## Thursday, July 12, 2012

### K - Reverse linked list program

Below is the C code for reversing each k-elements in the linked list.
Example: for k value 3
`Input:  1->2->3->4->4->5->6->7->8 `
`Output: 3->2->1->6->5->4->8->7 `
```#include<stdio.h>
#include<stdlib.h>
struct node
{
int info;
struct node *next;
};

typedef struct node Node;

Node* append(Node *h, int info)
{
Node *tempHead=h;
Node *newNode = (Node*)malloc(sizeof(Node));
newNode->info = info;
newNode->next = NULL;
if(tempHead == NULL) // if the list has zero elements. make new node as a head
{
h=newNode;
}
else if(tempHead->next==NULL)  // if the list is having only one node
{
tempHead->next = newNode;
}
else
{
Node *tempNode = h;
while(tempNode->next != NULL)  // if the list has more than one node, so moving to the last node
{
tempNode = tempNode->next;
}
tempNode->next = newNode; // appending the new node at the end
}
if(info == 101) // this is for making the list circular
newNode->next = h;
return h;
// printf("temp -> info is %d\n",temp->info);
}

/*****************************************************************************
for displaying the nodes in the list
*****************************************************************************/
void display(Node *h)
{
Node *temp = h;
while (temp->next!=NULL)
{
printf("%d->",temp->info);
temp = temp->next;
}
printf("%d\n",temp->info);
}

/*****************************************************************************
This function reverse each  K-nodes in the list. e.g for k=3
input:  1->2->3->4->4->5->6->7->8
result: 3->2->1->6->5->4->8->7
*****************************************************************************/
Node *kreverse(Node *head, int k)
{
Node *temp_head,*new_head,*tail,*temp;
int k_temp = k;
new_head = head;
head = head->next;
new_head->next = NULL;
tail = new_head;
while(head && k_temp-1)
{
temp = head->next;
head->next = new_head;
new_head = head;
head = temp;
k_temp--;
}
k_temp = k;
while(head)
{
temp = head->next;
head->next = NULL;
tail->next = head;
temp_head = tail;
tail = head;
head = temp;
while(head && k_temp-1)
{
temp = head->next;
head->next = temp_head->next;
temp_head->next = head;
head = temp;
k_temp--;
}
temp_head = tail;
k_temp = k;
}
return new_head;
}

main()
{
Node *head = NULL;
int i;
for (i=1;i<=20;i++)
{
head = append(head,i*10);
}
display(head);
head =  kreverse(head,3);
display(head);

}
```