Reverse search in a singly-linked list

[RAJASEKHAR KONDABALA]

Hi,

Does anybody know what the fastest way is to "search for a value in a
singly-linked list from its tail" as oposed to its head?

I am talking about a non-circular singly-linked list, i.e., head and tail
are not connected.

Of course, recursive function aproach to traverse the list is one way. But,
depending upon the list size, it could overrun the stack pretty fast.

-sekhar

 

[Derk Gwen]

# Of course, recursive function aproach to traverse the list is one way. But,
# depending upon the list size, it could overrun the stack pretty fast.

Your stack is too small.

If you want an iterative solution, reverse the list, search, reverse it again.
The time is linear in the size of the list, which is the same for searching
an unsorted set.

#define List(Type,eq) \
typedef struct List##Type {Type value; struct List##Type link;} List##Type; \
int searchList##Type(List##Type *list,Type value) { \
int i; List##Type *p; for (i=0,p=list; p; p=p->link,i++) \
if (eq(p->value,value)) return i; \
return -1;} \
List##Type *reverseList##Type(List##Type list) { \
List##Type *p,*c,*n; for (p=0,c=list; c; p=c,c=n) \
{n = c->link; c->link = p;} \
return p;} \
int reversesearchList##Type(List##Type *list,Type value) { \
List##Type *r = reverseList##Type(list); \
int n = searchList##Type(r,Type value);
reverseList##Type(r); \
return n;}

 

[Michael B Allen]

Hi,

Does anybody know what the fastest way is to "search for a value in a
singly-linked list from its tail" as oposed to its head?

Use a for loop with a function that can retrieve an element from the
array by index. Of course this will be something like an O(n*log n)
operation. It would be much faster to convert the list to an array first
for fast indexing. Better still, don't use a singularly linked list. Use
a doubly linked list.

Mike

 

[Christopher Benson-Manica]

Does anybody know what the fastest way is to "search for a value in a
singly-linked list from its tail" as oposed to its head?

(The below welcome text was originally written by Ben Pfaff)

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[Alexander Bartolich]

begin followup to RAJASEKHAR KONDABALA:

Does anybody know what the fastest way is to "search for a value
in a singly-linked list from its tail" as oposed to its head?

Do regular search from start to end, using a simple iterative loop.
Constant space complexity, linear time complexity.

const ListElem* find_last_occurence(const ListElem* first, int key)
{
const ListElem* result = 0;

while(first != 0)
{
if (first->key == key)
result = first;
first = first->next;
}

return result;
}

 

[Bertrand Mollinier Toublet]

Hi,

Does anybody know what the fastest way is to "search for a value in a
singly-linked list from its tail" as oposed to its head?

I am talking about a non-circular singly-linked list, i.e., head and tail
are not connected.

Given the usual definitions of the terms "singly-linked list" and
"tail", I'd be surprised that you could find any element in the list
besides the tail element itself if you are starting your search from the
tail. I must be missing something...

 

[CBFalconer]

Does anybody know what the fastest way is to "search for a value
in a singly-linked list from its tail" as oposed to its head?

I am talking about a non-circular singly-linked list, i.e., head
and tail are not connected.

Of course, recursive function aproach to traverse the list is
one way. But, depending upon the list size, it could overrun the
stack pretty fast.

Reverse the list, search from the head, reverse the list again (if
needed). Reversal takes n extremely simple operations, search
takes (on average) n/2 fairly complex operations. Pays off if
more than one search from tail is needed.

 

[xarax]

Reverse the list, search from the head, reverse the list again (if
needed). Reversal takes n extremely simple operations, search
takes (on average) n/2 fairly complex operations. Pays off if
more than one search from tail is needed.

Too much work. Just walk the list from head to tail
and each time you find a node that "matches" what
you're looking for, stuff the node pointer into a
variable (initialized to NULL, of course). At the
end of the walk, the variable will have the address
of the last node that matched. No need to re-order
the list, because you would have to walk the list
anyway, so just look the node with a single walk.

 

[Stig Brautaset]

Too much work. Just walk the list from head to tail and each time you
find a node that "matches" what you're looking for, stuff the node
pointer into a variable (initialized to NULL, of course). At the end
of the walk, the variable will have the address of the last node that
matched. No need to re-order the list, because you would have to walk
the list anyway, so just look the node with a single walk.

As CBFalconer said, reversing the list is probably going to be cheaper,
unless your test condition is extremely cheap. If we are comparing
strings, for example, reversing the list will almost certainly be
faster. Walking the list once we're forced to do N string compares for a
list with N nodes. If we reverse the list we can get away with N/2 such
compares on average.

This makes certain assumptions about the dataset that might not hold
true of course.

Stig

 

[Alexander Bartolich]

begin followup to Stig Brautaset:

If we are comparing strings, for example, reversing the list will
almost certainly be faster.
Nope.

Walking the list once we're forced to do N string compares for a
list with N nodes.

That's for walking an unsorted list.

If we reverse the list we can get away with N/2 such compares on
average.

Only if the list was sorted before.

Anyway, if it takes exactly N/2 comparisons to find the first
occurence of the mystical average element in a sorted list,
than search the last occurence of that element takes N/2 + M
comparisons, where M ist the number of matches.

 

[CBFalconer]

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Too much work. Just walk the list from head to tail and each time
you find a node that "matches" what you're looking for, stuff the
node pointer into a variable (initialized to NULL, of course). At
the end of the walk, the variable will have the address of the last
node that matched. No need to re-order the list, because you would
have to walk the list anyway, so just look the node with a single
walk.

You failed to read the complete paragraph, especially the portion
underlined above. Note that your method always requires n complex
operations.

 

[Stig Brautaset]

begin followup to Stig Brautaset:

Nope.

I believe so, yes, for unsorted lists. Of course, if you are going to
throw many searches at it you can do worse than sorting the list first.

That's for walking an unsorted list.

True, but whether the list was sorted or unsorted was not specified. I
assumed an unsorted list (and so did you, afaics, in your previous post
to the OP .

Only if the list was sorted before.

Are you sure? AFAICS you can get away with visiting N/2 nodes on
average, regardless of whether the list is sorted or not.

Stig

 

[Alexander Bartolich]

begin followup to Stig Brautaset:

Are you sure? AFAICS you can get away with visiting N/2 nodes on
average, regardless of whether the list is sorted or not.

On a sorted list you can stop the walk as soon as we find a

current_node->key > search_key

On an unsorted list you have look at all nodes, always.

 

[CBFalconer]

begin followup to Stig Brautaset:


On a sorted list you can stop the walk as soon as we find a

current_node->key > search_key

On an unsorted list you have look at all nodes, always.

Sorting does not enter into it. The original request was for a
search from the tail, implying that the result required was the
value nearest the tail. By reversing the list you have only to
search until something is found, or nothing is found. Without
reversal you have to scan the entire list at all times.

 

[xarax]

You failed to read the complete paragraph, especially the portion
underlined above. Note that your method always requires n complex
operations.

I was responding (primarily) to the original question
was did not specify whether the list was sorted nor
what the "value" was.

The underlined text in a responder's post is adding
an assumption that the "value" requires a non-scalar
comparison.

I could also add more requirements, such as the list
is currently being accessed and modified by multiple
threads. What's the most efficient way to find the
matching non-scalar value that is nearest to the end
of a LIFO singly-linked list being concurrently modified
by multiple threads? (That's a rhetorical question.)

Just taking the OP's question at face value, there are
no hidden "gotcha's", like non-scalar value (increases
the cost of comparison) or multithreading (cannot change
the ordering of the list). In that original case, there's
no reason to reorder the list (twice) to find the last
matching value.


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Farsight Systems Corporation
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LONGMONT, CO 80501-6906
http://www.farsight-systems.com
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Are ISV upgrade fees too high? Check our custom product development!

 

[CBFalconer]

I was responding (primarily) to the original question
was did not specify whether the list was sorted nor
what the "value" was.

The underlined text in a responder's post is adding
an assumption that the "value" requires a non-scalar
comparison.

No it doesn't. It assumes that finding an item involves a
comparison. List reversal does not, which is why it is so quick
and easy.

I could also add more requirements, such as the list
is currently being accessed and modified by multiple
threads. What's the most efficient way to find the
matching non-scalar value that is nearest to the end
of a LIFO singly-linked list being concurrently modified
by multiple threads? (That's a rhetorical question.)

And entirely off-topic in c.l.c. C has no threads, nor
concurrency constructs.

Just taking the OP's question at face value, there are
no hidden "gotcha's", like non-scalar value (increases
the cost of comparison) or multithreading (cannot change
the ordering of the list). In that original case, there's
no reason to reorder the list (twice) to find the last
matching value.

It only needs one reversal. The second is only if the original
list must be restored. Having put a singly linked list reversal
routine in your pocket, you will find it has many uses. It is
also simple and understandable, so you can build things quickly
out of proven components, WAPO (while avoiding premature
optimization).

Pseudo code:

list = revlist(list);
lastoftype1 = searchfirst(list, type1);
lastoftype2 = searchfirst(list, type2);
list = revlist(list); /* if needed */

 

[Alexander Bartolich]

begin followup to CBFalconer:

Sorting does not enter into it. The original request was for a
search from the tail, implying that the result required was the
value nearest the tail. By reversing the list you have only to
search until something is found, or nothing is found. Without
reversal you have to scan the entire list at all times.

Alright. Let's organize this and take all fun out of it.
Actually there are two independent aspects, giving four cases.
A search for the first occurrence of the average element:

successful failed
sorted | n/2 n/2
unsorted | n/2 n

N ist the number of list items. Result of the lookup is the required
number of comparisons. Now the search for the _last_ occurrence:

successful failed
sorted | n/2 + m n/2
unsorted | n n

Where m is the number of occurences.

So sorting does indeed enter the story. For a sorted list the
performance improvement of reversing the list upfront is minimal.

And on the other hand reversion is a probably a rather expensive
operation. You walk all elements of the list and insert them into
new list. This requires changing the next-pointer of every element,
as opposed to just reading the key element for comparison.

And things get worse if you need the keep the original, unreversed
list for later on. There is the issue of concurrency - you can have
many simultaneous readers, but only one writer. And of course the
magic spell of all performance discussions: invalidated cache lines.

 

[CBFalconer]

begin followup to CBFalconer:


Alright. Let's organize this and take all fun out of it.

.... snip stuff not germane to the problem ...

The fact that the OP wants the value nearest the tail implies both
an unsorted list and a list with possible multiple entries for any
given key.

 

[pete]

No it doesn't. It assumes that finding an item involves a
comparison. List reversal does not, which is why it is so quick
and easy.

And entirely off-topic in c.l.c. C has no threads, nor
concurrency constructs.

I think this is a comp.programming topic, but I don't know.

You seem to be viewing this problem
from two differents points of view:
1 It's a speed problem.
2 It's a last occurance search.

The first option occured to me.
It implies that the comparison takes considerable time,
and/or that the searched for node,
is known to be near the end of the list.
After further consideration,
I think that OP's interest in a last occurance search,
might be more likely.

I don't think that a sorted list was implied.
In either case, reversing the list first,
would be the simplest thing.

Suppose the last occurance, was the first node.
One way:
You reverse the list, search it's length, find the node,
and maybe reverse it back.
The other way, you search the length of the list,
keeping track of your first node.

So, the first way, you have the same operational intensity
as the second way, plus one or two list reversals.

Suppose the last occurance, was the last node.
One way:
You reverse the list, search one node and boom, you're done,
and maybe reverse it back.
The other way, you search the length of the list.

I like the list reversal, because it's simple,
and list reversal isn't too big of a deal,
as list operations go.

In order to know which way is faster,
you would have to know how time consuming the comparison is
and where the node is expected to be found.
If the searched for node,
is expected to be randomly located in the list,
then there will be on average,
half as many node comparisons using the reversal method,
but a lot more pointer assignments.

If the data is an int type,
and the comparisons are just "((a) > (b))" macros,
instead of functions, then not reversing might be faster.

Without knowing more, the reversal first way, appeals to me.

 

[Roger Willcocks]

I think this is a comp.programming topic, but I don't know.

You seem to be viewing this problem
from two differents points of view:
1 It's a speed problem.
2 It's a last occurance search.

The first option occured to me.
It implies that the comparison takes considerable time,
and/or that the searched for node,
is known to be near the end of the list.
After further consideration,
I think that OP's interest in a last occurance search,
might be more likely.

I don't think that a sorted list was implied.
In either case, reversing the list first,
would be the simplest thing.

Suppose the last occurance, was the first node.
One way:
You reverse the list, search it's length, find the node,
and maybe reverse it back.
The other way, you search the length of the list,
keeping track of your first node.

So, the first way, you have the same operational intensity
as the second way, plus one or two list reversals.

Suppose the last occurance, was the last node.
One way:
You reverse the list, search one node and boom, you're done,
and maybe reverse it back.
The other way, you search the length of the list.

I like the list reversal, because it's simple,
and list reversal isn't too big of a deal,
as list operations go.

In order to know which way is faster,
you would have to know how time consuming the comparison is
and where the node is expected to be found.
If the searched for node,
is expected to be randomly located in the list,
then there will be on average,
half as many node comparisons using the reversal method,
but a lot more pointer assignments.

If the data is an int type,
and the comparisons are just "((a) > (b))" macros,
instead of functions, then not reversing might be faster.

Without knowing more, the reversal first way, appeals to me.

Both solutions require that you touch every node in the list, so they're not
exactly memory friendly. The advantage of reversing the list is that
subsequent searches can drop out early when a matching value is found. Of
course, if the list doesn't contain the value you're looking for, you'll end
up visiting every node anyway.

My take on this is that if you find you need to search a linked list from
its tail, you've got your design wrong. The list should be chained the other
way, or it should be a doubly linked list. Get the design right and the
question just goes away.

One other point - scanning a linked list is inherently inefficient; if speed
is a factor some other data structure - a (balanced) tree, or hash table for
instance - would be favourite. If you;re stuck with the list, you could
perhaps add a search cache based on one of those techniques.

If you're really really stuck with a list (because for instance this is an
assignment question) the answer is 'it depends'...