Rearrange a given linked list in-place Last Updated : 04 Sep, 2024 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice Given a singly linked list L0 -> L1 -> … -> Ln-1 -> Ln. Rearrange the nodes in the list so that the newly formed list is : L0 -> Ln -> L1 -> Ln-1 -> L2 -> Ln-2 ... You are required to do this in place without altering the nodes' values. Examples: Input: 1 -> 2 -> 3 -> 4Output: 1 -> 4 -> 2 -> 3 Explanation: Here n = 4, so the correct order is L0->L3->L1->L2Input: 1 -> 2 -> 3 -> 4 -> 5 Output: 1 -> 5 -> 2 -> 4 -> 3Explanation: Here n = 4, so the correct order is L0->L4->L1->L3->L2[Naive Approach] Using two nested loops- O(n^2) Time and O(1) SpaceThe idea is to start traversing from head node. For each node, perform the following operations:Traverse the list to find the last node.Disconnect the last node from the list.Place the removed last node after the current node.Update current to its next node.[Efficient Approach] By Reversing Second Half - O(n) Time and O(1) SpaceFind the middle of the linked list using the fast and slow pointer method. This involves moving one pointer twice as fast as the other so that when the faster pointer reaches the end, the slower pointer will be at the middle. Reverse the second half of the list starting just after the middle node and split them in two parts.Merge the two halves together by alternating nodes from the first half with nodes from the reversed second half.Below is the implementation of the above approach : C++ // C++ code to rearrange a given linked list in-place #include <bits/stdc++.h> using namespace std; class Node { public: int data; Node *next; Node(int x) { data = x; next = NULL; } }; Node *reverselist(Node *head) { Node *prev = nullptr, *curr = head, *next; while (curr) { next = curr->next; curr->next = prev; prev = curr; curr = next; } return prev; } void printList(Node *curr) { while (curr != nullptr) { cout << curr->data << " "; curr = curr->next; } } // Function to rearrange a linked list Node *rearrange(Node *head) { // Find the middle point using tortoise and hare method Node *slow = head, *fast = slow->next; while (fast && fast->next) { slow = slow->next; fast = fast->next->next; } // Split the linked list into two halves Node *head1 = head; Node *head2 = slow->next; slow->next = NULL; // Reverse the second half head2 = reverselist(head2); // Merge alternate nodes head = new Node(0); Node *curr = head; while (head1 || head2) { // First add the element from the first list if (head1) { curr->next = head1; curr = curr->next; head1 = head1->next; } // Then add the element from the second list if (head2) { curr->next = head2; curr = curr->next; head2 = head2->next; } } // Return the head of the new list return head->next; } int main() { // singly linked list 1->2->3->4->5 Node *head = new Node(1); head->next = new Node(2); head->next->next = new Node(3); head->next->next->next = new Node(4); head->next->next->next->next = new Node(5); head = rearrange(head); printList(head); return 0; } C // C code to rearrange a given linked list in-place #include <stdio.h> #include <stdlib.h> struct Node { int data; struct Node *next; }; struct Node *reverselist(struct Node *head) { // Initialize prev and current pointers struct Node *prev = NULL, *curr = head, *next; while (curr) { next = curr->next; curr->next = prev; prev = curr; curr = next; } return prev; } void printList(struct Node *curr) { while (curr != NULL) { printf("%d ", curr->data); curr = curr->next; } } // Function to rearrange a linked list struct Node *rearrange(struct Node *head) { // Find the middle point using tortoise // and hare method struct Node *slow = head, *fast = slow->next; while (fast && fast->next) { slow = slow->next; fast = fast->next->next; } // Split the linked list into two halves // head1, head of first half // head2, head of second half struct Node *head1 = head; struct Node *head2 = slow->next; slow->next = NULL; // Reverse the second half head2 = reverselist(head2); // Merge alternate nodes struct Node dummy; struct Node *curr = &dummy; dummy.next = NULL; while (head1 || head2) { // First add the element from head1 if (head1) { curr->next = head1; curr = curr->next; head1 = head1->next; } // Then add the element from head2 if (head2) { curr->next = head2; curr = curr->next; head2 = head2->next; } } return dummy.next; } struct Node *newNode(int key) { struct Node *temp = (struct Node *)malloc(sizeof(struct Node)); temp->data = key; temp->next = NULL; return temp; } int main() { // singly linked list 1->2->3->4->5 struct Node *head = newNode(1); head->next = newNode(2); head->next->next = newNode(3); head->next->next->next = newNode(4); head->next->next->next->next = newNode(5); head = rearrange(head); printList(head); return 0; } Java // Java code to rearrange a given linked list in-place class Node { int data; Node next; Node(int d) { data = d; next = null; } } class GfG { static void printList(Node node) { if (node == null) { return; } while (node != null) { System.out.print(node.data + " "); node = node.next; } } static Node reverseList(Node node) { Node prev = null, curr = node, next; while (curr != null) { next = curr.next; curr.next = prev; prev = curr; curr = next; } return prev; } // Method to rearrange the linked list static Node rearrange(Node node) { // Check if the list is empty or has only one node if (node == null || node.next == null) { return node; } // Find the middle point using tortoise and hare // method Node slow = node, fast = node.next; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } // Split the linked list into two halves Node firstHalf = node; Node secondHalf = slow.next; slow.next = null; // Reverse the second half of the list secondHalf = reverseList(secondHalf); // Merge alternate nodes from the two halves Node dummy = new Node(0); Node curr = dummy; while (firstHalf != null || secondHalf != null) { if (firstHalf != null) { curr.next = firstHalf; curr = curr.next; firstHalf = firstHalf.next; } if (secondHalf != null) { curr.next = secondHalf; curr = curr.next; secondHalf = secondHalf.next; } } // Return the new head of the rearranged list return dummy.next; } public static void main(String[] args){ // singly linked list 1->2->3->4->5 Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); head = rearrange(head); printList(head); } } Python # Python program to rearrange link list in place class Node: def __init__(self, d): self.data = d self.next = None def printlist(node): if node is None: return while node is not None: print(node.data, end=" ") node = node.next def reverselist(node): prev = None curr = node next = None while curr is not None: next = curr.next curr.next = prev prev = curr curr = next node = prev return node def rearrange(node): # Find the middle point using the # tortoise and hare method slow = node fast = slow.next while fast is not None and fast.next is not None: slow = slow.next fast = fast.next.next # Split the linked list into two halves node1 = node node2 = slow.next slow.next = None # Reverse the second half node2 = reverselist(node2) # Merge alternate nodes node = Node(0) curr = node while node1 is not None or node2 is not None: # Add the element from the first list if node1 is not None: curr.next = node1 curr = curr.next node1 = node1.next # Add the element from the second list if node2 is not None: curr.next = node2 curr = curr.next node2 = node2.next # Return the head of the rearranged list return node.next # singly linked list 1->2->3->4->5 head = None head = Node(1) head.next = Node(2) head.next.next = Node(3) head.next.next.next = Node(4) head.next.next.next.next = Node(5) head = rearrange(head) printlist(head) C# // C# program to rearrange link list in place using System; public class Node { public int Data; public Node Next; public Node(int data) { Data = data; Next = null; } } class GfG { static void PrintList(Node node) { if (node == null) { Console.WriteLine("List is empty."); return; } while (node != null) { Console.Write(node.Data + " "); node = node.Next; } } static Node ReverseList(Node node) { Node prev = null; Node curr = node; Node next = null; while (curr != null) { next = curr.Next; curr.Next = prev; prev = curr; curr = next; } return prev; } // Method to rearrange the linked list in-place and // return the new head static Node Rearrange(Node node) { if (node == null || node.Next == null) return node; // Find the middle point using tortoise and hare // method Node slow = node; Node fast = node.Next; while (fast != null && fast.Next != null) { slow = slow.Next; fast = fast.Next.Next; } // Split the linked list into two halves Node firstHalf = node; Node secondHalf = slow.Next; slow.Next = null; // Reverse the second half secondHalf = ReverseList(secondHalf); // Merge alternate nodes Node dummy = new Node(0); Node curr = dummy; while (firstHalf != null || secondHalf != null) { if (firstHalf != null) { curr.Next = firstHalf; curr = curr.Next; firstHalf = firstHalf.Next; } if (secondHalf != null) { curr.Next = secondHalf; curr = curr.Next; secondHalf = secondHalf.Next; } } // Return the new head of the list return dummy.Next; } public static void Main(string[] args) { //singly linked list 1->2->3->4->5 Node head = new Node(1); head.Next = new Node(2); head.Next.Next = new Node(3); head.Next.Next.Next = new Node(4); head.Next.Next.Next.Next = new Node(5); head = Rearrange(head); PrintList(head); } } JavaScript // Javascript program to rearrange link list in place class Node { constructor(d) { this.data = d; this.next = null; } } function printList(node) { let result = ""; while (node != null) { result += node.data + " "; node = node.next; } console.log(result); } function reverseList(node) { let prev = null, curr = node, next; while (curr != null) { next = curr.next; curr.next = prev; prev = curr; curr = next; } return prev; } // Function to rearrange the linked list function rearrange(head) { if (head == null || head.next == null) { return head; } // Find the middle point using // tortoise and hare method let slow = head, fast = slow.next; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } // Split the linked list into two halves let node1 = head; let node2 = slow.next; slow.next = null; // Reverse the second half node2 = reverseList(node2); // Merge alternate nodes let dummy = new Node(0); let curr = dummy; while (node1 != null || node2 != null) { if (node1 != null) { curr.next = node1; curr = curr.next; node1 = node1.next; } if (node2 != null) { curr.next = node2; curr = curr.next; node2 = node2.next; } } return dummy.next; } // singly linked list 1->2->3->4->5 let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); head = rearrange(head); printList(head); Output1 5 2 4 3 Time Complexity: O(n) , where n are the number of nodes in linked list.Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Analysis of Algorithms K kartik Follow Improve Article Tags : Linked List DSA Amazon Practice Tags : AmazonLinked List Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. Itâs the heart of coding, enabling programmers to think, reason, and arrive at smart solutions just like we do.Here are some tips for improving your programming logic: Understand the pro 2 min read Analysis of AlgorithmsAnalysis of Algorithms is a fundamental aspect of computer science that involves evaluating performance of algorithms and programs. Efficiency is measured in terms of time and space.BasicsWhy is Analysis Important?Order of GrowthAsymptotic Analysis Worst, Average and Best Cases Asymptotic NotationsB 1 min read Data StructuresArray Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous 3 min read String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut 2 min read Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The 2 min read Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Hereâs the comparison of Linked List vs Arrays Linked List: 2 min read Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first 2 min read Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems 2 min read Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most 4 min read Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of 3 min read Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this 15+ min read AlgorithmsSearching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input 2 min read Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ 3 min read Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution 14 min read Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get 3 min read Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net 3 min read Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of 3 min read Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit 4 min read AdvancedSegment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree 3 min read Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i 2 min read GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br 2 min read Interview PreparationInterview Corner: All Resources To Crack Any Tech InterviewThis article serves as your one-stop guide to interview preparation, designed to help you succeed across different experience levels and company expectations. Here is what you should expect in a Tech Interview, please remember the following points:Tech Interview Preparation does not have any fixed s 3 min read GfG160 - 160 Days of Problem SolvingAre you preparing for technical interviews and would like to be well-structured to improve your problem-solving skills? Well, we have good news for you! GeeksforGeeks proudly presents GfG160, a 160-day coding challenge starting on 15th November 2024. In this event, we will provide daily coding probl 3 min read Practice ProblemGeeksforGeeks Practice - Leading Online Coding PlatformGeeksforGeeks Practice is an online coding platform designed to help developers and students practice coding online and sharpen their programming skills with the following features. GfG 160: This consists of most popular interview problems organized topic wise and difficulty with with well written e 6 min read Problem of The Day - Develop the Habit of CodingDo you find it difficult to develop a habit of Coding? If yes, then we have a most effective solution for you - all you geeks need to do is solve one programming problem each day without any break, and BOOM, the results will surprise you! Let us tell you how:Suppose you commit to improve yourself an 5 min read Like