Okay This is the code for insering a node into a linked list.
vec_store holds seq and size. Variable seq holds the vectors and a pointer. and vec_mag takes magnitude of vectors.
For some reason, the (vec_mag(v)<=vec_mag(temp2->next->data)) doesn't work which is the last condition.
Any1 can solve the problem? By the way this is C code.
vector last_vec(vec_store s){
node temp3;
temp3=s->seq;
while (temp3->next!=NULL)
{temp3 = temp3->next;
}
return temp3->data;
}
void insert_vec(vec_store s, vector v){
node temp1,temp2,temp4;
int i;
temp1 = malloc(sizeof (struct node_record));
if(s->seq==NULL){
s->seq=temp1;
temp1->next=NULL;
temp1->data=v;
s->size++;
printf("1\n");
}
else if(vec_mag(v)<=vec_mag(s->seq->data)){
s->size++;
temp2=s->seq;
temp1->data=v;
temp1->next=temp2;
s->seq=temp1;
printf("2\n");
}
else if(vec_mag(v)>=vec_mag(last_vec(s)))
{ s->size=s->size+1;
temp4=s->seq;
while (temp4->next!=NULL)
{temp4 = temp4->next;
}
temp1->next=NULL;
temp1->data=v;
temp4->next=temp1;
printf("3\n");
}
else{
temp2 = s->seq;
temp4 = s->seq;
for(i=0;i<s->size-1;i++){
if(vec_mag(v)<=vec_mag(temp2->next->data)){
temp1->data = v;
temp1->next = temp2->next;
temp2->next=temp1;
printf("4\n");
s->size++;
break;
}
}
}
}
The problem is that in that loop, you don't actually move along the list at all.
"Anon" is correct - you loop the variable i through the size of the list, you don't shift the pointers before the comparison.
But there are more issues here.
I'm not sure what your data structures look like since you haven't posted their source, but I'm going to assume that you mean the nodes (temp1 - temp4) to be node pointers instead of full instances of the structures.
This is a good effort, but there are excessive variables used, needless computations and unnecessary copy-by-value's. Nothing computationally wrong with that if you get the result you were looking for, but I don't think it's doing exactly what you'd want it to and it makes it a bit harder to trace/maintain. Sometimes it makes a world of difference to set things up in logical blocks with a couple of comments.
I haven't tried to compile this (try to just read the logic and comments), but you might have more luck with something like the following (apologies for the c++ comments in C code):
// construct the node and its data first (except for the next pointer)
// - it's going to be inserted no matter what the list is like
node* inserted = (node*) malloc(sizeof(struct node));
inserted->data = v;
// store the vector magnitude so you don't have to compute it on every comparison
double v_mag = vec_mag(v);
// Case 1 - empty list
if (s->seq == NULL)
{
inserted->next = NULL;
s->seq = inserted;
}
// Case 2 - in case there's only one element in the list
// (this is me being too lazy to work this step into the main logic in case 3)
else if (s->seq->next == NULL)
{
t1_mag = vec_mag(s->seq->data);
if (v_mag <= t1_mag)
{
//insert
inserted->next = s->seq;
s->seq = inserted;
}
else
{
//append
inserted->next = NULL;
s->seq = inserted;
}
}
// Case 3 - there are at least 2 elements in the list
else
{
// set the temporary nodes to the first 2
node* temp1 = s->seq;
node* temp2 = temp1->next;
// store their magnitudes
double t1_mag = vec_mag(temp1->data);
double t2_mag = vec_mag(temp2->data);
// while we aren't at the list, and we aren't at a spot where the node should be inserted
while (temp2 != NULL && !(v_mag >= t1_mag && v_mag <= t2_mag ))
{
// shift the two to the next in the line
temp1 = temp2;
// no need to recompute this magnitude from the last step - just copy it
t1_mag = t2_mag;
temp2 = temp2->next;
t2_mag = vec_mag(temp2->data);
}
// if we can trust the integrity of the list, either temp2 is null (at the end of the list),
// or another node (we found a suitable place to insert).
// Either way, just blindly insert the node.
inserted->next = temp2;
temp1->next = inserted;
}
// Node has been inserted
s->size++;
Related
I have a linked list which is cyclic and I want to find out the total number of elements in this list. How to achieve this?
One solution that I can think of is maintaining two pointers. First pointer (*start) will always point to the starting node, say Node A.
The other pointer (*current) will be initialized as: current = start->next.
Now, just iterate each node with current -> next until it points to start.
And keep incrementing a counter: numberOfNodes++;
The code will look like:
public int countNumberOfItems(Node* start){
Node* current = start -> next;
int numberOfNodes = 1; //Atleast the starting node is there.
while(current->next != start){
numberOfNodes++;
current = current->next;
}
return numberOfNodes;
}
Let's say the list has x nodes before the loop and y nodes in the loop. Run the Floyd cycle detection counting the number of slow steps, s. Once you detect a meet point, run around the loop once more to get y.
Now, starting from the list head, make s - y steps, getting to the node N. Finally, run two slow pointers from N and M until they meet, for t steps. Convince yourself (or better prove) that they meet where the initial part of the list enters the loop.
Therefore, the initial part has s - y + t + 1 nodes, and the loop is formed by y nodes, giving s + t + 1 total.
You just want to count the nodes in your linked list right? I've put an example below. But in your case there is a cycle so you also need to detect that in order not to count some of the nodes multiple times.
I've corrected my answer there is now an ordinary count and count in loop (with a fast and slow pointer).
static int count( Node n)
{
int res = 1;
Node temp = n;
while (temp.next != n)
{
res++;
temp = temp.next;
}
return res;
}
static int countInLoop( Node list)
{
Node s_pointer = list, f_pointer = list;
while (s_pointer !=null && f_pointer!=null && f_pointer.next!=null)
{
s_pointer = s_pointer.next;
f_pointer = f_pointer.next.next;
if (s_pointer == f_pointer)
return count(s_pointer);
}
return 0;
}
First find the cycle using Floyd Cycle Detection algorithm and also maintain count when you checking cycle once found loop then print count for the same.
function LinkedList() {
let length = 0;
let head = null;
let Node = function(element) {
this.element = element;
this.next = null;
}
this.head = function() {
return head;
};
this.add = function(element) {
let node = new Node(element);
if(head === null){
head = node;
} else {
let currentNode = head;
while(currentNode.next) {
currentNode = currentNode.next;
}
currentNode.next = node;
}
};
this.detectLoopWithCount = function() {
head.next.next.next.next.next.next.next.next = head; // make cycle
let fastPtr = head;
let slowPtr = head;
let count = 0;
while(slowPtr && fastPtr && fastPtr.next) {
count++;
slowPtr = slowPtr.next;
fastPtr = fastPtr.next.next;
if (slowPtr == fastPtr) {
console.log("\n Bingo :-) Cycle found ..!! \n ");
console.log('Total no. of elements = ', count);
return;
}
}
}
}
let mylist = new LinkedList();
mylist.add('list1');
mylist.add('list2');
mylist.add('list3');
mylist.add('list4');
mylist.add('list5');
mylist.add('list6');
mylist.add('list7');
mylist.add('list8');
mylist.detectLoopWithCount();
There is a "slow" pointer which moves one node at a time. There is a "fast" pointer which moves twice as fast, two nodes at a time.
A visualization as slow and fast pointers move through linked list with 10 nodes:
1: |sf--------|
2: |-s-f------|
3: |--s--f----|
4: |---s---f--|
5: |----s----f|
At this point one of two things are true: 1) the linked list does not loop (checked with fast != null && fast.next != null) or 2) it does loop. Let's continue visualization assuming it does loop:
6: |-f----s---|
7: |---f---s--|
8: |-----f--s-|
9: |-------f-s|
10: s == f
If the linked list is not looped, the fast pointer finishes the race at O(n/2) time; we can remove the constant and call it O(n). If the linked list does loop, the slow pointer moves through the whole linked list and eventually equals the faster pointer at O(n) time.
I am trying to write the code for the following question:
Insert an element(sum of neighbors) between every pair of consecutive elements?
Example: if input is
12 23 34 45 for n=4
Output should be:
12 35 23 57 34 79 45
The code I wrote is:
struct node *InsBet(node *head) {
node *i,*j,*t;
i=head;
while(i->next!=NULL) {
t = (node*)malloc(sizeof(node));
t->data = i->data + i->next->data;
i->next = t;t->prev = i;
t->next = i->next;i->next->prev = t;
i = i->next;
}
return head;
}
Upon printing the array it is crashing my terminal.
My print program is:
void PrintList(node *head) {
node *i;
i=head;
while(i!=NULL) {
printf("%d ",i->data);
i=i->next;
}
}
The first problem is that you're overriding i->next before copying it to t->next
Switch the order of
i->next = t;t->prev = i;
t->next = i->next;i->next->prev = t;
into
t->next = i->next; i->next->prev = t;
i->next = t; t->prev = i;
To elaborate, assume you have a chain of 2 elements in your list: A-->B, and you want to add the temporary element between, so you create t, but since the first thing you do is overwrite the forward pointer of the first element (A in this case), you lose any chance of ever accessing B again. Instead, you assign into the forward pointer of the temporary element the address of itselfm creating an infinite loop.
The second problem is that you advance the current pointer (i) by only one link, which means it would now point to the temporary element you've just added, and you would try to add an additional temporary element between t and B. This would cause an infinite loop - instead advance i by -
i = t->next;
The above answer explained it very well but just to give you a working code, here you go:
PS, you don't need to return the head pointer because its passed by reference and there is no use in returning it
void InsBet(node *head) {
node *i,*t;
i=head;
while(i->next!=NULL) {
t = (node*)malloc(sizeof(node));
t->data = i->data + i->next->data;
t->prev = i;
t->next = i->next;
i->next = i->next->next;
i->prev = t;
i = t->next;
}
}
I am asked to implement an algorithm based upon the data structure of a linkedList in the form of pseudocode.
Unfortunately I have a Python/Java background and thus no experience with pointers.
Could someone explain me how I would iterate over a doublyLinkedList, change and compare values of elements.
From what I have understood so far, i would do something like this.: to have an iteration over each element.
for L.head to L.tail
But how would I then access the current object in the list analogous to A[i] for i to L.length?
As the order of a linkedList is determined by pointers rather than indices in a linkedList can I simply do things like currentObj.prev.key = someVal or currentObj.key < currentObj.prev.key or is there some other wokflow to work with individual elements?
Again, I am obviously stuck as I lack an basic understanding on how to deal with pointers.
Cheers,
Andrew
So basically the data structures are:
Node:
node {
node next;//"single link"
node prev;//for "doubly"...
}
and List:
list {
node head;//for singly linked, this'd be enough
node tail;//..for "doubly" you "point" also to "tail"
int/*long*/ size; //can be of practical use
}
The (common) operations of a list:
Creation/Initialization:
list:list() {
head = null;
tail = null;
size = 0;
}
Add a node at the first position:
void:addFirst(node) {
if(isEmpty()) {
head = node;
tail = node;
size = 1;
} else {
head.prev = node;
node.next = head;
head = node;
size++;
}
}
// ..."add last" is quite analogous...
"is empty", can be implemented in different ways..as long as you keep the invariant
bool:isEmpty() {
return size==0;
//or return head==null ... or tail==null
}
"add a node at position i":
void:add(i, node) {
assert(i>=0 && i <=size);//!
if(isEmpty() || i == 0) {
addFirst(node);
} else if (i == size) {
addLast(node);
} else {//here we could decide, whether i is closer to head or tail, but for simplicity, we iterate "from head to tail".
int j = 1;
node cur = head.next; //<- a "cursor" node, we insert "before" it ;)
while(j < i) {//1 .. i-1
cur = cur.next;// move the cursor
j++;//count
}
//j == i, cur.next is not null, curr.prev is not null, cur is the node at currently i/j position
node.prev = cur.prev;
cur.prev.next = node;
cur.prev = node;
node.next = cur;
}
//don't forget the size:
size++;
}
Delete(node) is easy!
"Delete at position", "find node", "get node by position", etc. should use a similar loop (as add(i, node))...to find the correct index or node.
The strength/advantage of a doubly (comparing to a singly) linked list, is that it can iterate as "forth" as "back". To use this advantage (it is only advantageous on index-based operations, for "find(node)" e.g. you still don't know where to start/iterate best), you determine whether pos is closer to head(0) or to tail(size-1), and start&route your iteration accordingly.
...What else operations are you intereseted in (detail)?
I wanted to sort a linked list containing 0s, 1s or 2s. Now, this is clearly a variant of the Dutch National Flag Problem.
http://en.wikipedia.org/wiki/Dutch_national_flag_problem
The algorithm for the same as given in the link is:
"Have the top group grow down from the top of the array, the bottom group grow up from the bottom, and keep the middle group just above the bottom. The algorithm stores the locations just below the top group, just above the bottom, and just above the middle in three indexes. At each step, examine the element just above the middle. If it belongs to the top group, swap it with the element just below the top. If it belongs in the bottom, swap it with the element just above the bottom. If it is in the middle, leave it. Update the appropriate index. Complexity is Θ(n) moves and examinations."
And a C++ implementation given for the same is:
void threeWayPartition(int data[], int size, int low, int high) {
int p = -1;
int q = size;
for (int i = 0; i < q;) {
if (data[i] == low) {
swap(data[i], data[++p]);
++i;
} else if (data[i] >= high) {
swap(data[i], data[--q]);
} else {
++i;
}
}
}
My only question is how do we traverse back in a linked list like we are doing here in an array?
A standard singly-linked list doesn't allow you to move backwards given a linked list cell. However, you could use a doubly-linked list, where each cell stores a next and a previous pointer. That would let you navigate the list forwards and backwards.
However, for the particular problem you're trying to solve, I don't think this is necessary. One major difference between algorithms on arrays and on linked lists is that when working with linked lists, you can rearrange the cells in the list to reorder the elements in the list. Consequently, the algorithm you've detailed above - which works by changing the contents of the array - might not actually be the most elegant algorithm on linked lists.
If you are indeed working with linked lists, one possible way to solve this problem would be the following:
Create lists holding all values that are 0, 1, or 2.
Remove all cells from the linked list and distribute them into the list of elements that are equal to 0, 1, or 2.
Concatenate these three lists together.
This does no memory allocation and purely works by rearranging the linked list cells. It still runs in time Θ(n), which is another plus. Additionally, you can do this without ever having to walk backwards (i.e. this works on a singly-linked list).
I'll leave the complete implementation to you, but as an example, here's simple C++ code to distribute the linked list cells into the zero, one, and two lists:
struct Cell {
int value;
Cell* next;
}
/* Pointers to the heads of the three lists. */
Cell* lists[3] = { NULL, NULL, NULL };
/* Distribute the cells across the lists. */
while (list != NULL) {
/* Cache a pointer to the next cell in the list, since we will be
* rewiring this linked list.
*/
Cell* next = list->next;
/* Prepend this cell to the list it belongs to. */
list->next = lists[list->value];
lists[list->value] = list;
/* Advance to the next cell in the list. */
list = next;
}
Hope this helps!
As others have said, there is no way to "back up" in a linked list without reverse links. Though it's not exactly an answer to your question, the sort can be easily accomplished with three queues implementing a bucket sort with three buckets.
The advantage of queues (vice pushing on stacks) is that the sort is stable. That is, if there are data in the list nodes (other than the 0,1,2-valued keys), these will remain in the same order for each key.
This is only one of many cases where the canonical algorithm for arrays is not the best for lists.
There is a very slick, simple way to implement the queues: circularly linked lists where the first node, say p, is the tail of the queue and consequently p->next is is the head. With this, the code is concise.
#include <stdio.h>
#include <stdlib.h>
typedef struct node_s {
struct node_s *next;
int val;
int data;
} NODE;
// Add node to tail of queue q and return the new queue.
NODE *enqueue(NODE *q, NODE *node)
{
if (q) {
node->next = q->next;
q->next = node;
}
else node->next = node;
return node;
}
// Concatenate qa and qb and return the result.
NODE *cat(NODE *qa, NODE *qb)
{
NODE *head = qa->next;
qa->next = qb->next;
qb->next = head;
return qb;
}
// Sort a list where all values are 0, 1, or 2.
NODE *sort012(NODE *list)
{
NODE *next = NULL, *q[3] = { NULL, NULL, NULL};
for (NODE *p = list; p; p = next) {
next = p->next;
q[p->val] = enqueue(q[p->val], p);
}
NODE *result = cat(q[0], cat(q[1], q[2]));
// Now transform the circular queue to a simple linked list.
NODE *head = result->next;
result->next = NULL;
return head;
}
int main(void)
{
NODE *list = NULL;
int N = 100;
// Build a list of nodes for testing
for (int i = 0; i < N; ++i) {
NODE *p = malloc(sizeof(NODE));
p->val = rand() % 3;
p->data = N - i; // List ends up with data 1,2,3,..,N
p->next = list;
list = p;
}
list = sort012(list);
for (NODE *p = list; p; p = p->next)
printf("key val=%d, data=%d\n", p->val, p->data);
return 0;
}
This is now a complete simple test and it runs just fine.
This is untested. (I will try to test it if I get time.) But it ought to be at least very close to a solution.
Using a doubly linked list. If you have already implemented a linked list object and the related link list node object, and are able to traverse it in the forward direction it isn't a whole bunch more work to traverse in the reverse direction.
Assuming you have a Node object somewhat like:
public class Node
{
public Node Next;
public Object Value;
}
Then all you really need to do is change you Node class and you Insert method(s) up a little bit to keep track of of the Node that came previously:
public class Node
{
public Node Next;
public Node Previous;
public Object Value;
}
public void Insert(Node currentNode, Node insertedNode)
{
Node siblingNode = currentNode.Next;
insertedNode.Previous = currentNode;
insertedNode.Next = siblingNode;
if(siblingNode!= null)
siblingNode.previous = insertedNode;
currentNode.next = insertedNode;
}
PS Sorry, I didn't notice the edit that included the C++ stuff so it's more C#
Works for all cases by CHANGING NODES rather than NODE DATA.. Hoping its never too late!
METHOD(To throw some light on handling corner cases):
1. Keep three dummy nodes each for 0,1,2;
2. Iterate throught the list and add nodes to respective list.
3. Make the next of zero,one,two pointers as NULL.
4. Backup this last nodes of each list.
5. Now handle 8 different possible cases to join these list and Determine the HEAD.
zero one two
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
An implementation of this in C++.
Node* sortList(Node *head)
{
struct Node dummyzero,dummyone,dummytwo;
dummyzero.next = dummyone.next = dummytwo.next = NULL;
struct Node *zero =&dummyzero,*one = &dummyone,*two=&dummytwo;
Node *curr = head,*next=NULL;
while(curr)
{
next = curr->next;
if(curr->data==0)
{
zero->next = curr;
zero = zero->next;
}
else if(curr->data==1)
{
one->next = curr;
one = one->next;
}
else
{
two->next = curr;
two = two->next;
}
curr = next;
}
zero->next = one->next = two->next =NULL; //Since this dummynode, No segmentation fault here.
Node *zerolast = zero,*onelast = one,*twolast = two;
zero = dummyzero.next;
one = dummyone.next;
two = dummytwo.next;
if(zero==NULL)
{
if(one==NULL)
head = two;
else
{
head = one;
onelast->next = two;
}
}
else
{
head = zero;
if(one==NULL)
zerolast->next = two;
else
{
zerolast->next = one;
onelast->next = two;
}
}
return head;
}
The idea is to use dutch flag sorting algorithm, with a slight modification:
sort 0's and 1's as per dutch flag method,
But for 2's instead of adding them at the end of list, keep them in a separate linked list.
And finally append the 2's list to the sorted list of 0's and 1's.
Node * sort012_linked_list(Node * head) {
if (!head || !head->next)
return head;
Node * head_of_2s = NULL;
Node * prev = NULL;
Node * curr = head;
while (curr) {
if (curr->data == 0) {
if (prev == NULL || prev->data == 0) {
prev = curr;
curr = curr->next;
}
else {
prev->next = curr->next;
curr->next = head;
head = curr;
curr = prev->next;
}
}
else if (curr->data == 1) {
prev = curr;
curr = curr->next;
}
else { // curr->data == 2
if (prev == NULL) {
head = curr->next;
curr->next = head_of_2s;
head_of_2s = curr;
curr = head;
}
else {
prev->next = curr->next;
curr->next = head_of_2s;
head_of_2s = curr;
curr = prev->next;
}
}
}
if (prev)
prev->next = head_of_2s;
return head;
}
What is the most efficient way to remove duplicate items from an array under the constraint that axillary memory usage must be to a minimum, preferably small enough to not even require any heap allocations? Sorting seems like the obvious choice, but this is clearly not asymptotically efficient. Is there a better algorithm that can be done in place or close to in place? If sorting is the best choice, what kind of sort would be best for something like this?
I'll answer my own question since, after posting, I came up with a really clever algorithm to do this. It uses hashing, building something like a hash set in place. It's guaranteed to be O(1) in axillary space (the recursion is a tail call), and is typically O(N) time complexity. The algorithm is as follows:
Take the first element of the array, this will be the sentinel.
Reorder the rest of the array, as much as possible, such that each element is in the position corresponding to its hash. As this step is completed, duplicates will be discovered. Set them equal to sentinel.
Move all elements for which the index is equal to the hash to the beginning of the array.
Move all elements that are equal to sentinel, except the first element of the array, to the end of the array.
What's left between the properly hashed elements and the duplicate elements will be the elements that couldn't be placed in the index corresponding to their hash because of a collision. Recurse to deal with these elements.
This can be shown to be O(N) provided no pathological scenario in the hashing:
Even if there are no duplicates, approximately 2/3 of the elements will be eliminated at each recursion. Each level of recursion is O(n) where small n is the amount of elements left. The only problem is that, in practice, it's slower than a quick sort when there are few duplicates, i.e. lots of collisions. However, when there are huge amounts of duplicates, it's amazingly fast.
Edit: In current implementations of D, hash_t is 32 bits. Everything about this algorithm assumes that there will be very few, if any, hash collisions in full 32-bit space. Collisions may, however, occur frequently in the modulus space. However, this assumption will in all likelihood be true for any reasonably sized data set. If the key is less than or equal to 32 bits, it can be its own hash, meaning that a collision in full 32-bit space is impossible. If it is larger, you simply can't fit enough of them into 32-bit memory address space for it to be a problem. I assume hash_t will be increased to 64 bits in 64-bit implementations of D, where datasets can be larger. Furthermore, if this ever did prove to be a problem, one could change the hash function at each level of recursion.
Here's an implementation in the D programming language:
void uniqueInPlace(T)(ref T[] dataIn) {
uniqueInPlaceImpl(dataIn, 0);
}
void uniqueInPlaceImpl(T)(ref T[] dataIn, size_t start) {
if(dataIn.length - start < 2)
return;
invariant T sentinel = dataIn[start];
T[] data = dataIn[start + 1..$];
static hash_t getHash(T elem) {
static if(is(T == uint) || is(T == int)) {
return cast(hash_t) elem;
} else static if(__traits(compiles, elem.toHash)) {
return elem.toHash;
} else {
static auto ti = typeid(typeof(elem));
return ti.getHash(&elem);
}
}
for(size_t index = 0; index < data.length;) {
if(data[index] == sentinel) {
index++;
continue;
}
auto hash = getHash(data[index]) % data.length;
if(index == hash) {
index++;
continue;
}
if(data[index] == data[hash]) {
data[index] = sentinel;
index++;
continue;
}
if(data[hash] == sentinel) {
swap(data[hash], data[index]);
index++;
continue;
}
auto hashHash = getHash(data[hash]) % data.length;
if(hashHash != hash) {
swap(data[index], data[hash]);
if(hash < index)
index++;
} else {
index++;
}
}
size_t swapPos = 0;
foreach(i; 0..data.length) {
if(data[i] != sentinel && i == getHash(data[i]) % data.length) {
swap(data[i], data[swapPos++]);
}
}
size_t sentinelPos = data.length;
for(size_t i = swapPos; i < sentinelPos;) {
if(data[i] == sentinel) {
swap(data[i], data[--sentinelPos]);
} else {
i++;
}
}
dataIn = dataIn[0..sentinelPos + start + 1];
uniqueInPlaceImpl(dataIn, start + swapPos + 1);
}
Keeping auxillary memory usage to a minimum, your best bet would be to do an efficient sort to get them in order, then do a single pass of the array with a FROM and TO index.
You advance the FROM index every time through the loop. You only copy the element from FROM to TO (and increment TO) when the key is different from the last.
With Quicksort, that'll average to O(n-log-n) and O(n) for the final pass.
If you sort the array, you will still need another pass to remove duplicates, so the complexity is O(NN) in the worst case (assuming Quicksort), or O(Nsqrt(N)) using Shellsort.
You can achieve O(N*N) by simply scanning the array for each element removing duplicates as you go.
Here is an example in Lua:
function removedups (t)
local result = {}
local count = 0
local found
for i,v in ipairs(t) do
found = false
if count > 0 then
for j = 1,count do
if v == result[j] then found = true; break end
end
end
if not found then
count = count + 1
result[count] = v
end
end
return result, count
end
I don't see any way to do this without something like a bubblesort. When you find a dupe, you need to reduce the length of the array. Quicksort is not designed for the size of the array to change.
This algorithm is always O(n^2) but it also use almost no extra memory -- stack or heap.
// returns the new size
int bubblesqueeze(int* a, int size) {
for (int j = 0; j < size - 1; ++j) {
for (int i = j + 1; i < size; ++i) {
// when a dupe is found, move the end value to index j
// and shrink the size of the array
while (i < size && a[i] == a[j]) {
a[i] = a[--size];
}
if (i < size && a[i] < a[j]) {
int tmp = a[j];
a[j] = a[i];
a[i] = tmp;
}
}
}
return size;
}
Is you have two different var for traversing a datadet insted of just one then you can limit the output by dismissing all diplicates that currently are already in the dataset.
Obvious this example in C is not an efficiant sorting algorith but it is just an example on one way to look at the probkem.
You could also blindly sort the data first and then relocate the data for removing dups, but I'm not sure that would be faster.
#define ARRAY_LENGTH 15
int stop = 1;
int scan_sort[ARRAY_LENGTH] = {5,2,3,5,1,2,5,4,3,5,4,8,6,4,1};
void step_relocate(char tmp,char s,int *dataset)
{
for(;tmp<s;s--)
dataset[s] = dataset[s-1];
}
int exists(int var,int *dataset)
{
int tmp=0;
for(;tmp < stop; tmp++)
{
if( dataset[tmp] == var)
return 1;/* value exsist */
if( dataset[tmp] > var)
tmp=stop;/* Value not in array*/
}
return 0;/* Value not in array*/
}
void main(void)
{
int tmp1=0;
int tmp2=0;
int index = 1;
while(index < ARRAY_LENGTH)
{
if(exists(scan_sort[index],scan_sort))
;/* Dismiss all values currently in the final dataset */
else if(scan_sort[stop-1] < scan_sort[index])
{
scan_sort[stop] = scan_sort[index];/* Insert the value as the highest one */
stop++;/* One more value adde to the final dataset */
}
else
{
for(tmp1=0;tmp1<stop;tmp1++)/* find where the data shall be inserted */
{
if(scan_sort[index] < scan_sort[tmp1])
{
index = index;
break;
}
}
tmp2 = scan_sort[index]; /* Store in case this value is the next after stop*/
step_relocate(tmp1,stop,scan_sort);/* Relocated data already in the dataset*/
scan_sort[tmp1] = tmp2;/* insert the new value */
stop++;/* One more value adde to the final dataset */
}
index++;
}
printf("Result: ");
for(tmp1 = 0; tmp1 < stop; tmp1++)
printf( "%d ",scan_sort[tmp1]);
printf("\n");
system( "pause" );
}
I liked the problem so I wrote a simple C test prog for it as you can see above. Make a comment if I should elaborate or you see any faults.