Insertion Sort for Singly Linked List Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice Given a singly linked list, the task is to sort the list (in ascending order) using the insertion sort algorithm.Examples:Input: 5->4->1->3->2Output: 1->2->3->4->5Input: 4->3->2->1Output: 1->2->3->4The prerequisite is Insertion Sort on Array. The idea is to gradually build a sorted portion of the list within the same memory space as the original list.Step-by-step approach:Start with an initially empty "sorted" list, which will be built by rearranging nodes from the original list.Traverse the original linked list one node at a time.For each node, find its correct position within the "sorted" portion of the list. If the node should be placed at the beginning (i.e., it's smaller than the first node in the sorted list), it becomes the new head of the sorted list.Otherwise, traverse the sorted list to find the correct position and insert the node there.Continue this process until all nodes from the original list have been repositioned in the sorted orderReturn the head of sorted list.Below is the implementation of the above approach C++ // C++ program to sort linked list // using insertion sort #include <bits/stdc++.h> using namespace std; class Node { public: int val; struct Node* next; Node(int x) { val = x; next = NULL; } }; // Function to insert a new_node in // the result list. Node* sortedInsert(Node* newnode, Node* sorted) { // Special case for the head end if (sorted == NULL || sorted->val >= newnode->val) { newnode->next = sorted; sorted = newnode; } else { Node* curr = sorted; // Locate the node before the point // of insertion while (curr->next != NULL && curr->next->val < newnode->val) { curr = curr->next; } newnode->next = curr->next; curr->next = newnode; } return sorted; } Node* insertionSort(Node* head) { // Initialize sorted linked list Node* sorted = NULL; Node* curr = head; // Traverse the given linked list and insert // every node to sorted while (curr != NULL) { // Store next for next iteration Node* next = curr->next; // Insert current in sorted linked list sorted = sortedInsert(curr, sorted); // Update current curr = next; } return sorted; } void printList(Node* curr) { while (curr != NULL) { cout << " " << curr->val; curr = curr->next; } } int main() { // Create a hard-coded linked list: // 5->4->1->3->2 Node* head = new Node(5); head->next = new Node(4); head->next->next = new Node(1); head->next->next->next = new Node(3); head->next->next->next->next = new Node(2); head = insertionSort(head); printList(head); return 0; } C // C program to sort linked list using insertion sort #include <stdio.h> #include <stdlib.h> struct Node { int val; struct Node* next; }; // Function to insert a new_node in the result list. struct Node* sortedInsert(struct Node* createNode, struct Node* sorted) { // Special case for the head end if (sorted == NULL || sorted->val >= createNode->val) { createNode->next = sorted; sorted = createNode; } else { struct Node* curr = sorted; // Locate the node before the point of insertion while (curr->next != NULL && curr->next->val < createNode->val) { curr = curr->next; } createNode->next = curr->next; curr->next = createNode; } return sorted; } struct Node* insertionSort(struct Node* head) { // Initialize sorted linked list struct Node* sorted = NULL; struct Node* curr = head; // Traverse the given linked list and insert // every node to sorted while (curr != NULL) { // Store next for next iteration struct Node* next = curr->next; // Insert current in sorted linked list sorted = sortedInsert(curr, sorted); // Update current curr = next; } return sorted; } void printList(struct Node* curr) { while (curr != NULL) { printf(" %d", curr->val); curr = curr->next; } } struct Node* createNode(int x) { struct Node* node = (struct Node*)malloc(sizeof(struct Node)); node->val = x; node->next = NULL; return node; } int main() { // Create a hard-coded linked list: // 5->4->1->3->2 struct Node* head = createNode(5); head->next = createNode(4); head->next->next = createNode(1); head->next->next->next = createNode(3); head->next->next->next->next = createNode(2); head = insertionSort(head); printList(head); return 0; } Java // Java program to sort linked list // using insertion sort class Node { int val; Node next; Node(int x) { val = x; next = null; } } public class GfG { // Function to insert a new_node in // the result list. static Node sortedInsert(Node newnode, Node sorted) { // Special case for the head end if (sorted == null || sorted.val >= newnode.val) { newnode.next = sorted; sorted = newnode; } else { Node curr = sorted; // Locate the node before the // point of insertion while (curr.next != null && curr.next.val < newnode.val) { curr = curr.next; } newnode.next = curr.next; curr.next = newnode; } return sorted; } static Node insertionSort(Node head) { // Initialize sorted linked list Node sorted = null; Node curr = head; // Traverse the given linked list and // insert every node to sorted while (curr != null) { // Store next for next iteration Node next = curr.next; // Insert current in sorted linked list sorted = sortedInsert(curr, sorted); // Update current curr = next; } return sorted; } static void printList(Node curr) { while (curr != null) { System.out.print(" " + curr.val); curr = curr.next; } } public static void main(String[] args) { // Create a hard-coded linked list: // 5->4->1->3->2 Node head = new Node(5); head.next = new Node(4); head.next.next = new Node(1); head.next.next.next = new Node(3); head.next.next.next.next = new Node(2); head = insertionSort(head); printList(head); } } Python # Python program to sort linked list # using insertion sort class Node: def __init__(self, x): self.val = x self.next = None def sorted_insert(newnode, sorted_head): # Special case for the head end if sorted_head is None or sorted_head.val >= newnode.val: newnode.next = sorted_head return newnode else: curr = sorted_head # Locate the node before the point of insertion while curr.next is not None and curr.next.val < newnode.val: curr = curr.next newnode.next = curr.next curr.next = newnode return sorted_head def insertion_sort(head): # Initialize sorted linked list sorted_head = None curr = head # Traverse the given linked list and # insert every node to sorted while curr is not None: next_node = curr.next # Insert current node in sorted linked list sorted_head = sorted_insert(curr, sorted_head) # Update current curr = next_node return sorted_head def print_list(curr): while curr is not None: print(f" {curr.val}", end="") curr = curr.next print() if __name__ == "__main__": # Create a hard-coded linked list: # 5->4->1->3->2 head = Node(5) head.next = Node(4) head.next.next = Node(1) head.next.next.next = Node(3) head.next.next.next.next = Node(2) head = insertion_sort(head) print_list(head) C# // C# program to sort linked list // using insertion sort using System; class Node { public int Val; public Node Next; public Node(int x) { Val = x; Next = null; } } class GfG { // Function to insert a new node in // the result list. static Node SortedInsert(Node newNode, Node sorted) { // Special case for the head end if (sorted == null || sorted.Val >= newNode.Val) { newNode.Next = sorted; sorted = newNode; } else { Node curr = sorted; // Locate the node before the // point of insertion while (curr.Next != null && curr.Next.Val < newNode.Val) { curr = curr.Next; } newNode.Next = curr.Next; curr.Next = newNode; } return sorted; } static Node InsertionSort(Node head) { // Initialize sorted linked list Node sorted = null; Node curr = head; // Traverse the given linked list and // insert every node to sorted while (curr != null) { // Store next for next iteration Node next = curr.Next; // Insert current in sorted linked list sorted = SortedInsert(curr, sorted); // Update current curr = next; } return sorted; } static void PrintList(Node curr) { while (curr != null) { Console.Write(" " + curr.Val); curr = curr.Next; } Console.WriteLine(); } static void Main() { // Create a hard-coded linked list: // 5->4->1->3->2 Node head = new Node(5); head.Next = new Node(4); head.Next.Next = new Node(1); head.Next.Next.Next = new Node(3); head.Next.Next.Next.Next = new Node(2); head = InsertionSort(head); PrintList(head); } } JavaScript // JavaScript program to sort linked list // using insertion sort class Node { constructor(val) { this.val = val; this.next = null; } } // Function to insert a new_node in // the result list. function sortedInsert(newnode, sorted) { // Special case for the head end if (sorted === null || sorted.val >= newnode.val) { newnode.next = sorted; sorted = newnode; } else { let curr = sorted; // Locate the node before // the point of insertion while (curr.next !== null && curr.next.val < newnode.val) { curr = curr.next; } newnode.next = curr.next; curr.next = newnode; } return sorted; } function insertionSort(head) { // Initialize sorted linked list let sorted = null; let curr = head; // Traverse the given linked list // and insert every node to sorted while (curr !== null) { // Store next for next iteration let next = curr.next; // Insert current in sorted linked list sorted = sortedInsert(curr, sorted); // Update current curr = next; } return sorted; } function printList(curr) { while (curr !== null) { console.log(" " + curr.val); curr = curr.next; } console.log(); } // Create a hard-coded linked list: // 5->4->1->3->2 let head = new Node(5); head.next = new Node(4); head.next.next = new Node(1); head.next.next.next = new Node(3); head.next.next.next.next = new Node(2); head = insertionSort(head); printList(head); Output 1 2 3 4 5Time Complexity: O(n2), In the worst case, we might have to traverse all nodes of the sorted list for inserting a node. 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