[algogeeks] Re: Reverse a List with Recursion
I am looking for a solution with no 'tail' node available/static types
used etc.
Also with algorithms given without above two conditions, i don't see
they doing the right thing for me. Maybe i need to check more.
Have you guys tried to see if this is working for you?
On Jul 17, 4:42 pm, bharath kannan bharathgo...@gmail.com wrote:
node * reverse(node *head)
{
if(head-next)
{
node * temp=reverse(head-next);
head-next-next=head;
head-next=NULL;
return temp;
}
return head;
}
On Sun, Jul 17, 2011 at 4:57 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
*node *reverse(node *head)
{
if(head==NULL)
return head;
if(head-next==NULL)
return head;
node *temp = reverse(head-next);
head-next-next = head;
head-next = NULL;
return (temp);
}*
On Sun, Jul 17, 2011 at 4:30 PM, Anika Jain anika.jai...@gmail.comwrote:
initial call to this will be rev(head);
On Sun, Jul 17, 2011 at 4:28 PM, Anika Jain anika.jai...@gmail.comwrote:
node *listing::rev(node *p)
{
if(p-next==NULL)
{
head=p;
return p;
}
else
{
node *t=rev(p-next);
t-next=p;
p-next=NULL;
tail=p;
return p;
}
}
On Sun, Jul 17, 2011 at 3:21 PM, Nishant Mittal
mittal.nishan...@gmail.com wrote:
void rev_recursion(NODE **head)
{
if(*head==NULL)
return;
NODE *first, *rest;
first=*head;
rest=first-next;
if(!rest)
return;
rev_recursion(rest);
first-next-next=first;
first-next=NULL;
*head=rest;
}
On Sun, Jul 17, 2011 at 2:53 PM, vaibhav shukla
vaibhav200...@gmail.com wrote:
struct node *reverse_recurse(struct node *start)
{
if(start-next)
{
reverse_recurse(start-next);
start-next-next=start;
return(start);
}
else
{
head=start;
}
}
in main
if(head)
{
temp = reverse_recurse(head);
temp-next =NULL;
}
head and temp are global
On Sun, Jul 17, 2011 at 2:42 PM, Navneet Gupta
navneetn...@gmail.comwrote:
Hi,
I was trying to accomplish this task with the following call , header
= ReverseList(header)
I don't want to pass tail pointer or anything and just want that i get
a reversed list with new header properly assigned after this call. I
am getting issues in corner conditions like returning the correct node
to be assigned to header.
Can anyone give an elegant solution with above requirement? Since it
is with recursion, please test for multiple scenarios (empty list, one
node list, twe nodes
list etc) before posting your solution. In case
of empty list, the procedure should report that.
--
Regards,
Navneet
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best wishes!!
Vaibhav Shukla
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To
Re: [algogeeks] Re: Reverse a List with Recursion
@Navneet:
Its not possible to solve this without a single tail pointer. I have come
across this question saying it requires a single pointer and that pointer is
the tail pointer. So along with head you need a tail pointer or a global
pointer. Otherwise cannot keep track of header for reversed list.
On Sun, Jul 17, 2011 at 6:57 PM, Navneet navneetn...@gmail.com wrote:
I am looking for a solution with no 'tail' node available/static types
used etc.
Also with algorithms given without above two conditions, i don't see
they doing the right thing for me. Maybe i need to check more.
Have you guys tried to see if this is working for you?
On Jul 17, 4:42 pm, bharath kannan bharathgo...@gmail.com wrote:
node * reverse(node *head)
{
if(head-next)
{
node * temp=reverse(head-next);
head-next-next=head;
head-next=NULL;
return temp;
}
return head;
}
On Sun, Jul 17, 2011 at 4:57 PM, Piyush Sinha ecstasy.piy...@gmail.com
wrote:
*node *reverse(node *head)
{
if(head==NULL)
return head;
if(head-next==NULL)
return head;
node *temp = reverse(head-next);
head-next-next = head;
head-next = NULL;
return (temp);
}*
On Sun, Jul 17, 2011 at 4:30 PM, Anika Jain anika.jai...@gmail.com
wrote:
initial call to this will be rev(head);
On Sun, Jul 17, 2011 at 4:28 PM, Anika Jain anika.jai...@gmail.com
wrote:
node *listing::rev(node *p)
{
if(p-next==NULL)
{
head=p;
return p;
}
else
{
node *t=rev(p-next);
t-next=p;
p-next=NULL;
tail=p;
return p;
}
}
On Sun, Jul 17, 2011 at 3:21 PM, Nishant Mittal
mittal.nishan...@gmail.com wrote:
void rev_recursion(NODE **head)
{
if(*head==NULL)
return;
NODE *first, *rest;
first=*head;
rest=first-next;
if(!rest)
return;
rev_recursion(rest);
first-next-next=first;
first-next=NULL;
*head=rest;
}
On Sun, Jul 17, 2011 at 2:53 PM, vaibhav shukla
vaibhav200...@gmail.com wrote:
struct node *reverse_recurse(struct node *start)
{
if(start-next)
{
reverse_recurse(start-next);
start-next-next=start;
return(start);
}
else
{
head=start;
}
}
in main
if(head)
{
temp = reverse_recurse(head);
temp-next =NULL;
}
head and temp are global
On Sun, Jul 17, 2011 at 2:42 PM, Navneet Gupta
navneetn...@gmail.comwrote:
Hi,
I was trying to accomplish this task with the following call ,
header
= ReverseList(header)
I don't want to pass tail pointer or anything and just want that i
get
a reversed list with new header properly assigned after this call.
I
am getting issues in corner conditions like returning the correct
node
to be assigned to header.
Can anyone give an elegant solution with above requirement? Since
it
is with recursion, please test for multiple scenarios (empty list,
one
node list, twe nodes
list etc) before posting your solution. In case
of empty list, the procedure should report that.
--
Regards,
Navneet
--
You received this message because you are subscribed to the Google
Groups Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
best wishes!!
Vaibhav Shukla
DU-MCA
--
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*Piyush Sinha*
*IIIT, Allahabad*
*+91-7483122727*
*
