Given a singly linked list L0 -> L1 -> … -> Ln-1 -> Ln. Rearrange the nodes in the list so that the new 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 -> 4
Output: 1 -> 4 -> 2 -> 3
Input: 1 -> 2 -> 3 -> 4 -> 5
Output: 1 -> 5 -> 2 -> 4 -> 3
Simple Solution
1) Initialize current node as head.
2) While next of current node is not null, do following
a) Find the last node, remove it from the end and insert it as next
of the current node.
b) Move current to next of current
The time complexity of the above simple solution is O(n2) where n is the number of nodes in the linked list.
Better Solution:
- Copy contents of the given linked list to a vector.
- Rearrange the given vector by swapping nodes from both ends.
- Copy the modified vector back to the linked list.
Implementation of this approach: https://ide.geeksforgeeks.org/1eGSEy
Efficient Solution:
1) Find the middle point using tortoise and hare method.
2) Split the linked list into two halves using found middle point in step 1.
3) Reverse the second half.
4) Do alternate merge of first and second halves.
Below is the implementation of this method.
C++
// C++ program to rearrange a linked list in-place#include <bits/stdc++.h>using namespace std;// Linkedlist Node structurestruct Node { int data; struct Node* next;};// Function to create newNode in a linkedlistNode* newNode(int key){ Node* temp = new Node; temp->data = key; temp->next = NULL; return temp;}// Function to reverse the linked listvoid reverselist(Node** head){ // Initialize prev and current pointers Node *prev = NULL, *curr = *head, *next; while (curr) { next = curr->next; curr->next = prev; prev = curr; curr = next; } *head = prev;}// Function to print the linked listvoid printlist(Node* head){ while (head != NULL) { cout << head->data << " "; if (head->next) cout << "-> "; head = head->next; } cout << endl;}// Function to rearrange a linked listvoid rearrange(Node** head){ // 1) 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; } // 2) Split the linked list in two halves // head1, head of first half 1 -> 2 // head2, head of second half 3 -> 4 Node* head1 = *head; Node* head2 = slow->next; slow->next = NULL; // 3) Reverse the second half, i.e., 4 -> 3 reverselist(&head2); // 4) Merge alternate nodes *head = newNode(0); // Assign dummy Node // curr is the pointer to this dummy Node, which will // be used to form the new list Node* curr = *head; while (head1 || head2) { // First add the element from 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; } } // Assign the head of the new list to head pointer *head = (*head)->next;}// Driver programint main(){ 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); printlist(head); // Print original list rearrange(&head); // Modify the list printlist(head); // Print modified list return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program to rearrange link list in place// Linked List Classclass LinkedList { static Node head; // head of the list /* Node Class */ static class Node { int data; Node next; // Constructor to create a new node Node(int d) { data = d; next = null; } } void printlist(Node node) { if (node == null) { return; } while (node != null) { System.out.print(node.data + " -> "); node = node.next; } } Node reverselist(Node node) { Node prev = null, curr = node, next; while (curr != null) { next = curr.next; curr.next = prev; prev = curr; curr = next; } node = prev; return node; } void rearrange(Node node) { // 1) Find the middle point using tortoise and hare // method Node slow = node, fast = slow.next; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } // 2) Split the linked list in two halves // node1, head of first half 1 -> 2 -> 3 // node2, head of second half 4 -> 5 Node node1 = node; Node node2 = slow.next; slow.next = null; // 3) Reverse the second half, i.e., 5 -> 4 node2 = reverselist(node2); // 4) Merge alternate nodes node = new Node(0); // Assign dummy Node // curr is the pointer to this dummy Node, which // will be used to form the new list Node curr = node; while (node1 != null || node2 != null) { // First add the element from first list if (node1 != null) { curr.next = node1; curr = curr.next; node1 = node1.next; } // Then add the element from second list if (node2 != null) { curr.next = node2; curr = curr.next; node2 = node2.next; } } // Assign the head of the new list to head pointer node = node.next; } public static void main(String[] args) { LinkedList list = new LinkedList(); list.head = new Node(1); list.head.next = new Node(2); list.head.next.next = new Node(3); list.head.next.next.next = new Node(4); list.head.next.next.next.next = new Node(5); list.printlist(head); // print original list list.rearrange(head); // rearrange list as per ques System.out.println(""); list.printlist(head); // print modified list }}// This code has been contributed by Mayank Jaiswal |
Python3
# Python program to rearrange link list in place# Node Classclass Node: # Constructor to create a new node def __init__(self, d): self.data = d self.next = None def printlist(node): if(node == None): return while(node != None): print(node.data," -> ", end = "") node = node.nextdef reverselist(node): prev = None curr = node next=None while (curr != None): next = curr.next curr.next = prev prev = curr curr = next node = prev return nodedef rearrange(node): # 1) Find the middle point using tortoise and hare # method slow = node fast = slow.next while (fast != None and fast.next != None): slow = slow.next fast = fast.next.next # 2) Split the linked list in two halves # node1, head of first half 1 -> 2 -> 3 # node2, head of second half 4 -> 5 node1 = node node2 = slow.next slow.next = None # 3) Reverse the second half, i.e., 5 -> 4 node2 = reverselist(node2) # 4) Merge alternate nodes node = Node(0) #Assign dummy Node # curr is the pointer to this dummy Node, which # will be used to form the new list curr = node while (node1 != None or node2 != None): # First add the element from first list if (node1 != None): curr.next = node1 curr = curr.next node1 = node1.next # Then add the element from second list if(node2 != None): curr.next = node2 curr = curr.next node2 = node2.next # Assign the head of the new list to head pointer node = node.nexthead = Nonehead = Node(1)head.next = Node(2)head.next.next = Node(3)head.next.next.next = Node(4)head.next.next.next.next = Node(5)printlist(head) #print original listrearrange(head) #rearrange list as per quesprint()printlist(head) #print modified list# This code is contributed by ab2127 |
C#
// C# program to rearrange link list in placeusing System;// Linked List Classpublic class LinkedList { Node head; // head of the list /* Node Class */ class Node { public int data; public Node next; // Constructor to create a new node public Node(int d) { data = d; next = null; } } void printlist(Node node) { if (node == null) { return; } while (node != null) { Console.Write(node.data + " -> "); node = node.next; } } Node reverselist(Node node) { Node prev = null, curr = node, next; while (curr != null) { next = curr.next; curr.next = prev; prev = curr; curr = next; } node = prev; return node; } void rearrange(Node node) { // 1) Find the middle point using // tortoise and hare method Node slow = node, fast = slow.next; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } // 2) Split the linked list in two halves // node1, head of first half 1 -> 2 -> 3 // node2, head of second half 4 -> 5 Node node1 = node; Node node2 = slow.next; slow.next = null; // 3) Reverse the second half, i.e., 5 -> 4 node2 = reverselist(node2); // 4) Merge alternate nodes node = new Node(0); // Assign dummy Node // curr is the pointer to this dummy Node, which // will be used to form the new list Node curr = node; while (node1 != null || node2 != null) { // First add the element from first list if (node1 != null) { curr.next = node1; curr = curr.next; node1 = node1.next; } // Then add the element from second list if (node2 != null) { curr.next = node2; curr = curr.next; node2 = node2.next; } } // Assign the head of the new list to head pointer node = node.next; } // Driver code public static void main(String[] args) { LinkedList list = new LinkedList(); list.head = new Node(1); list.head.next = new Node(2); list.head.next.next = new Node(3); list.head.next.next.next = new Node(4); list.head.next.next.next.next = new Node(5); list.printlist(list.head); // print original list list.rearrange( list.head); // rearrange list as per ques Console.WriteLine(""); list.printlist(list.head); // print modified list }}/* This code is contributed PrinciRaj1992 */ |
Javascript
<script>// Javascript program to rearrange link list in place// Linked List Classvar head; // head of the list /* Node Class */ class Node {// Constructor to create a new nodeconstructor(d) { this.data = d; this.next = null;} } function printlist(node) { if (node == null) { return; } while (node != null) { document.write(node.data + " -> "); node = node.next; } } function reverselist(node) { var prev = null, curr = node, next; while (curr != null) { next = curr.next; curr.next = prev; prev = curr; curr = next; } node = prev; return node; } function rearrange(node) { // 1) Find the middle point using tortoise and hare // method var slow = node, fast = slow.next; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } // 2) Split the linked list in two halves // node1, head of first half 1 -> 2 -> 3 // node2, head of second half 4 -> 5 var node1 = node; var node2 = slow.next; slow.next = null; // 3) Reverse the second half, i.e., 5 -> 4 node2 = reverselist(node2); // 4) Merge alternate nodes node = new Node(0); // Assign dummy Node // curr is the pointer to this dummy Node, which // will be used to form the new list var curr = node; while (node1 != null || node2 != null) { // First add the element from first list if (node1 != null) { curr.next = node1; curr = curr.next; node1 = node1.next; } // Then add the element from second list if (node2 != null) { curr.next = node2; curr = curr.next; node2 = node2.next; } } // Assign the head of the new list to head pointer node = node.next; } 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); printlist(head); // print original list rearrange(head); // rearrange list as per ques document.write("<br/>"); printlist(head); // print modified list// This code contributed by gauravrajput1 </script> |
C
// C program to rearrange a linked list in-place#include <stdio.h>#include <stdlib.h>// Linkedlist Node structuretypedef struct Node { int data; struct Node* next;} Node;// Function to create newNode in a linkedlistNode* newNode(int key){ Node* temp = (Node*)malloc(sizeof(Node)); temp->data = key; temp->next = NULL; return temp;}// Function to reverse the linked listvoid reverselist(Node** head){ // Initialize prev and current pointers Node *prev = NULL, *curr = *head, *next; while (curr) { next = curr->next; curr->next = prev; prev = curr; curr = next; } *head = prev;}// Function to print the linked listvoid printlist(Node* head){ while (head != NULL) { printf("%d ", head->data); if (head->next) printf("-> "); head = head->next; } printf("\n");}// Function to rearrange a linked listvoid rearrange(Node** head){ // 1) 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; } // 2) Split the linked list in two halves // head1, head of first half 1 -> 2 // head2, head of second half 3 -> 4 Node* head1 = *head; Node* head2 = slow->next; slow->next = NULL; // 3) Reverse the second half, i.e., 4 -> 3 reverselist(&head2); // 4) Merge alternate nodes *head = newNode(0); // Assign dummy Node // curr is the pointer to this dummy Node, which will // be used to form the new list Node* curr = *head; while (head1 || head2) { // First add the element from 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; } } // Assign the head of the new list to head pointer *head = (*head)->next;}// Driver programint main(){ 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); printlist(head); // Print original list rearrange(&head); // Modify the list printlist(head); // Print modified list return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Output
1 -> 2 -> 3 -> 4 -> 5 1 -> 5 -> 2 -> 4 -> 3
Time Complexity: O(n)
Auxiliary Space: O(1)
Another Approach: (Using recursion)
- Hold a pointer to the head node and go till the last node using recursion
- Once the last node is reached, start swapping the last node to the next of head node
- Move the head pointer to the next node
- Repeat this until the head and the last node meet or come adjacent to each other
- Once the Stop condition met, we need to discard the left nodes to fix the loop created in the list while swapping nodes.
C++
// C++ implementation#include <bits/stdc++.h>using namespace std;// Creating the structure for nodestruct Node { int data; struct Node* next;};// Function to create newNode in a linkedlistNode* newNode(int key){ Node* temp = new Node; temp->data = key; temp->next = NULL; return temp;}// Function to print the listvoid printlist(Node* head){ while (head) { cout << head->data; if (head->next) cout << "->"; head = head->next; } cout << endl;}// Function to rearrangevoid rearrange(Node** head, Node* last){ if (!last) return; // Recursive call rearrange(head, last->next); // (*head)->next will be set to NULL after // rearrangement. Need not do any operation further Just // return here to come out of recursion if (!(*head)->next) return; // Rearrange the list until both head and last meet or // next to each other. if ((*head) != last && (*head)->next != last) { Node* tmp = (*head)->next; (*head)->next = last; last->next = tmp; *head = tmp; } else { if ((*head) != last) *head = (*head)->next; (*head)->next = NULL; }}// Drivers Codeint main(){ 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); // Print original list printlist(head); Node* tmp = head; // Modify the list rearrange(&tmp, head); // Print modified list printlist(head); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java implementationimport java.io.*;// Creating the structure for nodeclass Node { int data; Node next; // Function to create newNode in a linkedlist Node(int key) { data = key; next = null; }}class GFG { Node left = null; // Function to print the list void printlist(Node head) { while (head != null) { System.out.print(head.data + " "); if (head.next != null) { System.out.print("->"); } head = head.next; } System.out.println(); } // Function to rearrange void rearrange(Node head) { if (head != null) { left = head; reorderListUtil(left); } } void reorderListUtil(Node right) { if (right == null) { return; } reorderListUtil(right.next); // we set left = null, when we reach stop condition, // so no processing required after that if (left == null) { return; } // Stop condition: odd case : left = right, even // case : left.next = right if (left != right && left.next != right) { Node temp = left.next; left.next = right; right.next = temp; left = temp; } else { // stop condition , set null to left nodes if (left.next == right) { left.next.next = null; // even case left = null; } else { left.next = null; // odd case left = null; } } } // Drivers Code public static void main(String[] args) { 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); GFG gfg = new GFG(); // Print original list gfg.printlist(head); // Modify the list gfg.rearrange(head); // Print modified list gfg.printlist(head); }}// This code is contributed by Vishal Singh |
Python3
# Python3 implementationclass Node: def __init__(self, key): self.data = key self.next = Noneleft = None# Function to print the listdef printlist(head): while (head != None): print(head.data, end = " ") if (head.next != None): print("->", end = "") head = head.next print()# Function to rearrangedef rearrange(head): global left if (head != None): left = head reorderListUtil(left)def reorderListUtil(right): global left if (right == None): return reorderListUtil(right.next) # We set left = null, when we reach stop # condition, so no processing required # after that if (left == None): return # Stop condition: odd case : left = right, even # case : left.next = right if (left != right and left.next != right): temp = left.next left.next = right right.next = temp left = temp else: # Stop condition , set null to left nodes if (left.next == right): # Even case left.next.next = None left = None else: # Odd case left.next = None left = None# Driver codehead = Node(1)head.next = Node(2)head.next.next = Node(3)head.next.next.next = Node(4)head.next.next.next.next = Node(5)# Print original listprintlist(head)# Modify the listrearrange(head)# Print modified listprintlist(head)# This code is contributed by patel2127 |
C#
// C# implementationusing System;// Creating the structure for nodepublic class Node { public int data; public Node next; // Function to create newNode // in a linkedlist public Node(int key) { data = key; next = null; }}class GFG{ Node left = null;// Function to print the listvoid printlist(Node head){ while (head != null) { Console.Write(head.data + " "); if (head.next != null) { Console.Write("->"); } head = head.next; } Console.WriteLine();}// Function to rearrangevoid rearrange(Node head){ if (head != null) { left = head; reorderListUtil(left); }}void reorderListUtil(Node right){ if (right == null) { return; } reorderListUtil(right.next); // We set left = null, when we reach stop // condition, so no processing required // after that if (left == null) { return; } // Stop condition: odd case : left = right, even // case : left.next = right if (left != right && left.next != right) { Node temp = left.next; left.next = right; right.next = temp; left = temp; } else { // Stop condition , set null to left nodes if (left.next == right) { // Even case left.next.next = null; left = null; } else { // Odd case left.next = null; left = null; } }}// Driver Codestatic public void Main(){ 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); GFG gfg = new GFG(); // Print original list gfg.printlist(head); // Modify the list gfg.rearrange(head); // Print modified list gfg.printlist(head);}}// This code is contributed by rag2127 |
Javascript
<script>// javascript implementation// Creating the structure for nodeclass Node { // Function to create newNode in a linkedlist constructor(val) { this.data = val; this.next = null; }} var left = null; // Function to print the list function printlist(head) { while (head != null) { document.write(head.data + " "); if (head.next != null) { document.write("->"); } head = head.next; } document.write("<br/>"); } // Function to rearrange function rearrange(head) { if (head != null) { left = head; reorderListUtil(left); } } function reorderListUtil(right) { if (right == null) { return; } reorderListUtil(right.next); // we set left = null, when we reach stop condition, // so no processing required after that if (left == null) { return; } // Stop condition: odd case : left = right, even // case : left.next = right if (left != right && left.next != right) { var temp = left.next; left.next = right; right.next = temp; left = temp; } else { // stop condition , set null to left nodes if (left.next == right) { left.next.next = null; // even case left = null; } else { left.next = null; // odd case left = null; } } } // Drivers Code var 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); // Print original list printlist(head); // Modify the list rearrange(head); // Print modified list printlist(head);// This code contributed by aashish1995</script> |
C
// C implementation#include <stdio.h>#include <stdlib.h>// Creating the structure for nodetypedef struct Node { int data; struct Node* next;}Node;// Function to create newNode in a linkedlistNode* newNode(int key){ Node* temp = malloc(sizeof(Node)); temp->data = key; temp->next = NULL; return temp;}// Function to print the listvoid printlist(Node* head){ while (head) { printf("%d ", head->data); if (head->next) printf("->"); head = head->next; } printf("\n");}// Function to rearrangevoid rearrange(Node** head, Node* last){ if (!last) return; // Recursive call rearrange(head, last->next); // (*head)->next will be set to NULL // after rearrangement. // Need not do any operation further // Just return here to come out of recursion if (!(*head)->next) return; // Rearrange the list until both head // and last meet or next to each other. if ((*head) != last && (*head)->next != last) { Node* tmp = (*head)->next; (*head)->next = last; last->next = tmp; *head = tmp; } else { if ((*head) != last) *head = (*head)->next; (*head)->next = NULL; }}// Drivers Codeint main(){ 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); // Print original list printlist(head); Node* tmp = head; // Modify the list rearrange(&tmp, head); // Print modified list printlist(head); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Output
1 ->2 ->3 ->4 ->5 1 ->5 ->2 ->4 ->3
Time complexity: O(N) where N is no of nodes of linked list
Auxiliary Space: O(1) since using constant space
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