Reverse alternate elements and append to end of the list - algorithm

Question

Given a linked list as a->x->b->y->c->z , we need to reverse alternate element and append to end of list. That is , output it as a->b->c->z->y->x.
I have an O(n) solution but it takes extra memory , we take 2 lists and fill it with alternate elements respectively , so the two lists are a b c and x y z and then we will reverse the second list and append it to the tail of first so that it becomes a b c z y x .
My question is can we do it in place ? Or is there any other algorithm for the same ?

Answer 1

The basic idea:
Store x.
Make a point to b.
Make y point to the stored element (x).
Make b point to c.
etc.
At the end, make the last element at an odd position point to the stored element.
Pseudo-code: (simplified end-of-list check for readability)
current = startOfList
stored = NULL
while !endOfList
temp = current.next
current.next = current.next.next
temp.next = stored
stored = temp
current = current.next
current.next = stored
Complexity:
O(n) time, O(1) space.

Answer 2

Here is logic in recursion mode
public static Node alRev(Node head)
{
if (head == null) return head;
if (head.next != null)
{
if (head.next.next != null)
{
Node n = head.next;
head.next = head.next.next;
n.next = null;
Node temp = alRev(head.next);
if (temp != null){
temp.next = n;
return n;
}
}
else
return head.next;
}
else
return head;
return null;
}

Answer 3

This is a recent question from amazon interview, the Idea looks good and there seems to be no trick in it.

Answer 4

Java code with comments:
static void change(Node n)
{
if(n == null)
return;
Node current = n;
Node next = null, prev = null;
while(current != null && current.next != null)
{
// One of the alternate node which is to be reversed.
Node temp = current.next;
current.next = temp.next;
// Reverse the alternate node by changing its next pointer.
temp.next = next;
next = temp;
// This node will be used in the final step
// outside the loop to attach reversed nodes.
prev = current;
current = current.next;
}
// If there are odd number of nodes in the linked list.
if(current != null)
prev = current;
// Attach the reversed list to the unreversed list.
prev.next = next;
}

Answer 5

here the c code which don't uses any extra space for doing this..enjoy and have fun
in case of any doubt feel free to ask
#include<stdio.h>
#include<stdlib.h>
int n;
struct link
{
int val;
struct link *next;
};
void show(struct link *);
void addatbeg(struct link **p,int num)
{
struct link *temp,*help;
help=*p;
temp=(struct link *)malloc(sizeof(struct link));
temp->val=num;
temp->next=NULL;
if(help==NULL)
{
*p=temp;
}
else
{
temp->next=help;
*p=temp;
}
n++;
show(*p);
}
void revapp(struct link **p)
{
struct link *temp,*help,*q,*r;
r=NULL;
temp=*p;
help=*p;
while(temp->next!=NULL)
{
temp=temp->next;
q=r; //this portion will revrse the even position numbers
r=temp;
temp=temp->next;
//for making a connection between odd place numbers
if(help->next->next!=NULL)
{
help->next=temp;
help=help->next;
r->next=q;
}
else
{
r->next=q;
help->next=r;
show(*p);
return;
}
}
}
void show(struct link *q)
{
struct link *temp=q;
printf("\t");
while(q!=NULL )
{
printf("%d ->",q->val);
q=q->next;
if(q==temp)
{
printf("NULL\n");
return;
}
}
printf("NULL\n");
}
int main()
{
n=0;
struct link *p;
p=NULL;
// you can take user defined input but here i am solving it on predefined list
addatbeg(&p,8);
addatbeg(&p,7);
addatbeg(&p,6);
addatbeg(&p,5);
addatbeg(&p,4);
addatbeg(&p,3);
addatbeg(&p,2);
addatbeg(&p,1);
revapp(&p);
return 0;
}`