Given a singly linked list and an integer K, the task is to reverse every K nodes of the given linked list.
Examples:
Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL, K = 3
Output: 3 2 1 6 5 4 8 7Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL, K = 5
Output: 5 4 3 2 1 8 7 6
Approach: Two different approaches to solve this problem have been discussed in Set 1 and Set 2 of this article. In this article, an approach based on deque will be discussed.
- Create a deque.
- Store the address of the first k nodes in the deque.
- Pop first and the last value from the deque and swap the data values at those addresses.
- Repeat step 3 till the deque is not empty.
- Repeat step 2 for the next k nodes and till the end of the linked list is not reached.
Below is the implementation of the above approach:
C++14
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Link list nodestruct node { int data; struct node* next;};// Function to insert a node at// the head of the linked listvoid push(node** head_ref, int new_data){ /* Allocate node */ node* new_node = new node(); /* Put in the data */ new_node->data = new_data; /* Link the old list of the new node */ new_node->next = (*head_ref); /* Move the head to point to the new node */ (*head_ref) = new_node;}// Function to print the linked listvoid printList(node* head){ while (head != NULL) { cout << head->data << " "; head = head->next; }}/* Function to reverse the linked list in groups of size k and return the pointer to the new head node. */struct node* reverse(struct node* head, int k){ if (head == NULL) return head; // Create deque to store the address // of the nodes of the linked list deque<node*> q; // Store head pointer in current to // traverse the linked list node* current = head; int i; // Iterate through the entire linked // list by moving the current while (current != NULL) { i = 1; // Store addresses of the k // nodes in the deque while (i <= k) { if (current == NULL) break; q.push_back(current); current = current->next; i++; } /* pop first and the last value from the deque and swap the data values at those addresses Do this till there exist an address in the deque or deque is not empty*/ while (!q.empty()) { node* front = q.front(); node* last = q.back(); swap(front->data, last->data); // pop from the front if // the deque is not empty if (!q.empty()) q.pop_front(); // pop from the back if // the deque is not empty if (!q.empty()) q.pop_back(); } } return head;}// Driver codeint main(){ // Start with the empty list node* head = NULL; // Created Linked list is // 1->2->3->4->5->6->7->8->9->10 push(&head, 10); push(&head, 9); push(&head, 8); push(&head, 7); push(&head, 6); push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 2); push(&head, 1); int k = 2; // Get the new head after reversing the // linked list in groups of size k head = reverse(head, k); printList(head); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class LinkedList{ static Node head; // Creating node classstatic class Node{ int data; Node next; Node(int d) { data = d; next = null; } }// Inserts a new Node at front of the list. public void push(int new_data) { // Allocate the Node & // Put in the data Node new_node = new Node(new_data); // Make next of new Node as head new_node.next = head; // Move the head to point to new Node head = new_node; }// Prints content of linked list void printList(Node node) { while (node != null) { System.out.print(node.data + " "); node = node.next; } } // Function to reverse the linked list // in groups of size k and return the // pointer to the new head node.Node reverse(Node head, int k){ if (head == null) return head; // Create deque to store the address // of the nodes of the linked list Deque<Node> q = new ArrayDeque<Node>(); // Store head pointer in current to // traverse the linked list Node current = head; int i; // Iterate through the entire linked // list by moving the current while (current != null) { i = 1; // Store addresses of the k // nodes in the deque while (i <= k) { if (current == null) break; q.addLast(current); current = current.next; i++; } // pop first and the last value from // the deque and swap the data values at // those addresses // Do this till there exist an address in // the deque or deque is not empty while (!q.isEmpty()) { Node front = q.peekFirst(); Node last = q.peekLast(); // Swapping front and // last nodes int temp = front.data; front.data = last.data; last.data = temp; // pop from the front if // the deque is not empty if (!q.isEmpty()) q.removeFirst(); // pop from the back if // the deque is not empty if (!q.isEmpty()) q.removeLast(); } } return head;}// Driver codepublic static void main(String[] args) { LinkedList list = new LinkedList(); // Created Linked list is // 1->2->3->4->5->6->7->8->9->10 list.push(10); list.push(9); list.push(8); list.push(7); list.push(6); list.push(5); list.push(4); list.push(3); list.push(2); list.push(1); int k = 2; // Get the new head after reversing the // linked list in groups of size k head = list.reverse(head, k); list.printList(head); } } // This code is contributed by Bindu Madhav |
Python3
# Python3 implementation of the approach# Link list nodeclass Node: def __init__(self): self.data = 0 self.next = None# Function to insert a node at# the head of the linked listdef push(head_ref, new_data): # Allocate node new_node = Node() # Put in the data new_node.data = new_data # Link the old list of the new node new_node.next = (head_ref) # Move the head to point to the new node (head_ref) = new_node return head_ref# Function to print the linked listdef printList( head): while (head != None) : print( head.data, end = " ") head = head.next # Function to reverse the linked list in groups of # size k and return the pointer to the new head node. def reverse( head, k): if (head == None): return head # Create deque to store the address # of the nodes of the linked list q = [] # Store head pointer in current to # traverse the linked list current = head i = 0 # Iterate through the entire linked # list by moving the current while (current != None) : i = 1 # Store addresses of the k # nodes in the deque while (i <= k) : if (current == None): break q.append(current) current = current.next i = i + 1 # pop first and the last value from # the deque and swap the data values at # those addresses # Do this till there exist an address in # the deque or deque is not empty while (len(q) > 0): front = q[-1] last = q[0] temp = front.data front.data = last.data last.data = temp # pop from the front if # the deque is not empty if (len(q) > 0): q.pop() # pop from the back if # the deque is not empty if (len(q)): q.pop(0) return head# Driver code# Start with the empty listhead = None# Created Linked list is# 1.2.3.4.5.6.7.8.9.10head = push(head, 10)head = push(head, 9)head = push(head, 8)head = push(head, 7)head = push(head, 6)head = push(head, 5)head = push(head, 4)head = push(head, 3)head = push(head, 2)head = push(head, 1)k = 2# Get the new head after reversing the# linked list in groups of size khead = reverse(head, k)printList(head)# This code is contributed by Arnab Kundu |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic;class List{ static Node head; // Creating node classclass Node{ public int data; public Node next; public Node(int d) { data = d; next = null; } }// Inserts a new Node at front of the list. public void Push(int new_data) { // Allocate the Node & // Put in the data Node new_node = new Node(new_data); // Make next of new Node as head new_node.next = head; // Move the head to point to new Node head = new_node; }// Prints content of linked list void printList(Node node) { while (node != null) { Console.Write(node.data + " "); node = node.next; } } // Function to reverse the linked list // in groups of size k and return the // pointer to the new head node.Node reverse(Node head, int k){ if (head == null) return head; // Create deque to store the address // of the nodes of the linked list List<Node> q = new List<Node>(); // Store head pointer in current to // traverse the linked list Node current = head; int i; // Iterate through the entire linked // list by moving the current while (current != null) { i = 1; // Store addresses of the k // nodes in the deque while (i <= k) { if (current == null) break; q.Add(current); current = current.next; i++; } // pop first and the last value from // the deque and swap the data values at // those addresses // Do this till there exist an address in // the deque or deque is not empty while (q.Count != 0) { Node front = q[0]; Node last = q[q.Count - 1]; // Swapping front and // last nodes int temp = front.data; front.data = last.data; last.data = temp; // pop from the front if // the deque is not empty if (q.Count != 0) q.RemoveAt(0); // pop from the back if // the deque is not empty if (q.Count != 0) q.RemoveAt(q.Count - 1); } } return head;}// Driver codepublic static void Main(String[] args) { List list = new List(); // Created Linked list is // 1->2->3->4->5->6->7->8->9->10 list.Push(10); list.Push(9); list.Push(8); list.Push(7); list.Push(6); list.Push(5); list.Push(4); list.Push(3); list.Push(2); list.Push(1); int k = 2; // Get the new head after reversing the // linked list in groups of size k head = list.reverse(head, k); list.printList(head); } } // This code is contributed by todaysgaurav |
Javascript
<script>// Javascript implementation of the above approachvar head = null; // Creating node classclass Node{ constructor(d) { this.data = d; this.next = null; }}// Inserts a new Node at front of the list. function Push(new_data) { // Allocate the Node & // Put in the data var new_node = new Node(new_data); // Make next of new Node as head new_node.next = head; // Move the head to point to new Node head = new_node; }// Prints content of linked list function printList(node) { while (node != null) { document.write(node.data + " "); node = node.next; } } // Function to reverse the linked list // in groups of size k and return the // pointer to the new head node.function reverse(head, k){ if (head == null) return head; // Create deque to store the address // of the nodes of the linked list var q = []; // Store head pointer in current to // traverse the linked list var current = head; var i; // Iterate through the entire linked // list by moving the current while (current != null) { i = 1; // Store addresses of the k // nodes in the deque while (i <= k) { if (current == null) break; q.push(current); current = current.next; i++; } // pop first and the last value from // the deque and swap the data values at // those addresses // Do this till there exist an address in // the deque or deque is not empty while (q.length != 0) { var front = q[0]; var last = q[q.length - 1]; // Swapping front and // last nodes var temp = front.data; front.data = last.data; last.data = temp; // pop from the front if // the deque is not empty if (q.Count != 0) q.shift(); // pop from the back if // the deque is not empty if (q.Count != 0) q.pop(); } } return head;}// Driver code// Created Linked list is// 1->2->3->4->5->6->7->8->9->10Push(10); Push(9); Push(8); Push(7);Push(6); Push(5); Push(4); Push(3);Push(2); Push(1); var k = 2;// Get the new head after reversing the// linked list in groups of size khead = reverse(head, k); printList(head); // This code is contributed by itsok</script> |
Output:
2 1 4 3 6 5 8 7 10 9
Time Complexity: O(N)
Auxiliary Space: O(N)
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