Given a linked list, write a function to reverse every k nodes (where k is an input to the function).
Example:
Input: 1->2->3->4->5->6->7->8->NULL, K = 3
Output: 3->2->1->6->5->4->8->7->NULL
Input: 1->2->3->4->5->6->7->8->NULL, K = 5
Output: 5->4->3->2->1->8->7->6->NULL
Algorithm: reverse(head, k)
- Reverse the first sub-list of size k. While reversing keep track of the next node and previous node. Let the pointer to the next node be next and pointer to the previous node be prev. See this post for reversing a linked list.
- head->next = reverse(next, k) ( Recursively call for rest of the list and link the two sub-lists )
- Return prev ( prev becomes the new head of the list (see the diagrams of an iterative method of this post )
Below is image shows how the reverse function works:
Below is the implementation of the above approach:
C++
// CPP program to reverse a linked list // in groups of given size #include <bits/stdc++.h> using namespace std;
/* Link list node */class Node {
public:
int data;
Node* next;
}; /* Reverses the linked list in groups of size k and returns the pointer to the new head node. */Node* reverse(Node* head, int k)
{ // base case
if (!head)
return NULL;
Node* current = head;
Node* next = NULL;
Node* prev = NULL;
int count = 0;
/*reverse first k nodes of the linked list */
while (current != NULL && count < k) {
next = current->next;
current->next = prev;
prev = current;
current = next;
count++;
}
/* next is now a pointer to (k+1)th node
Recursively call for the list starting from current.
And make rest of the list as next of first node */
if (next != NULL)
head->next = reverse(next, k);
/* prev is new head of the input list */
return prev;
} /* UTILITY FUNCTIONS *//* Function to push a node */void 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 linked list */void printList(Node* node)
{ while (node != NULL) {
cout << node->data << " ";
node = node->next;
}
} /* Driver code*/int main()
{ /* Start with the empty list */
Node* head = NULL;
/* Created Linked list
is 1->2->3->4->5->6->7->8->9 */
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);
cout << "Given linked list \n";
printList(head);
head = reverse(head, 3);
cout << "\nReversed Linked list \n";
printList(head);
return (0);
} // This code is contributed by rathbhupendra |
C
// C program to reverse a linked list in groups of given size #include<stdio.h> #include<stdlib.h> /* Link list node */struct Node
{ int data;
struct Node* next;
}; /* Reverses the linked list in groups of size k and returns the pointer to the new head node. */
struct Node *reverse (struct Node *head, int k)
{ if (!head)
return NULL;
struct Node* current = head;
struct Node* next = NULL;
struct Node* prev = NULL;
int count = 0;
/*reverse first k nodes of the linked list */ while (current != NULL && count < k)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
count++;
}
/* next is now a pointer to (k+1)th node
Recursively call for the list starting from current.
And make rest of the list as next of first node */
if (next != NULL)
head->next = reverse(next, k);
/* prev is new head of the input list */
return prev;
} /* UTILITY FUNCTIONS *//* Function to push a node */void push(struct Node** head_ref, int new_data)
{ /* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct 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 linked list */void printList(struct Node *node)
{ while (node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
} /* Driver code*/int main(void)
{ /* Start with the empty list */
struct Node* head = NULL;
/* Created Linked list is 1->2->3->4->5->6->7->8->9 */
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);
printf("\nGiven linked list \n");
printList(head);
head = reverse(head, 3);
printf("\nReversed Linked list \n");
printList(head);
return(0);
} |
Java
// Java program to reverse a linked list in groups of // given size class LinkedList {
Node head; // head of list
/* Linked list Node*/
class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
Node reverse(Node head, int k)
{
if(head == null)
return null;
Node current = head;
Node next = null;
Node prev = null;
int count = 0;
/* Reverse first k nodes of linked list */
while (count < k && current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
count++;
}
/* next is now a pointer to (k+1)th node
Recursively call for the list starting from
current. And make rest of the list as next of
first node */
if (next != null)
head.next = reverse(next, k);
// prev is now head of input list
return prev;
}
/* Utility functions */
/* Inserts a new Node at front of the list. */
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
/* Function to print linked list */
void printList()
{
Node temp = head;
while (temp != null) {
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
/* Driver program to test above functions */
public static void main(String args[])
{
LinkedList llist = new LinkedList();
/* Constructed Linked List is 1->2->3->4->5->6->
7->8->8->9->null */
llist.push(9);
llist.push(8);
llist.push(7);
llist.push(6);
llist.push(5);
llist.push(4);
llist.push(3);
llist.push(2);
llist.push(1);
System.out.println("Given Linked List");
llist.printList();
llist.head = llist.reverse(llist.head, 3);
System.out.println("Reversed list");
llist.printList();
}
} /* This code is contributed by Rajat Mishra */ |
Python3
# Python program to reverse a # linked list in group of given size # Node class class Node:
# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
# Function to initialize head
def __init__(self):
self.head = None
def reverse(self, head, k):
if head == None:
return None
current = head
next = None
prev = None
count = 0
# Reverse first k nodes of the linked list
while(current is not None and count < k):
next = current.next
current.next = prev
prev = current
current = next
count += 1
# next is now a pointer to (k+1)th node
# recursively call for the list starting
# from current. And make rest of the list as
# next of first node
if next is not None:
head.next = self.reverse(next, k)
# prev is new head of the input list
return prev
# Function to insert a new node at the beginning
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
# Utility function to print the linked LinkedList
def printList(self):
temp = self.head
while(temp):
print(temp.data,end=' ')
temp = temp.next
# Driver program llist = LinkedList()
llist.push(9)
llist.push(8)
llist.push(7)
llist.push(6)
llist.push(5)
llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)
print("Given linked list")
llist.printList() llist.head = llist.reverse(llist.head, 3)
print ("\nReversed Linked list")
llist.printList() # This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
// C# program to reverse a linked list // in groups of given size using System;
public class LinkedList {
Node head; // head of list
/* Linked list Node*/
class Node {
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
Node reverse(Node head, int k)
{
if(head == null)
return null;
Node current = head;
Node next = null;
Node prev = null;
int count = 0;
/* Reverse first k nodes of linked list */
while (count < k && current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
count++;
}
/* next is now a pointer to (k+1)th node
Recursively call for the list starting from
current. And make rest of the list as next of
first node */
if (next != null)
head.next = reverse(next, k);
// prev is now head of input list
return prev;
}
/* Utility functions */
/* Inserts a new Node at front of the list. */
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
/* Function to print linked list */
void printList()
{
Node temp = head;
while (temp != null) {
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
/* Driver code*/
public static void Main(String[] args)
{
LinkedList llist = new LinkedList();
/* Constructed Linked List is 1->2->3->4->5->6->
7->8->8->9->null */
llist.push(9);
llist.push(8);
llist.push(7);
llist.push(6);
llist.push(5);
llist.push(4);
llist.push(3);
llist.push(2);
llist.push(1);
Console.WriteLine("Given Linked List");
llist.printList();
llist.head = llist.reverse(llist.head, 3);
Console.WriteLine("Reversed list");
llist.printList();
}
} // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript program to reverse a // linked list in groups of // given size var head; // head of list
/* Linked list Node */
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
function reverse(head , k) {
if (head == null)
return null;
var current = head;
var next = null;
var prev = null;
var count = 0;
/* Reverse first k nodes of linked list */
while (count < k && current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
count++;
}
/*
next is now a pointer to (k+1)th node
Recursively call for the list starting
from current. And make rest of the list
as next of first node
*/
if (next != null)
head.next = reverse(next, k);
// prev is now head of input list
return prev;
}
/* Utility functions */
/* Inserts a new Node at front of the list. */
function push(new_data) {
/*
1 & 2: Allocate the Node & Put in the data
*/
new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
/* Function to print linked list */
function printList() {
temp = head;
while (temp != null) {
document.write(temp.data + " ");
temp = temp.next;
}
document.write("<br/>");
}
/* Driver program to test above functions */
/*
Constructed Linked List is
1->2->3->4->5->6-> 7->8->8->9->null
*/
push(9);
push(8);
push(7);
push(6);
push(5);
push(4);
push(3);
push(2);
push(1);
document.write("Given Linked List<br/>");
printList();
head = reverse(head, 3);
document.write("Reversed list<br/>");
printList();
// This code contributed by gauravrajput1 </script> |
Output
Given linked list 1 2 3 4 5 6 7 8 9 Reversed Linked list 3 2 1 6 5 4 9 8 7
Complexity Analysis:
-
Time Complexity: O(n).
Traversal of list is done only once and it has ‘n’ elements. -
Auxiliary Space: O(n/k).
For each Linked List of size n, n/k or (n/k)+1 calls will be made during the recursion.
We can solve this question in O(1) Space Complexity.
Approach – 2 Space Optimized – Iterative
The following steps are required for this Algorithm:
- Create a dummy node and point it to the head of input i.e dummy->next = head.
- Calculate the length of the linked list which takes O(N) time, where N is the length of the linked list.
- Initialize three-pointers prev, curr, next to reverse k elements for every group.
- Iterate over the linked lists till next!=NULL.
- Points curr to the prev->next and next to the curr next.
- Then, Using the inner for loop reverse the particular group using these four steps:
- curr->next = next->next
- next->next = prev->next
- prev->next = next
- next = curr->next
7. This for loop runs for k-1 times for all groups except the last remaining element, for the last remaining element it runs for the remaining length of the linked list – 1.
8. Decrement count after for loop by k count -= k, to determine the length of the remaining linked list.
9. Change prev position to curr, prev = curr.
Here is the code for the above algorithm.
C++
// CPP program to reverse a linked list // in groups of given size #include <bits/stdc++.h> using namespace std;
/* Link list node */class Node {
public:
int data;
Node* next;
}; /* Reverses the linked list in groups of size k and returns the pointer to the new head node. */Node* reverse(Node* head, int k)
{ // If head is NULL or K is 1 then return head
if (!head || k == 1)
return head;
Node* dummy = new Node(); // creating dummy node
dummy->data = -1;
dummy->next = head;
// Initializing three points prev, curr, next
Node *prev = dummy, *curr = dummy, *next = dummy;
// Calculating the length of linked list
int count = 0;
while (curr) {
curr = curr->next;
count++;
}
// Iterating till next is not NULL
while (next) {
// Curr position after every reverse group
curr = prev->next;
// Next will always next to curr
next = curr->next;
// toLoop will set to count - 1 in case of remaining
// element
int toLoop = count > k ? k : count - 1;
for (int i = 1; i < toLoop; i++) {
// 4 steps as discussed above
curr->next = next->next;
next->next = prev->next;
prev->next = next;
next = curr->next;
}
// Setting prev to curr
prev = curr;
// Update count
count -= k;
}
// dummy -> next will be our new head for output linked
// list
return dummy->next;
} /* UTILITY FUNCTIONS *//* Function to push a node */void 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 linked list */void printList(Node* node)
{ while (node != NULL) {
cout << node->data << " ";
node = node->next;
}
} /* Driver code*/int main()
{ /* Start with the empty list */
Node* head = NULL;
/* Created Linked list
is 1->2->3->4->5->6->7->8->9 */
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);
cout << "Given linked list \n";
printList(head);
head = reverse(head, 3);
cout << "\nReversed Linked list \n";
printList(head);
return (0);
} // This code is contributed by Sania Kumari Gupta // (kriSania804) |
C
// C program to reverse a linked list // in groups of given size #include <stdio.h> #include <stdlib.h> /* Link list node */typedef struct Node {
int data;
struct Node* next;
} Node; // Reverses the linked list in groups of size k and returns // the pointer to the new head node. Node* reverse(Node* head, int k)
{ // If head is NULL or K is 1 then return head
if (!head || k == 1)
return head;
// creating dummy node
Node* dummy = (Node*)malloc(sizeof(Node));
dummy->data = -1;
dummy->next = head;
// Initializing three points prev, curr, next
Node *prev = dummy, *curr = dummy, *next = dummy;
// Calculating the length of linked list
int count = 0;
while (curr) {
curr = curr->next;
count++;
}
// Iterating till next is not NULL
while (next) {
// Curr position after every reverse group
curr = prev->next;
// Next will always next to curr
next = curr->next;
// toLoop will set to count - 1 in case of remaining
// element
int toLoop = count > k ? k : count - 1;
for (int i = 1; i < toLoop; i++) {
// 4 steps as discussed above
curr->next = next->next;
next->next = prev->next;
prev->next = next;
next = curr->next;
}
// Setting prev to curr
prev = curr;
// Update count
count -= k;
}
// dummy -> next will be our new head for output linked
// list
return dummy->next;
} /* UTILITY FUNCTIONS *//* Function to push a node */void push(Node** head_ref, int new_data)
{ /* allocate node */
Node* new_node = (Node*)malloc(sizeof(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 linked list */void printList(Node* node)
{ while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
} /* Driver code*/int main()
{ /* Start with the empty list */
Node* head = NULL;
/* Created Linked list
is 1->2->3->4->5->6->7->8->9 */
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);
printf("Given linked list \n");
printList(head);
head = reverse(head, 3);
printf("\nReversed Linked list \n");
printList(head);
return (0);
} // This code is contributed by Sania Kumari Gupta // (kriSania804) |
Java
import java.util.*;
// Linked List Node class Node {
int data;
Node next;
Node(int a)
{
data = a;
next = null;
}
} class GFG {
// utility function to insert node in the list
static Node push(Node head, int val)
{
Node newNode = new Node(val);
if (head == null) {
head = newNode;
return head;
}
Node temp = head;
while (temp.next != null)
temp = temp.next;
temp.next = newNode;
return head;
}
// utility function to reverse k nodes in the list
static Node reverse(Node head, int k)
{
// If head is NULL or K is 1 then return head
if (head == null || head.next == null)
return head;
// creating dummy node
Node dummy = new Node(-1);
dummy.next = head;
// Initializing three points prev, curr, next
Node prev = dummy;
Node curr = dummy;
Node next = dummy;
// Calculating the length of linked list
int count = 0;
while (curr != null) {
count++;
curr = curr.next;
}
// Iterating till next is not NULL
while (next != null) {
curr = prev.next; // Curr position after every
// reverse group
next = curr.next; // Next will always next to
// curr
int toLoop
= count > k
? k
: count - 1; // toLoop will set to
// count - 1 in case of
// remaining element
for (int i = 1; i < toLoop; i++) {
// 4 steps as discussed above
curr.next = next.next;
next.next = prev.next;
prev.next = next;
next = curr.next;
}
prev = curr; // Setting prev to curr
count -= k; // Update count
}
return dummy.next; // dummy -> next will be our new
// head for output linked
// list
}
// utility function to print the list
static void print(Node head)
{
while (head.next != null) {
System.out.print(head.data + " ");
head = head.next;
}
System.out.println(head.data);
}
public static void main(String args[])
{
Node head = null;
int k = 3;
head = push(head, 1);
head = push(head, 2);
head = push(head, 3);
head = push(head, 4);
head = push(head, 5);
head = push(head, 6);
head = push(head, 7);
head = push(head, 8);
head = push(head, 9);
System.out.println("Given Linked List");
print(head);
System.out.println("Reversed list");
Node newHead = reverse(head, k);
print(newHead);
}
} |
Python3
# Python program to reverse a linked list in groups of given size class Node:
def __init__(self, data):
self.data = data
self.next = None
# Reverses the linked list in groups # of size k and returns the pointer # to the new head node. def reverse(head, k):
# If head is NULL or K is 1 then return head
if not head or k == 1:
return head
dummy = Node(-1) # creating dummy node
dummy.next = head
# Initializing three points prev, curr, next
prev = dummy
curr = dummy
next = dummy
count = 0
toLoop = 0
i = 0
# Calculating the length of linked list
while curr:
curr = curr.next
count += 1
# Iterating till next is not none
while next:
curr = prev.next # Curr position after every reversed group
next = curr.next # Next will always next to curr
# toLoop will set to count - 1 in case of remaining element
toLoop = count > k and k or count - 1
for i in range(1, toLoop):
# 4 steps as discussed above
curr.next = next.next
next.next = prev.next
prev.next = next
next = curr.next
# Setting prev to curr
prev = curr
# Update count
count -= k
# dummy -> next will be our new head for output linked list
return dummy.next
# Function to print linked list def printList(node):
while node is not None:
print(node.data, end=" ")
node = node.next
# Created Linked list is 1->2->3->4->5->6->7->8->9 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.next.next.next.next.next = Node(6)
head.next.next.next.next.next.next = Node(7)
head.next.next.next.next.next.next.next = Node(8)
head.next.next.next.next.next.next.next.next = Node(9)
print("Given linked list")
printList(head) head = reverse(head, 3)
print("\nReversed Linked list")
printList(head) # This code is contributed by Tapesh(tapeshdua420) |
C#
// C# program to reverse a linked list // in groups of given size using System;
/* Link list node */class Node {
public int data;
public Node next;
public Node(int a)
{
data = a;
next = null;
}
} class GFG {
/* UTILITY FUNCTIONS */
/* Function to push a node */
static Node push(Node head, int val)
{
Node newNode = new Node(val);
if (head == null) {
head = newNode;
return head;
}
Node temp = head;
while (temp.next != null)
temp = temp.next;
temp.next = newNode;
return head;
}
// utility function to reverse k nodes in the list
static Node reverse(Node head, int k)
{
// If head is NULL or K is 1 then return head
if (head == null || head.next == null)
return head;
// creating dummy node
Node dummy = new Node(-1);
dummy.next = head;
// Initializing three points prev, curr, next
Node prev = dummy;
Node curr = dummy;
Node next = dummy;
// Calculating the length of linked list
int count = 0;
while (curr != null) {
count++;
curr = curr.next;
}
// Iterating till next is not NULL
while (next != null) {
curr = prev.next; // Curr position after every
// reverse group
next = curr.next; // Next will always next to
// curr
int toLoop
= count > k
? k
: count - 1; // toLoop will set to
// count - 1 in case of
// remaining element
for (int i = 1; i < toLoop; i++) {
// 4 steps as discussed above
curr.next = next.next;
next.next = prev.next;
prev.next = next;
next = curr.next;
}
prev = curr; // Setting prev to curr
count -= k; // Update count
}
return dummy.next; // dummy -> next will be our new
// head for output linked list
}
// utility function to print the list
static void print(Node head)
{
while (head.next != null) {
Console.Write(head.data + " ");
head = head.next;
}
Console.WriteLine(head.data);
}
public static void Main()
{
Node head = null;
int K = 3;
head = push(head, 1);
head = push(head, 2);
head = push(head, 3);
head = push(head, 4);
head = push(head, 5);
head = push(head, 6);
head = push(head, 7);
head = push(head, 8);
head = push(head, 9);
Console.WriteLine("Given linked list");
print(head);
Console.WriteLine("Reversed Linked list");
Node newHead = reverse(head, K);
print(newHead);
}
} // This code is contributed by Tapesh (tapeshdua420) |
Javascript
// JavaScript program to reverse a linked list // in groups of given size // Linked List Node class Node{ constructor(a){
this.data = a;
this.next = null;
}
} function push(head, val){
newNode = new Node(val);
if(head==null){
head = newNode;
return head;
}
temp = head;
while(temp.next!=null){
temp = temp.next;
}
temp.next = newNode;
return head;
} // utility function to reverse k nodes in the list function reverse(head, k){
// If head is NULL or K is 1 then return head
if(head==null || head.next==null){
return head;
}
// creating dummy node
var dummy = new Node(-1);
dummy.next = head;
// Initializing three points prev, curr, next
var prev = dummy;
var curr = dummy;
var next = dummy;
// Calculating the length of linked list
let count = 0;
while(curr!=null){
count++;
curr = curr.next;
}
// Iterating till next is not NULL
while(next!=null){
curr = prev.next; // Curr position after every
// reverse group
next = curr.next; // Next will always next to
// curr
// toLoop will set to count - 1 in case of
// remaining element
let toLoop = count > k ? k : count - 1;
for(let i=1;i<toLoop;i++){
// 4 steps as discussed above
curr.next = next.next;
next.next = prev.next;
prev.next = next;
next = curr.next;
}
prev = curr; // Setting prev to curr
count -= k; // Update count
}
return dummy.next; // dummy.next will be our new
// head for output linked list
} // utility function to print the list function print(head){
while(head.next!=null){
console.log(head.data + " ");
head = head.next;
}
console.log(head.data+"<br>");
} var head = null;
let k = 3; head = push(head, 1); head = push(head, 2); head = push(head, 3); head = push(head, 4); head = push(head, 5); head = push(head, 6); head = push(head, 7); head = push(head, 8); head = push(head, 9); console.log("Given linked list<br>");
print(head); console.log("Reversed linked list<br>");
var newHead = reverse(head, k);
print(newHead); // This code is contributed by lokesh. |
Output
Given linked list 1 2 3 4 5 6 7 8 9 Reversed Linked list 3 2 1 6 5 4 9 8 7
Complexity Analysis
Time Complexity: O(N) : While loop takes O(N/K) time and inner for loop takes O(K) time. So N/K * K = N. Therefore TC O(N)
Space Complexity: O(1) : No extra space is used.
Please write comments if you find the above code/algorithm incorrect, or find other ways to solve the same problem.