedTail = nullptr;          while (curr) {             left = curr;             right = split(left, size);              curr = split(right, size);               // Merge the two halves             Node* merged = merge(left, right);              // Set the head for the first merged sublist             if (!sortedHead) {                 sortedHead = merged;                 sortedTail = sortedHead;             }              else {                 sortedTail->next = merged;             }              // Move the tail pointer to the end of the merged list             while (sortedTail->next) {                 sortedTail = sortedTail->next;             }         }          // Update the head to the sorted list after          // merging all sublists of current size         head = sortedHead;     }      return head; }  void printList(Node* curr) {     while (curr != nullptr) {         cout << " " << curr->data;         curr = curr->next;     } }  int main() {        // Create a hard-coded linked list:     // 40 -> 20 -> 60 -> 10 -> 50 -> 30     Node* head = new Node(40);     head->next = new Node(20);     head->next->next = new Node(60);     head->next->next->next = new Node(10);     head->next->next->next->next = new Node(50);     head->next->next->next->next->next = new Node(30);      head = mergeSort(head);     printList(head);      return 0; }
Java 
// Iterative Java program to sort a linked // list using merge sort class Node {     int data;     Node next;      Node(int x) {        data = x;        next = null;     } }  class GfG {        // Function to merge two sorted sublists     static Node merge(Node left, Node right) {          // Handle the base case when either          // left or right is null         if (left == null) return right;         if (right == null) return left;          Node result = null;         Node curr = null;          // Initialize result to the smaller node          // and set the current pointer         if (left.data <= right.data) {             result = left;             left = left.next;         } else {             result = right;             right = right.next;         }         curr = result;          // Merge the nodes in sorted order         while (left != null && right != null) {             if (left.data <= right.data) {                 curr.next = left;                 left = left.next;             } else {                 curr.next = right;                 right = right.next;             }             curr = curr.next;         }          // Append the remaining nodes from either list         if (left != null) curr.next = left;         else curr.next = right;          return result;     }      // Function to split the list into     // sublists of given size     static Node split(Node head, int size) {         for (int i = 1; head != null && i < size; i++) {             head = head.next;         }         if (head == null) return null;          Node second = head.next;         head.next = null;          return second;     }      // Iterative merge sort function for linked list     static Node mergeSort(Node head) {         if (head == null || head.next == null) return head;          // Calculate the total length of the list         int len = 0;         for (Node temp = head; temp != null; temp = temp.next) len++;          Node left, right;         Node sortedHead = null;         Node sortedTail = null;          // Bottom-up merge sort, iterating over sublist sizes         for (int size = 1; size < len; size *= 2) {             Node curr = head;             sortedHead = null;             sortedTail = null;              while (curr != null) {                 left = curr;                 right = split(left, size);                  curr = split(right, size);                   // Merge the two halves                 Node merged = merge(left, right);                  // Set the head for the first merged sublist                 if (sortedHead == null) {                     sortedHead = merged;                     sortedTail = sortedHead;                 }                  else {                     sortedTail.next = merged;                 }                  // Move the tail pointer to the end                 // of the merged list                 while (sortedTail.next != null) {                     sortedTail = sortedTail.next;                 }             }              // Update the head to the sorted list after              // merging all sublists of current size             head = sortedHead;         }          return head;     }     static void printList(Node curr) {         while (curr != null) {             System.out.print(" " + curr.data);             curr = curr.next;         }     }      public static void main(String[] args) {          // Create a hard-coded linked list:         // 40 -> 20 -> 60 -> 10 -> 50 -> 30         Node head = new Node(40);         head.next = new Node(20);         head.next.next = new Node(60);         head.next.next.next = new Node(10);         head.next.next.next.next = new Node(50);         head.next.next.next.next.next = new Node(30);          head = mergeSort(head);         printList(head);     } }
Python 
# Iterative Python program to sort a linked  # list using merge sort class Node:     def __init__(self, x):         self.data = x         self.next = None  # Function to merge two sorted sublists def merge(left, right):          # Handle the base case when either      # left or right is null     if not left:         return right     if not right:         return left      result = None     curr = None      # Initialize result to the smaller node      # and set the current pointer     if left.data <= right.data:         result = left         left = left.next     else:         result = right         right = right.next          curr = result      # Merge the nodes in sorted order     while left and right:         if left.data <= right.data:             curr.next = left             left = left.next         else:             curr.next = right             right = right.next         curr = curr.next      # Append the remaining nodes      # from either list     if left:         curr.next = left     else:         curr.next = right      return result  # Function to split the list into # sublists of given size def split(head, size):     for i in range(1, size):         if not head:             return None         head = head.next     if not head:         return None      second = head.next     head.next = None     return second  # Iterative merge sort function for # linked list def mergeSort(head):     if not head or not head.next:         return head      # Calculate the total length of the list     length = 0     temp = head     while temp:         length += 1         temp = temp.next      left, right = None, None     sorted_head, sorted_tail = None, None      # Bottom-up merge sort, iterating      # over sublist sizes     size = 1     while size < length:         curr = head         sorted_head = None         sorted_tail = None          while curr:             left = curr             right = split(left, size)              curr = split(right, size)              # Merge the two halves             merged = merge(left, right)              # Set the head for the first             # merged sublist             if not sorted_head:                 sorted_head = merged                 sorted_tail = sorted_head             else:                 sorted_tail.next = merged              # Move the tail pointer to the end              # of the merged list             while sorted_tail.next:                 sorted_tail = sorted_tail.next          # Update the head to the sorted list after          # merging all sublists of current size         head = sorted_head         size *= 2      return head    def print_list(curr):     while curr:         print(f" {curr.data}", end="")         curr = curr.next     print()  if __name__ == "__main__":      # Create a hard-coded linked list:     # 40 -> 20 -> 60 -> 10 -> 50 -> 30     head = Node(40)     head.next = Node(20)     head.next.next = Node(60)     head.next.next.next = Node(10)     head.next.next.next.next = Node(50)     head.next.next.next.next.next = Node(30)      head = mergeSort(head)     print_list(head)
C# 
// Iterative C# program to sort a linked  // list using merge sort using System;  class Node {     public int data;     public Node next;      public Node(int x) {         data = x;         next = null;     } }  class GfG {      // Function to merge two sorted sublists     static Node merge(Node left, Node right) {          // Handle the base case when either          // left or right is null         if (left == null) return right;         if (right == null) return left;          Node result = null;         Node curr = null;          // Initialize result to the smaller node          // and set the current pointer         if (left.data <= right.data) {             result = left;             left = left.next;         } else {             result = right;             right = right.next;         }         curr = result;          // Merge the nodes in sorted order         while (left != null && right != null) {             if (left.data <= right.data) {                 curr.next = left;                 left = left.next;             } else {                 curr.next = right;                 right = right.next;             }             curr = curr.next;         }          // Append the remaining nodes from       	// either list         if (left != null) curr.next = left;         else curr.next = right;          return result;     }      // Function to split the list into     // sublists of given size     static Node split(Node head, int size) {         for (int i = 1; head != null && i < size; i++) {             head = head.next;         }         if (head == null) return null;          Node second = head.next;         head.next = null;         return second;     }      // Iterative merge sort function for    	// linked list     static Node mergeSort(Node head) {         if (head == null || head.next == null) return head;          // Calculate the total length of        	// the list         int length = 0;         for (Node temp = head; temp != null;	              		temp = temp.next) length++;          Node left, right;         Node sortedHead = null;         Node sortedTail = null;          // Bottom-up merge sort, iterating over        	// sublist sizes         for (int size = 1; size < length; size *= 2) {             Node curr = head;             sortedHead = null;             sortedTail = null;              while (curr != null) {                 left = curr;                 right = split(left, size);                  curr = split(right, size);                   // Merge the two halves                 Node merged = merge(left, right);                  // Set the head for the first                	// merged sublist                 if (sortedHead == null) {                     sortedHead = merged;                     sortedTail = sortedHead;                 } else {                     sortedTail.next = merged;                 }                  // Move the tail pointer to the end of               	// the merged list                 while (sortedTail.next != null) {                     sortedTail = sortedTail.next;                 }             }              // Update the head to the sorted list after              // merging all sublists of current size             head = sortedHead;         }          return head;     }      static void PrintList(Node curr) {         while (curr != null) {             Console.Write(" " + curr.data);             curr = curr.next;         }         Console.WriteLine();     }      static void Main(string[] args) {          // Create a hard-coded linked list:         // 40 -> 20 -> 60 -> 10 -> 50 -> 30         Node head = new Node(40);         head.next = new Node(20);         head.next.next = new Node(60);         head.next.next.next = new Node(10);         head.next.next.next.next = new Node(50);         head.next.next.next.next.next = new Node(30);          head = mergeSort(head);         PrintList(head);     } }
JavaScript 
// Iterative JavaScript program to sort a linked  // list using merge sort class Node {     constructor(x) {         this.data = x;         this.next = null;     } }  // Function to merge two sorted sublists function merge(left, right) {          // Handle the base case when either      // left or right is null     if (!left) return right;     if (!right) return left;      let result = null;     let curr = null;      // Initialize result to the smaller node      // and set the current pointer     if (left.data <= right.data) {         result = left;         left = left.next;     } else {         result = right;         right = right.next;     }     curr = result;      // Merge the nodes in sorted order     while (left && right) {         if (left.data <= right.data) {             curr.next = left;             left = left.next;         } else {             curr.next = right;             right = right.next;         }         curr = curr.next;     }      // Append the remaining nodes      // from either list     if (left) curr.next = left;     else curr.next = right;      return result; }  // Function to split the list into // sublists of given size function split(head, size) {     for (let i = 1; head && i < size; i++) {         head = head.next;     }     if (!head) return null;      const second = head.next;     head.next = null;     return second; }  // Iterative merge sort function for linked list function mergeSort(head) {     if (!head || !head.next) return head;      // Calculate the total length of the list     let length = 0;     for (let temp = head; temp;      			temp = temp.next) length++;      let left, right;     let sortedHead = null;     let sortedTail = null;      // Bottom-up merge sort, iterating     // over sublist sizes     for (let size = 1; size < length; size *= 2) {         let curr = head;         sortedHead = null;         sortedTail = null;          while (curr) {             left = curr;             right = split(left, size);              curr = split(right, size);               // Merge the two halves             const merged = merge(left, right);              // Set the head for the first             // merged sublist             if (!sortedHead) {                 sortedHead = merged;                 sortedTail = sortedHead;             } else {                 sortedTail.next = merged;             }              // Move the tail pointer to the              // end of the merged list             while (sortedTail.next) {                 sortedTail = sortedTail.next;             }         }          // Update the head to the sorted list after          // merging all sublists of current size         head = sortedHead;     }      return head; }  function printList(curr) {     while (curr !== null) {         console.log(` ${curr.data}`);         curr = curr.next;     }     console.log(); }  // Create a hard-coded linked list: // 40 -> 20 -> 60 -> 10 -> 50 -> 30 let head = new Node(40); head.next = new Node(20); head.next.next = new Node(60); head.next.next.next = new Node(10); head.next.next.next.next = new Node(50); head.next.next.next.next.next = new Node(30);  head = mergeSort(head); printList(head);

Output
 10 20 30 40 50 60

Time complexity: O(n log n), where n is the number of nodes in the linked list, due to repeatedly dividing the list and merging.
Auxiliary Space: O(1), since it does not use additional data structures for storing nodes during the merge process.



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Merge two sorted lists (in-place)

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Article Tags :
  • DSA
  • Linked List
  • Linked-List-Sorting
  • Merge Sort
Practice Tags :
  • Linked List
  • Merge Sort

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