Rotate an array and a linked list

Suppose we have an array A having n elements. You have to rotate the array by k positions.
For e.g. if we have the array A={1,2,3,4,5} and k=2 then the resulting array should be {3,4,5,1,2}. Try to post an in-place solution. The array need not be sorted originally. The question has multiple solutions. Try to discuss all.

Now what will be the approach if you want to rotate a singly linked list instead of an array.

# include<iostream>
using namespace std;
main()
{int i,n,k;
cout<<"enter size of array";
cin>>n;
int a[n];
for(i=0;i<n;++i)
cin>>a[i];
cout<<"\nenter k";
cin>>k;
k=k%n;
for(i=0;i<n;++i)
{if(i+k<n)
cout<<a[i+k]<<" ";
else
cout<<a[i+k-n]<<" ";
}
// system("pause");
}

You dont have to print the answer ...
you have to change the original array ..

printf("enter the size and k");
scanf("%d%d",&n,&k);
for(j=0;j<k;j++)
{
c=a[0];
for(i=0;i<n-1;i++)
a[i]=a[i+1];
a[n-1]=c;
}

now print array a[];

for rotating the list , first break the list at pont k(measured from front),then attach it to the last point.point of attachment should be tail->head.

this should work...
1->get to the kth element from the last
2->point this element to the head
3->reverse the list beyond k

tiwari ji
please explain the working by taking an example, i am unable to get it

void rotate(int n)
{
node *r;
r=start;
while(r->next!=NULL)
{
r=r->next;

}
while(n--)
{
r->next=start;
r=start;
start=start->next;
cout<<r->n<<endl;
r->next=NULL;
}
}