Merge Sort for Linked Lists Last Updated : 19 Sep, 2024 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice Given a singly linked list, The task is to sort the linked list in non-decreasing order using merge sort.Examples: Input: 40 -> 20 -> 60 -> 10 -> 50 -> 30 -> NULLOutput: 10 -> 20 -> 30 -> 40 -> 50 -> 60 -> NULLInput: 9 -> 5 -> 2 -> 8 -> NULLOutput: 2 -> 5 -> 8 -> 9 -> NULL Approach:The prerequisite for this problem is Merge Sort. Here we have to maintain a MergeSort function that sorts the list in three steps:Split the List into Two Halves: Use two pointers, fast and slow, starting at the head. Move fast two steps and slow one step. When fast reaches the end, slow is at the midpoint. Split the list into two halves: the first half from head to just before slow, and the second from slow->next to the end. Set slow->next to NULL.Apply MergeSort Recursively: Recursively call MergeSort() on both halves. The base case is when the list is empty (head == NULL) or has one node (head->next == NULL), in which case return the list as is.Merge the Two Sorted Halves: After sorting both halves, call merge() to merge them by comparing nodes and linking accordingly. Append any remaining nodes from the exhausted half. Finally, returns the new head of the sorted list. C++ // C++ program for merge sort on singly linked list #include <iostream> using namespace std; class Node { public: int data; Node *next; Node(int x) { data = x; next = nullptr; } }; // Function to split the singly linked list into two halves Node *split(Node *head) { Node *fast = head; Node *slow = head; // Move fast pointer two steps and slow pointer // one step until fast reaches the end while (fast != nullptr && fast->next != nullptr) { fast = fast->next->next; if (fast != nullptr) { slow = slow->next; } } // Split the list into two halves Node *temp = slow->next; slow->next = nullptr; return temp; } // Function to merge two sorted singly linked lists Node *merge(Node *first, Node *second) { // If either list is empty, return the other list if (first == nullptr) return second; if (second == nullptr) return first; // Pick the smaller value between first and second nodes if (first->data < second->data) { // Recursively merge the rest of the lists and // link the result to the current node first->next = merge(first->next, second); return first; } else { // Recursively merge the rest of the lists // and link the result to the current node second->next = merge(first, second->next); return second; } } // Function to perform merge sort on a singly linked list Node *MergeSort(Node *head) { // Base case: if the list is empty or has only one node, // it's already sorted if (head == nullptr || head->next == nullptr) return head; // Split the list into two halves Node *second = split(head); // Recursively sort each half head = MergeSort(head); second = MergeSort(second); // Merge the two sorted halves return merge(head, second); } void printList(Node *head) { Node *curr = head; while (curr != nullptr) { cout << curr->data << " "; curr = curr->next; } cout << endl; } int main() { // Create a hard-coded singly linked list: // 9 -> 8 -> 5 -> 2 Node *head = new Node(9); head->next = new Node(8); head->next->next = new Node(5); head->next->next->next = new Node(2); head = MergeSort(head); printList(head); return 0; } C // C program for merge sort on singly linked list #include <stdio.h> #include <stdlib.h> struct Node { int data; struct Node* next; }; // Function to split the singly linked list into two halves struct Node* split(struct Node* head) { struct Node* fast = head; struct Node* slow = head; // Move fast pointer two steps and slow pointer // one step until fast reaches the end while (fast != NULL && fast->next != NULL) { fast = fast->next->next; if (fast != NULL) { slow = slow->next; } } // Split the list into two halves struct Node* temp = slow->next; slow->next = NULL; return temp; } // Function to merge two sorted singly linked lists struct Node* merge(struct Node* first, struct Node* second) { // If either list is empty, return the other list if (first == NULL) return second; if (second == NULL) return first; // Pick the smaller value between first and second nodes if (first->data < second->data) { // Recursively merge the rest of the lists and // link the result to the current node first->next = merge(first->next, second); return first; } else { // Recursively merge the rest of the lists // and link the result to the current node second->next = merge(first, second->next); return second; } } // Function to perform merge sort on a singly linked list struct Node* MergeSort(struct Node* head) { // Base case: if the list is empty or has only one node, // it's already sorted if (head == NULL || head->next == NULL) { return head; } // Split the list into two halves struct Node* second = split(head); // Recursively sort each half head = MergeSort(head); second = MergeSort(second); // Merge the two sorted halves return merge(head, second); } void printList(struct Node* head) { struct Node* curr = head; while (curr != NULL) { printf("%d ", curr->data); curr = curr->next; } printf("\n"); } // Function to create a new node struct Node* createNode(int x) { struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); newNode->data = x; newNode->next = NULL; return newNode; } int main() { // Create a hard-coded singly linked list: // 9 -> 8 -> 5 -> 2 struct Node* head = createNode(9); head->next = createNode(8); head->next->next = createNode(5); head->next->next->next = createNode(2); head = MergeSort(head); printList(head); return 0; } Java // Java program for merge sort on singly linked list class Node { int data; Node next; Node(int x) { data = x; next = null; } } // Function to split the singly linked list into two halves class GfG { static Node split(Node head) { Node fast = head; Node slow = head; // Move fast pointer two steps and slow pointer // one step until fast reaches the end while (fast != null && fast.next != null) { fast = fast.next.next; if (fast != null) { slow = slow.next; } } // Split the list into two halves Node temp = slow.next; slow.next = null; return temp; } // Function to merge two sorted singly linked lists static Node merge(Node first, Node second) { // If either list is empty, return the other list if (first == null) return second; if (second == null) return first; // Pick the smaller value between first and second nodes if (first.data < second.data) { // Recursively merge the rest of the lists and // link the result to the current node first.next = merge(first.next, second); return first; } else { // Recursively merge the rest of the lists // and link the result to the current node second.next = merge(first, second.next); return second; } } // Function to perform merge sort on a singly linked list static Node mergeSort(Node head) { // Base case: if the list is empty or has only one node, // it's already sorted if (head == null || head.next == null) { return head; } // Split the list into two halves Node second = split(head); // Recursively sort each half head = mergeSort(head); second = mergeSort(second); // Merge the two sorted halves return merge(head, second); } static void printList(Node head) { Node curr = head; while (curr != null) { System.out.print(curr.data + " "); curr = curr.next; } System.out.println(); } public static void main(String[] args) { // Create a hard-coded singly linked list: // 9 -> 8 -> 5 -> 2 Node head = new Node(9); head.next = new Node(8); head.next.next = new Node(5); head.next.next.next = new Node(2); head = mergeSort(head); printList(head); } } Python # Python program for merge sort on singly linked list class Node: def __init__(self, x): self.data = x self.next = None def split(head): fast = head slow = head # Move fast pointer two steps and slow pointer # one step until fast reaches the end while fast and fast.next: fast = fast.next.next if fast: slow = slow.next # Split the list into two halves second = slow.next slow.next = None return second def merge(first, second): # If either list is empty, return the other list if not first: return second if not second: return first # Pick the smaller value between first and second nodes if first.data < second.data: first.next = merge(first.next, second) return first else: second.next = merge(first, second.next) return second def merge_sort(head): # Base case: if the list is empty or has only one node, # it's already sorted if not head or not head.next: return head # Split the list into two halves second = split(head) # Recursively sort each half head = merge_sort(head) second = merge_sort(second) # Merge the two sorted halves return merge(head, second) def print_list(head): curr = head while curr: print(curr.data, end=" ") curr = curr.next print() if __name__ == "__main__": # Create a hard-coded singly linked list: # 9 -> 8 -> 5 -> 2 head = Node(9) head.next = Node(8) head.next.next = Node(5) head.next.next.next = Node(2) head = merge_sort(head) print_list(head) C# // C# program for merge sort on singly linked list using System; class Node { public int data; public Node next; public Node(int x) { data = x; next = null; } } // Function to split the singly linked list into two halves class GfG { static Node Split(Node head) { Node fast = head; Node slow = head; // Move fast pointer two steps and slow pointer // one step until fast reaches the end while (fast != null && fast.next != null) { fast = fast.next.next; if (fast != null) { slow = slow.next; } } // Split the list into two halves Node temp = slow.next; slow.next = null; return temp; } // Function to merge two sorted singly linked lists static Node Merge(Node first, Node second) { // If either list is empty, return the other list if (first == null) return second; if (second == null) return first; // Pick the smaller value between first and second // nodes if (first.data < second.data) { // Recursively merge the rest of the lists and // link the result to the current node first.next = Merge(first.next, second); return first; } else { // Recursively merge the rest of the lists // and link the result to the current node second.next = Merge(first, second.next); return second; } } // Function to perform merge sort on a singly linked // list static Node MergeSort(Node head) { // Base case: if the list is empty or has only one // node, it's already sorted if (head == null || head.next == null) return head; // Split the list into two halves Node second = Split(head); // Recursively sort each half head = MergeSort(head); second = MergeSort(second); // Merge the two sorted halves return Merge(head, second); } static void PrintList(Node head) { Node curr = head; while (curr != null) { Console.Write(curr.data + " "); curr = curr.next; } Console.WriteLine(); } public static void Main() { // Create a hard-coded singly linked list: // 9 -> 8 -> 5 -> 2 Node head = new Node(9); head.next = new Node(8); head.next.next = new Node(5); head.next.next.next = new Node(2); head = MergeSort(head); PrintList(head); } } JavaScript // JavaScript program for merge sort on singly linked list class Node { constructor(x) { this.data = x; this.next = null; } } // Function to split the singly linked list into two halves function split(head) { let fast = head; let slow = head; // Move fast pointer two steps and slow pointer // one step until fast reaches the end while (fast && fast.next) { fast = fast.next.next; if (fast) { slow = slow.next; } } // Split the list into two halves let second = slow.next; slow.next = null; return second; } // Function to merge two sorted singly linked lists function merge(first, second) { // If either list is empty, return the other list if (!first) return second; if (!second) return first; // Pick the smaller value between first and second nodes if (first.data < second.data) { first.next = merge(first.next, second); return first; } else { second.next = merge(first, second.next); return second; } } // Function to perform merge sort on a singly linked list function mergeSort(head) { // Base case: if the list is empty or has only one node, // it's already sorted if (!head || !head.next) return head; // Split the list into two halves let second = split(head); // Recursively sort each half head = mergeSort(head); second = mergeSort(second); // Merge the two sorted halves return merge(head, second); } function printList(head) { let curr = head; while (curr) { console.log(curr.data + " "); curr = curr.next; } console.log(); } // Create a hard-coded singly linked list: // 9 -> 8 -> 5 -> 2 let head = new Node(9); head.next = new Node(8); head.next.next = new Node(5); head.next.next.next = new Node(2); head = mergeSort(head); printList(head); Output2 5 8 9 Time Complexity: O(n*log(n))Auxiliary Space: O(logn) Comment More infoAdvertise with us Next Article Merge Sort for Doubly Linked List kartik Follow Improve Article Tags : Linked List Sorting DSA Microsoft Adobe Paytm Accolite MAQ Software Veritas Merge Sort Linked-List-Sorting +7 More Practice Tags : AccoliteAdobeMAQ SoftwareMicrosoftPaytmVeritasLinked ListMerge SortSorting +5 More Similar Reads Merge Sort - Data Structure and Algorithms Tutorials Merge sort is a popular sorting algorithm known for its efficiency and stability. 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Here we handle arrays of unequal si 15+ min read Merge k Sorted ArraysGiven K sorted arrays, merge them and print the sorted output.Examples:Input: K = 3, arr = { {1, 3, 5, 7}, {2, 4, 6, 8}, {0, 9, 10, 11}}Output: 0 1 2 3 4 5 6 7 8 9 10 11 Input: k = 4, arr = { {1}, {2, 4}, {3, 7, 9, 11}, {13} }Output: 1 2 3 4 7 9 11 13Table of ContentNaive - Concatenate all and SortU 15+ min read Merge K sorted arrays of different sizes | ( Divide and Conquer Approach )Given k sorted arrays of different length, merge them into a single array such that the merged array is also sorted.Examples: Input : {{3, 13}, {8, 10, 11} {9, 15}} Output : {3, 8, 9, 10, 11, 13, 15} Input : {{1, 5}, {2, 3, 4}} Output : {1, 2, 3, 4, 5} Let S be the total number of elements in all th 8 min read Merge K sorted linked listsGiven k sorted linked lists of different sizes, the task is to merge them all maintaining their sorted order.Examples: Input: Output: Merged lists in a sorted order where every element is greater than the previous element.Input: Output: Merged lists in a sorted order where every element is greater t 15+ min read Union and Intersection of two Linked List using Merge SortGiven two singly Linked Lists, create union and intersection lists that contain the union and intersection of the elements present in the given lists. Each of the two lists contains distinct node values.Note: The order of elements in output lists doesn't matter.Examples:Input: head1: 10 -> 15 - 15+ min read Sorting by combining Insertion Sort and Merge Sort algorithmsInsertion sort: The array is virtually split into a sorted and an unsorted part. Values from the unsorted part are picked and placed at the correct position in the sorted part.Advantages: Following are the advantages of insertion sort: If the size of the list to be sorted is small, insertion sort ru 2 min read Find array with k number of merge sort callsGiven two numbers n and k, find an array containing values in [1, n] and requires exactly k calls of recursive merge sort function. Examples: Input : n = 3 k = 3 Output : a[] = {2, 1, 3} Explanation: Here, a[] = {2, 1, 3} First of all, mergesort(0, 3) will be called, which then sets mid = 1 and call 6 min read Difference of two Linked Lists using Merge sortGiven two Linked List, the task is to create a Linked List to store the difference of Linked List 1 with Linked List 2, i.e. the elements present in List 1 but not in List 2.Examples: Input: List1: 10 -> 15 -> 4 ->20, List2: 8 -> 4 -> 2 -> 10 Output: 15 -> 20 Explanation: In the 14 min read Like