Reverse a sublist of linked list

We are given a linked list and positions m and n. We need to reverse the linked list from position m to n.

Examples:  

Input : 10->20->30->40->50->60->70->NULL
        m = 3, n = 6
Output : 10->20->60->50->40->30->70->NULL

Input :  1->2->3->4->5->6->NULL 
         m = 2, n = 4
Output : 1->4->3->2->5->6->NULL

To reverse the linked list from position m to n, we find addresses of start and end position of the linked list by running a loop, and then we unlink this part from the rest of the list and then use the normal linked list reverse function which we have earlier used for reversing the complete linked list, and use it to reverse the portion of the linked list which need to be reversed. After reversal, we again attach the portion reversed to the main list.

C++

// C++ program to reverse a linked list
// from position m to position n
#include <bits/stdc++.h>
using namespace std;

// Linked list node
struct Node {
    int data;
    struct Node* next;
};

// function used to reverse a linked list
struct Node* reverse(struct Node* head)
{
    struct Node* prev = NULL;
    struct Node* curr = head;
    while (curr) {
        struct Node* next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

// function used to reverse a linked list from position m to n
Node* reverseBetween(Node* head, int m, int n)
{
    if (m == n)
        return head;

    // revs and revend is start and end respectively of the
    // portion of the linked list which need to be reversed.
    // revs_prev is previous of starting position and
    // revend_next is next of end of list to be reversed.
    Node *revs = NULL, *revs_prev = NULL;
    Node *revend = NULL, *revend_next = NULL;

    // Find values of above pointers.
    int i = 1;
    Node* curr = head;
    while (curr && i <= n) {
        if (i < m)
            revs_prev = curr;
        if (i == m)
            revs = curr;
        if (i == n) {
            revend = curr;
            revend_next = curr->next;
        }
        curr = curr->next;
        i++;
    }
    revend->next = NULL;
    // Reverse linked list starting with revs.
    revend = reverse(revs);
    // If starting position was not head
    if (revs_prev)
        revs_prev->next = revend;
    // If starting position was head
    else
        head = revend;
    revs->next = revend_next;
    return head;
}

void print(struct Node* head)
{
    while (head != NULL) {
        cout<<head->data<<" ";
        head = head->next;
    }
    cout<<endl;
}

// function to add a new node at the
// beginning of the list
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

// Driver code
int main()
{
    struct Node* head = NULL;
    push(&head, 70);
    push(&head, 60);
    push(&head, 50);
    push(&head, 40);
    push(&head, 30);
    push(&head, 20);
    push(&head, 10);
    reverseBetween(head, 3, 6);
    print(head);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C program to reverse a linked list
// from position m to position n
#include <stdio.h>
#include <stdlib.h>

// Linked list node
typedef struct Node {
    int data;
    struct Node* next;
} Node;

// function used to reverse a linked list
Node* reverse(Node* head)
{
    Node* prev = NULL;
    Node* curr = head;
    while (curr) {
        Node* next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

// function used to reverse a linked list from position m to n
Node* reverseBetween(Node* head, int m, int n)
{
    if (m == n)
        return head;

    // revs and revend is start and end respectively of the
    // portion of the linked list which need to be reversed.
    // revs_prev is previous of starting position and
    // revend_next is next of end of list to be reversed.
    Node *revs = NULL, *revs_prev = NULL;
    Node *revend = NULL, *revend_next = NULL;

    // Find values of above pointers.
    int i = 1;
    Node* curr = head;
    while (curr && i <= n) {
        if (i < m)
            revs_prev = curr;
        if (i == m)
            revs = curr;
        if (i == n) {
            revend = curr;
            revend_next = curr->next;
        }
        curr = curr->next;
        i++;
    }
    revend->next = NULL;
    // Reverse linked list starting with revs.
    revend = reverse(revs);
    // If starting position was not head
    if (revs_prev)
        revs_prev->next = revend;
    // If starting position was head
    else
        head = revend;
    revs->next = revend_next;
    return head;
}

void print(Node* head)
{
    while (head != NULL) {
        printf("%d ", head->data);
        head = head->next;
    }
    printf("\n");
}

// function to add a new node at the beginning of the list
void push(Node** head_ref, int new_data)
{
    Node* new_node = (Node*)malloc(sizeof(Node));
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

// Driver code
int main()
{
    Node* head = NULL;
    push(&head, 70);
    push(&head, 60);
    push(&head, 50);
    push(&head, 40);
    push(&head, 30);
    push(&head, 20);
    push(&head, 10);
    reverseBetween(head, 3, 6);
    print(head);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program to reverse a linked list
// from position m to position n

class LinkedList {

    static Node head;

    static class Node {

        int data;
        Node next;

        Node(int d)
        {
            data = d;
            next = null;
        }
    }

    /* Function to reverse the linked list */
    static Node reverse(Node node)
    {
        Node prev = null;
        Node current = node;

        while (current != null) {
            Node next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        node = prev;
        return node;
    }
    // function used to reverse a linked list from position m to n
static  Node reverseBetween(Node head, int m, int n)
{
    if (m == n)
        return head;

    // revs and revend is start and end respectively of the
    // portion of the linked list which need to be reversed.
    // revs_prev is previous of starting position and
    // revend_next is next of end of list to be reversed.
    Node revs = null, revs_prev = null;
    Node revend = null, revend_next = null;

    // Find values of above pointers.
    int i = 1;
    Node curr = head;
    while (curr!=null && i <= n) {
        if (i < m)
            revs_prev = curr;
        if (i == m)
            revs = curr;
        if (i == n) {
            revend = curr;
            revend_next = curr.next;
        }
        curr = curr.next;
        i++;
    }
    revend.next = null;
    // Reverse linked list starting with revs.
    revend = reverse(revs);
    // If starting position was not head
    if (revs_prev!=null)
        revs_prev.next = revend;
    // If starting position was head
    else
        head = revend;
    revs.next = revend_next;
    return head;
}
    // prints content of double linked list
    void printList(Node node)
    {
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        LinkedList list = new LinkedList();
        list.head = new Node(10);
        list.head.next = new Node(20);
        list.head.next.next = new Node(30);
        list.head.next.next.next = new Node(40);
        list.head.next.next.next.next = new Node(50);
          list.head.next.next.next.next.next = new Node(60);
        list.head.next.next.next.next.next.next = new Node(70);
        reverseBetween(head,3,6);
        list.printList(head);

    }
}

// This code has been contributed by Abhijeet Kumar(abhijeet19403)

Python3

# Python3 program to reverse a linked list
# from position m to position n

# Linked list node
class Node:

    def __init__(self, data):

        self.data = data
        self.next = None

# The standard reverse function used
# to reverse a linked list
def reverse(head):

    prev = None
    curr = head

    while (curr):
        next = curr.next
        curr.next = prev
        prev = curr
        curr = next

    return prev

# Function used to reverse a linked list
# from position m to n which uses reverse
# function
def reverseBetween(head, m, n):

    if (m == n):
        return head

    # revs and revend is start and end respectively
    # of the portion of the linked list which
    # need to be reversed. revs_prev is previous
    # of starting position and revend_next is next
    # of end of list to be reversed.
    revs = None
    revs_prev = None
    revend = None
    revend_next = None

    # Find values of above pointers.
    i = 1
    curr = head

    while (curr and i <= n):
        if (i < m):
            revs_prev = curr

        if (i == m):
            revs = curr

        if (i == n):
            revend = curr
            revend_next = curr.next

        curr = curr.next
        i += 1

    revend.next = None

    # Reverse linked list starting with
    # revs.
    revend = reverse(revs)

    # If starting position was not head
    if (revs_prev):
        revs_prev.next = revend

    # If starting position was head
    else:
        head = revend

    revs.next = revend_next
    return head

def prints(head):

    while (head != None):
        print(head.data, end = ' ')
        head = head.next

    print()

# Function to add a new node at the
# beginning of the list
def push(head_ref, new_data):

    new_node = Node(new_data)
    new_node.data = new_data
    new_node.next = (head_ref)
    (head_ref) = new_node
    return head_ref

# Driver code
if __name__=='__main__':

    head = None
    head = push(head, 70)
    head = push(head, 60)
    head = push(head, 50)
    head = push(head, 40)
    head = push(head, 30)
    head = push(head, 20)
    head = push(head, 10)

    reverseBetween(head, 3, 6)

    prints(head)

# This code is contributed by rutvik_56

C#

// C# program to reverse a linked list
// from position m to position n

using System;

class LinkedList {
    static Node head;

    public class Node {
        public int data;
        public Node next;

        public Node(int d)
        {
            data = d;
            next = null;
        }
    }

    // function to add a new node at
    // the end of the list
    public void AddNode(Node node)
    {
        if (head == null)
            head = node;
        else {
            Node temp = head;
            while (temp.next != null) {
                temp = temp.next;
            }
            temp.next = node;
        }
    }

    // function to reverse the list
    static public Node reverse(Node head)
    {
        Node prev = null, current = head, next = null;
        while (current != null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        return prev;
    }
    // function used to reverse a linked list from position m to n
    public  void reverseBetween( int m, int n)
    {
        if (m == n)
            return ;

        // revs and revend is start and end respectively of the
        // portion of the linked list which need to be reversed.
        // revs_prev is previous of starting position and
        // revend_next is next of end of list to be reversed.
        Node revs = null, revs_prev = null;
        Node revend = null, revend_next = null;

        // Find values of above pointers.
        int i = 1;
        Node curr = head;
        while (curr!=null && i <= n) {
            if (i < m)
                revs_prev = curr;
               if (i == m)
                revs = curr;
            if (i == n) {
                revend = curr;
                revend_next = curr.next;
            }
            curr = curr.next;
            i++;
        }
        revend.next = null;
        // Reverse linked list starting with revs.
        revend = reverse(revs);
        // If starting position was not head
        if (revs_prev!=null)
            revs_prev.next = revend;
        // If starting position was head
        else
            head = revend;
        revs.next = revend_next;
    }


    // function to print the list data
    public void PrintList()
    {
        Node current = head;
        while (current != null) {
            Console.Write(current.data + " ");
            current = current.next;
        }
        Console.WriteLine();
    }
}
class GFG{
    // Driver Code
    public static void Main(string[] args)
    {
        LinkedList list = new LinkedList();
        list.AddNode(new LinkedList.Node(10));
        list.AddNode(new LinkedList.Node(20));
        list.AddNode(new LinkedList.Node(30));
        list.AddNode(new LinkedList.Node(40));
        list.AddNode(new LinkedList.Node(50));
          list.AddNode(new LinkedList.Node(60));
          list.AddNode(new LinkedList.Node(70));

        list.reverseBetween(3,6);

        list.PrintList();
    }
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)

Javascript

<script>
// JavaScript program to reverse a linked list
// from position m to position n

// Linked list node
class Node{

    constructor(data){

        this.data = data
        this.next = null
    }
}

// The standard reverse function used
// to reverse a linked list
function reverse(head){

    let prev = null
    let curr = head

    while (curr){
        let next = curr.next
        curr.next = prev
        prev = curr
        curr = next
    }

    return prev
}

// Function used to reverse a linked list
// from position m to n which uses reverse
// function
function reverseBetween(head, m, n){

    if (m == n)
        return head

    // revs and revend is start and end respectively
    // of the portion of the linked list which
    // need to be reversed. revs_prev is previous
    // of starting position and revend_next is next
    // of end of list to be reversed.
    let revs = null
    let revs_prev = null
    let revend = null
    let revend_next = null

    // Find values of above pointers.
    let i = 1
    let curr = head

    while (curr && i <= n){
        if (i < m)
            revs_prev = curr

        if (i == m)
            revs = curr

        if (i == n){
            revend = curr
            revend_next = curr.next
        }

        curr = curr.next
        i += 1
    }

    revend.next = null

    // Reverse linked list starting with
    // revs.
    revend = reverse(revs)

    // If starting position was not head
    if (revs_prev)
        revs_prev.next = revend

    // If starting position was head
    else
        head = revend

    revs.next = revend_next
    return head
}

function prints(head){

    while (head != null){
        document.write(head.data,' ')
        head = head.next
    }

    document.write("</br>")
}

// Function to add a new node at the
// beginning of the list
function push(head_ref, new_data){

    let new_node = new Node(new_data)
    new_node.data = new_data
    new_node.next = head_ref
    head_ref = new_node
    return head_ref
}

// Driver code

let head = null
head = push(head, 70)
head = push(head, 60)
head = push(head, 50)
head = push(head, 40)
head = push(head, 30)
head = push(head, 20)
head = push(head, 10)

reverseBetween(head, 3, 6)

prints(head)

// This code is contributed by shinjanpatra

</script>

Output

10 20 60 50 40 30 70 

Time Complexity: O(N), Here N is the number of nodes in the linked list. In the worst case we need to traverse the list twice.
Auxiliary Space: O(1), As constant extra space is used.

Method 2: (single traversal)

In this method we need to traverse the list only once in the worst case. The algorithm makes used of the idea to reverse a normal linked list. Below is the algorithm:

• Get the pointer to the head and tail of the reversed linked list.
• Get the pointer to the node before mth and node after nth node.
• Reverse the list as discussed in this  post.
• Connect back the links properly.

Implementation of the above approach:

C++

// C++ program to reverse a linked list
// from position m to position n
#include <bits/stdc++.h>
using namespace std;

// Linked list node
struct Node {
    int data;
    struct Node* next;
};

// function used to reverse a linked list from position m to
// n
Node* reverseBetween(Node* head, int m, int n)
{
    // First move the current pointer to the node from where
    // we have to reverse the linked list
    Node *curr = head, *prev = NULL;
    // prev points to the node before mth node
    int i;
    for (i = 1; i < m; i++) {
        prev = curr;
        curr = curr->next;
    }
    // This pointer stores the pointer to the head of the
    // reversed linkedlist
    Node* rtail = curr;
    // This pointer stores the pointer to the tail of the
    // reversed linkedlist
    Node* rhead = NULL;
    // Now reverse the linked list from m to n nodes
    while (i <= n) {
        Node* next = curr->next;
        curr->next = rhead;
        rhead = curr;
        curr = next;
        i++;
    }
    // if prev is not null it means that some of the nodes
    // exits before m  or we can say m!=1
    if (prev != NULL)
        prev->next = rhead;
    else
        head = rhead;
    // at this point curr will point to the next of nth
    // node where we will connect the tail of the reversed
    // linked list
    rtail->next = curr;
    // at the end return the new head.
    return head;
}

void print(struct Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
    cout << endl;
}

// function to add a new node at the
// beginning of the list
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

// Driver code
int main()
{
    struct Node* head = NULL;
    push(&head, 70);
    push(&head, 60);
    push(&head, 50);
    push(&head, 40);
    push(&head, 30);
    push(&head, 20);
    push(&head, 10);
    struct Node* nhead = reverseBetween(head, 3, 6);
    print(nhead);
    return 0;
}

// This code is contributed by Abhijeet Kumar(abhijeet19403)

Java

// Java program to reverse a linked list
// from position m to position n

class LinkedList {

    static Node head;

    static class Node {

        int data;
        Node next;

        Node(int d)
        {
            data = d;
            next = null;
        }
    }
    // function used to reverse a linked list from position
    // m to n

    static Node reverseBetween(Node head, int m, int n)
    {

        // First move the current pointer to the node from
        // where we have to reverse the linked list
        Node curr = head, prev = null;
        // prev points to the node before mth node
        int i;
        for (i = 1; i < m; i++) {
            prev = curr;
            curr = curr.next;
        }
        // This pointer stores the pointer to the head of
        // the reversed linkedlist
        Node rtail = curr;
        // This pointer stores the pointer to the tail of
        // the reversed linkedlist
        Node rhead = null;
        // Now reverse the linked list from m to n nodes
        while (i <= n) {
            Node next = curr.next;
            curr.next = rhead;
            rhead = curr;
            curr = next;
            i++;
        }
        // if prev is not null it means that some of the
        // nodes exits before m ( or if m!=1)
        if (prev != null)
            prev.next = rhead;
        else
            head = rhead;
        // at this point curr will point to the next of nth
        // node where we will connect the tail of the
        // reversed linked list
        rtail.next = curr;
        // at the end return the new head.
        return head;
    }

    // prints content of double linked list
    void printList(Node node)
    {
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        LinkedList list = new LinkedList();
        list.head = new Node(10);
        list.head.next = new Node(20);
        list.head.next.next = new Node(30);
        list.head.next.next.next = new Node(40);
        list.head.next.next.next.next = new Node(50);
        list.head.next.next.next.next.next = new Node(60);
        list.head.next.next.next.next.next.next
            = new Node(70);
        reverseBetween(head, 3, 6);
        list.printList(head);
    }
}

// This code has been contributed by Abhijeet Kumar(abhijeet19403)

Python3

# Python3 program to reverse a linked list
# from position m to position n

# Linked list node
class Node:

    def __init__(self, data):

        self.data = data
        self.next = None

def reverse(head):

    prev = None
    curr = head

    while (curr):
        next = curr.next
        curr.next = prev
        prev = curr
        curr = next

    return prev

# Function used to reverse a linked list
# from position m to n
def reverseBetween(head, m, n):

          # First move the current pointer to the node from
        # where we have to reverse the linked list
        curr = head
        prev = None
        # prev points to the node before mth node
        i = 1
        while i<m:
            prev = curr
            curr = curr.next
            i+=1

        # This pointer stores the pointer to the head of
        # the reversed linkedlist
        rtail = curr
        # This pointer stores the pointer to the tail of
        # the reversed linkedlist
        rhead = None
        # Now reverse the linked list from m to n nodes
        while (i <= n):
            temp = curr.next
            curr.next = rhead
            rhead = curr
            curr = temp
            i+=1

        # if prev is not null it means that some of the
        # nodes exits before m ( or if m!=1)
        if prev:
            prev.next = rhead
        else:
            head = rhead
        # at this point curr will point to the next of nth
        # node where we will connect the tail of the
        # reversed linked list
        rtail.next = curr
        # at the end return the new head.
        return head

def prints(head):

    while (head != None):
        print(head.data, end = ' ')
        head = head.next

    print()

# Function to add a new node at the
# beginning of the list
def push(head_ref, new_data):

    new_node = Node(new_data)
    new_node.data = new_data
    new_node.next = (head_ref)
    (head_ref) = new_node
    return head_ref

# Driver code
if __name__=='__main__':

    head = None
    head = push(head, 70)
    head = push(head, 60)
    head = push(head, 50)
    head = push(head, 40)
    head = push(head, 30)
    head = push(head, 20)
    head = push(head, 10)

    reverseBetween(head, 3, 6)

    prints(head)

# This code is contributed by Abhijeet Kumar(abhijeet19403)

C#

// C# program to reverse a linked list
// from position m to position n

using System;

class LinkedList {
    static Node head;

    public class Node {
        public int data;
        public Node next;

        public Node(int d)
        {
            data = d;
            next = null;
        }
    }

    // function to add a new node at
    // the end of the list
    public void AddNode(Node node)
    {
        if (head == null)
            head = node;
        else {
            Node temp = head;
            while (temp.next != null) {
                temp = temp.next;
            }
            temp.next = node;
        }
    }

    // function used to reverse a linked list from position m to n
    public void reverseBetween(int m, int n)
    {

        // First move the current pointer to the node from
        // where we have to reverse the linked list
        Node curr = head, prev = null;
        // prev points to the node before mth node
        int i;
        for (i = 1; i < m; i++) {
            prev = curr;
            curr = curr.next;
        }
        // This pointer stores the pointer to the head of
        // the reversed linkedlist
        Node rtail = curr;
        // This pointer stores the pointer to the tail of
        // the reversed linkedlist
        Node rhead = null;
        // Now reverse the linked list from m to n nodes
        while (i <= n) {
            Node next = curr.next;
            curr.next = rhead;
            rhead = curr;
            curr = next;
            i++;
        }
        // if prev is not null it means that some of the
        // nodes exits before m ( or if m!=1)
        if (prev != null)
            prev.next = rhead;
        else
            head = rhead;
        // at this point curr will point to the next of nth
        // node where we will connect the tail of the
        // reversed linked list
        rtail.next = curr;

    }

    // function to print the list data
    public void PrintList()
    {
        Node current = head;
        while (current != null) {
            Console.Write(current.data + " ");
            current = current.next;
        }
        Console.WriteLine();
    }
}
class GFG{
    // Driver Code
    public static void Main(string[] args)
    {
        LinkedList list = new LinkedList();
        list.AddNode(new LinkedList.Node(10));
        list.AddNode(new LinkedList.Node(20));
        list.AddNode(new LinkedList.Node(30));
        list.AddNode(new LinkedList.Node(40));
        list.AddNode(new LinkedList.Node(50));
          list.AddNode(new LinkedList.Node(60));
          list.AddNode(new LinkedList.Node(70));

        list.reverseBetween(3,6);

        list.PrintList();
    }
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)

Javascript

<script>
// JavaScript program to reverse a linked list
// from position m to position n

// Linked list node
class Node{

    constructor(data){

        this.data = data
        this.next = null
    }
}

// Function used to reverse a linked list
// from position m to n
function reverseBetween(head, m, n){
        // First move the current pointer to the node from
        // where we have to reverse the linked list
        let curr = head;
        let prev = null;
        // prev points to the node before mth node
        let i;
        for (i = 1; i < m; i++) {
            prev = curr;
            curr = curr.next;
        }
        // This pointer stores the pointer to the head of
        // the reversed linkedlist
        let rtail = curr;
        // This pointer stores the pointer to the tail of
        // the reversed linkedlist
        let rhead = null;
        // Now reverse the linked list from m to n nodes
        while (i <= n) {
            Node next = curr.next;
            curr.next = rhead;
            rhead = curr;
            curr = next;
            i++;
        }
        // if prev is not null it means that some of the
        // nodes exits before m ( or if m!=1)
        if (prev != null)
            prev.next = rhead;
        else
            head = rhead;
        // at this point curr will point to the next of nth
        // node where we will connect the tail of the
        // reversed linked list
        rtail.next = curr;
}

function prints(head){

    while (head != null){
        document.write(head.data,' ')
        head = head.next
    }

    document.write("</br>")
}

// Function to add a new node at the
// beginning of the list
function push(head_ref, new_data){

    let new_node = new Node(new_data)
    new_node.data = new_data
    new_node.next = head_ref
    head_ref = new_node
    return head_ref
}

// Driver code

let head = null
head = push(head, 70)
head = push(head, 60)
head = push(head, 50)
head = push(head, 40)
head = push(head, 30)
head = push(head, 20)
head = push(head, 10)

reverseBetween(head, 3, 6)

prints(head)

// This code is contributed by Abhijeet Kumar(abhijeet19403)

</script>

Output

10 20 60 50 40 30 70 

Time Complexity: O(n), Here n is the position n till which we have to reverse the linked list. In the worst case we need to traverse the list once when n is equal to the size of the linked list and in best case time complexity can go upto O(1).
Auxiliary Space: O(1), As constant extra space is used.

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