To reverse K nodes in a linked list

input

4
1
2
3
4
5
6
7
8
9

selected_reversal.c

#include <stdio.h>
#include <stdlib.h>

struct node
{
	int val;
	struct node* next;
};

void enqueue(struct node**,struct node **,struct node **);
struct node* dequeue(struct node**,struct node **);
void traverse(struct node*);
void push(struct node**,struct node**);
struct node* pop(struct node**);
void listrev(struct node**,struct node**,int);
void free_queue(struct node*);
int isstackempty(struct node*);
struct node* makenode(int);
int listlength(struct node*);
struct node* filndlast(struct node*);

int main()
{
	int key,k;
	struct node *front=NULL,*rear=NULL;
	FILE *fp;
	fp=fopen("input","r");
	fscanf(fp,"%d",&k);
	fscanf(fp,"%d",&key);
	while(!feof(fp))
	{
		struct node* newnode=makenode(key);
		enqueue(&front,&rear,&newnode);
		fscanf(fp,"%d",&key);
	}
	fclose(fp);
	fp=fopen("output","w");
	fclose(fp);
	traverse(front);
	listrev(&front,&rear,k);
	traverse(front);
	free_queue(front);
	return 0;
}

void enqueue(struct node** front,struct node** rear,struct node** newnode)
{
	(*newnode)->next=NULL;
	if(*front==NULL)
		*front=*rear=*newnode;
	else
	{
		(*rear)->next=*newnode;
		(*rear)=*newnode;
	}
}

struct node* dequeue(struct node** front,struct node** rear)
{
	struct node* temp=NULL;
	if(*front==*rear)
	{
		temp=*front;
		*front=*rear=NULL;
	}
	else if(*front!=NULL)
	{
		temp=*front;
		*front=(*front)->next;
	}
	if(temp!=NULL)
		temp->next=NULL;
	return temp;
}

void free_queue(struct node* front)
{
	while(front!=NULL)
	{
		struct node* save=front->next;
		free(front);
		front=save;
	}
}

void traverse(struct node* head)
{
	struct node* temp=head;
	while(temp!=NULL)
	{
		printf("%d",temp->val);
		if(temp->next!=NULL)
			printf("--->");
		else
			printf("\n");
		temp=temp->next;
	}
}

void push(struct node** top,struct node** newnode)
{
	(*newnode)->next=NULL;
	if(*top==NULL)
		*top=*newnode;
	else
	{
		(*newnode)->next=*top;
		*top=*newnode;
	}
}

struct node* pop(struct node** top)
{
	struct node* temp=NULL;
	if(!isstackempty(*top))
	{
		temp=(*top);		
		*top=(*top)->next;
	}
	temp->next=NULL;
	return temp;
}

int isstackempty(struct node* top)
{
		if(top==NULL)
			return 1;
		else
			return 0;
}

struct node* makenode(int key)
{
	struct node* newnode=malloc(sizeof(struct node));
	newnode->val=key;
	newnode->next=NULL;
	return newnode;
}

int listlength(struct node* head)
{
	int count=0;
	struct node* temp=head;
	while(temp!=NULL)
	{
		count++;
		temp=temp->next;
	}
	return count;
}

struct node* findlast(struct node* head)
{
	if(head!=NULL)
		while(head->next!=NULL)
			head=head->next;
	return head;
}

void listrev(struct node** front,struct node** rear,int k)
{
	struct node *temp1=NULL,*temp2=NULL,*top=NULL,*head=NULL,*save=NULL;
	int i=0,flag=0;
	temp1=head=*front;
	*front=*rear=NULL;
	save=temp1;
	int len=listlength(head);
	int times=len/k;
	while(times--)
	{
		temp1=head;
		while(temp1!=NULL&&i++<k)
		{
			temp2=temp1->next;
			push(&top,&temp1);
			temp1=temp2;
		}
		head=temp1;
		if(!isstackempty(top))
			temp1=pop(&top);
		if(flag>0)
			save->next=temp1;
		if(flag==0)
			*front=temp1;
		while(!isstackempty(top))
		{
			temp2=pop(&top);
			temp1->next=temp2;
			temp1=temp2;
		}
		save=temp1;
		i=0;
		flag++;
	}
	if(head!=NULL)	
		save->next=head;
	else
		save->next=NULL;
	*rear=findlast(*front);
}

Reverse linked list by k nodes each
for eg
for k=4
input
4
list:1---->2---->3---->4---->5---->6---->7---->8---->9
o/p:4---->3---->2---->1---->8---->7---->6---->5---->9

No additional space required, no new nodes are created, uses a stack for reversal