Given a singly linked list, Reverse this linked list from position M to N. Do it in one-pass. You should try to do it in place. The program should run in O(nodes) time complexity code example
Example 1: reverse a singly linked list in c
/**
* C program to reverse a Singly Linked List
*/
#include <stdio.h>
#include <stdlib.h>
/* Structure of a node */
struct node {
int data; //Data part
struct node *next; //Address part
}*head;
/* Functions used in the program */
void createList(int n);
void reverseList();
void displayList();
int main()
{
int n, choice;
/*
* Create a singly linked list of n nodes
*/
printf("Enter the total number of nodes: ");
scanf("%d", &n);
createList(n);
printf("\nData in the list \n");
displayList();
/*
* Reverse the list
*/
printf("\nPress 1 to reverse the order of singly linked list\n");
scanf("%d", &choice);
if(choice == 1)
{
reverseList();
}
printf("\nData in the list\n");
displayList();
return 0;
}
/*
* Create a list of n nodes
*/
void createList(int n)
{
struct node *newNode, *temp;
int data, i;
if(n <= 0)
{
printf("List size must be greater than zero.\n");
return;
}
head = (struct node *)malloc(sizeof(struct node));
/*
* If unable to allocate memory for head node
*/
if(head == NULL)
{
printf("Unable to allocate memory.");
}
else
{
/*
* Read data of node from the user
*/
printf("Enter the data of node 1: ");
scanf("%d", &data);
head->data = data; // Link the data field with data
head->next = NULL; // Link the address field to NULL
temp = head;
/*
* Create n nodes and adds to linked list
*/
for(i=2; i<=n; i++)
{
newNode = (struct node *)malloc(sizeof(struct node));
/* If memory is not allocated for newNode */
if(newNode == NULL)
{
printf("Unable to allocate memory.");
break;
}
else
{
printf("Enter the data of node %d: ", i);
scanf("%d", &data);
newNode->data = data; // Link the data field of newNode with data
newNode->next = NULL; // Link the address field of newNode with NULL
temp->next = newNode; // Link previous node i.e. temp to the newNode
temp = temp->next;
}
}
printf("SINGLY LINKED LIST CREATED SUCCESSFULLY\n");
}
}
/*
* Reverse the order of nodes of a singly linked list
*/
void reverseList()
{
struct node *prevNode, *curNode;
if(head != NULL)
{
prevNode = head;
curNode = head->next;
head = head->next;
prevNode->next = NULL; // Make first node as last node
while(head != NULL)
{
head = head->next;
curNode->next = prevNode;
prevNode = curNode;
curNode = head;
}
head = prevNode; // Make last node as head
printf("SUCCESSFULLY REVERSED LIST\n");
}
}
/*
* Display entire list
*/
void displayList()
{
struct node *temp;
/*
* If the list is empty i.e. head = NULL
*/
if(head == NULL)
{
printf("List is empty.");
}
else
{
temp = head;
while(temp != NULL)
{
printf("Data = %d\n", temp->data); // Print the data of current node
temp = temp->next; // Move to next node
}
}
}
Example 2: reverse a linked list
class recursion {
static Node head; // head of list
static class Node {
int data;
Node next;
Node(int d)
{ data = d;
next = null; } }
static Node reverse(Node head)
{
if (head == null || head.next == null)
return head;
/* reverse the rest list and put the first element
at the end */
Node rest = reverse(head.next);
head.next.next = head;
/* tricky step -- see the diagram */
head.next = null;
/* fix the head pointer */
return rest;
} /* Function to print linked list */
static void print()
{
Node temp = head;
while (temp != null) {
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
static void push(int data)
{
Node temp = new Node(data);
temp.next = head;
head = temp;
} /* Driver program to test above function*/
public static void main(String args[])
{
/* Start with the empty list */
push(20);
push(4);
push(15);
push(85);
System.out.println("Given linked list");
print();
head = reverse(head);
System.out.println("Reversed Linked list");
print();
} } // This code is contributed by Prakhar Agarwal