Indexed singly linked list - c++

I want to create a singly linked list with. I already have an implementation that does not permit indexing on it.
I would like to add this feature but I am not sure how to do that.
So basically the head should have index "1". But I can't figure it out myself what am I supposed to do in If statement to increase the index by 1 each step.
void AddNode (int addData)
{
nodePtr NewNode = new node;
NewNode->next = NULL;
NewNode->data = addData;
if (head != NULL)
{
curr = head;
while(curr->next != NULL)
{
curr = curr->next;
}
curr->next = NewNode;
NewNode->index;
}
else
{
head = NewNode;
NewNode->index = 1;
}
}

You mean for the ability to do something like get a linked list node via a get(index)?
Also, your head should not have index 1, it should be 0. This does not comply with standard practice of zero-based indexing.
A linked list does not technically have indexes nor should it store them. It seems like a linked list might not be the right data structure for what you are doing, however you can treat it like so with a loop like this (excuse if my c++ syntax is rusty)
int get(int index) {
Node current = head;
for (int x = 0; x < index; x++) {
if (current->next == null) {
//some sort of error handling for index out of bounds
} else {
current = current->next;
}
}
return current->data
}
get(2) would return the 3rd element of the list.

Graph for the structure
Why do you want to add it to the end of the list? Just simply add the new node to the front.
I don't think it is necessary to follow the order of 1,2,3.... Instead, you can do it reversely
Before you add a new node, you visit the head and find the index(i)of it. When you add the new one, the index of this one will be i+1.
Two advantages:
- It doses not change anything when you loop through this list.
- You know how many you have added into this list.

So what you really want to do is to add the ability to index elements of a linked list.
There is no need to actually store the index anywhere (as you don't really store the index of an array/vector element as the type and the address of the first element of the array is everything you need to retrieve the i-th element).
The only information you want to keep is the length of the list since computing it is costly because you have to traverse the list from head to tail every time you need it. Once you have this information the addNode should only update the size of the list.
As you already know accessing the i-th elements of a linked list is also costly (compared to a vector), but it is easy to code.
Something like the following should work:
void get(Node* head, size_t pos) {
while (head && pos--)
head = head->next;
return pos<=0 ? head : nullptr ;
}
It traverses the list from the head until it either reaches the end (head is nullptr) or pos is <=0.
Once you are out of the loop if pos>0 means that the list is shorter than pos otherwise you can return head (which will point to the pos-th element)

Related

Creating a second singly-linked list in reverse order of another list

I need to create a function that takes a normal singly-linked list and creates another list in reverse order of the first list, the first element in the new list will be the last element in the original list and so on.
My function for some reason only ever returns the same number over and over for the entirety of the new list. Therefore, if the last number in my original list is e.g. '50', the new list will be made up entirely of '50's.
This is my code, what am I doing wrong? If anyone wants me to post the whole program for more clarity or context give me a shout.
void invert() {
node *list1=top,*newlist,*temp,*prev;
while (list1->next!=NULL) {
list1=list1->next;
}
newlist=new node;
newlist->num=list1->num;
newlist->next=NULL;
if (top2==NULL) {
top2=newlist;
}
else {
for (temp=top2;temp!=NULL;temp=temp->next) {
prev=temp;
}
prev->next=newlist;
}
list1->next=NULL;
}
Your code cannot be correct as it creates only a single new node and then modifies the next link in existing nodes. From your requirements, you have to create a new list which means cloning all nodes and linking them in the reverse order.
Following the suggestion of #user4581301, I came up with the following:
node* invert(node* list)
{
node* inverted = NULL;
// run through original in order
for (node* p = list; p != NULL; p = p->next)
{
// clone the node
node* newNode = new node();
newNode->num = p->num;
// and link it so that the predecessor in the original list
// (which has been processed in the previous iteration) is
// linked as next node
newNode->next = inverted;
inverted = newNode;
}
return inverted;
}

Using an array of pointers-to-pointers to manipulate the pointers it points to (C++)

I've been doing this as an exercise on my own to get better at C++ (messing around with a linked list I wrote). What I want to do is to reverse the list by twisting the pointers around, rather than just 'printing' the data out in reverse (which is relatively straightforward).
I have an array of pointers-to-pointers, each pointing to a node in a linked list. But this is less a question about linked-list dynamics (which I understand), and more about pointer magick.
A node looks like this,
template<class T>
struct node {
T data;
node *next;
node(T value) : data(value), next(nullptr) {}
};
And the code in question,
node<T> **reverseArr[listLength];
node<T> *parser = root;
for (auto i : reverseArr) {
i = &parser;
parser = parser->next;
}
root = *(reverseArr[listLength - 1]);
for (int ppi = listLength - 1; ppi >= 0; --ppi) {
if (ppi == 0) {
(*reverseArr[ppi])->next = nullptr;
//std::cout << "ppi is zero!" << "\t";
}
else {
(*reverseArr[ppi])->next = (*reverseArr[ppi - 1]);
//std::cout << "ppi, 'tis not zero!" << "\t";
}
}
My logic:
The new root is the last element of the list,
Iterate through the array in reverse,
Set the current node's next pointer to the previous one by setting the current node's nextNode to the next node in the loop.
What's happening:
If I leave the debug print statements commented, nothing. The function's called but the linked list remains unchanged (not reversed)
If I uncomment the debug prints, the program seg-faults (which doesn't make a whole lot of sense to me but seems to indicate a flaw in my code)
I suspect there's something I'm missing that a fresh pair of eyes might catch. Am I, perhaps, mishandling the array (not accounting for the decay to a pointer or something)?
You're overthinking the problem. The correct way to reverse a single-linked list is much simpler than you think, and does not involve arrays at all.
All you need to do is walk through the list setting each node's next pointer to the head of the list, then set the head of the list to that node. Essentially, you are unlinking each node and inserting it at the start of the list. Once you reach the end, your list is reversed.
It just requires a bit of care, because the order that you do things is important. Something like this should do it:
template <class T>
node<T> * reverse( node<T> * head )
{
node<T> *current = head;
head = NULL;
while( current != NULL )
{
// store remainder of list
node<T> *remain = current->next;
// re-link current node at the head
current->next = head;
head = current;
// continue iterating remainder of list
current = remain;
}
return head;
}
The operation has a linear time complexity. You would invoke it by passing your list's head node as follows:
root = reverse( root );
It should go without saying that it would be a bad idea to call this function with any node that is not the head of a list, or to pass in a list that contains cycles.

Return a pointer to a sorted list; linked list in C++

i want this function sortPair to take 2 Node pointers and return a pointer to a list of the 2 elements sorted alphabetically. The code below is what I have so far. If someone could let me know where I went wrong, that would be great.
struct Node{
string val;
Node* next;
};
Node* sortPair (Node* p1, Node* p2){
//Assert that neither pointer is null
assert(p1!=NULL);
assert(p2!=NULL);
Node* head=NULL;
Node* current=NULL;
Node* last = NULL;
current = new Node();
if(p1-> val >p2-> val) //If p1->val comes before p2->val in the alphabet
{
current->val = p1->val;
head = current;
last = current;
current = new Node();
current -> val = p2->val;
last = current;
last ->next = NULL;
}
else
{
current->val = p2->val;
head = current;
last = current;
current = new Node();
current -> val = p1->val;
last = current;
last ->next = NULL;
}
return head;
}
A linked list is just a series of nodes that are linked by each element having a pointer to the next one.
From your code, it seems like you do not want to make a list of the two nodes, but rather insert a node into a list that already exists.
If all you want to do is to make a linked list of the two nodes that are there, then set the one with the lower or higher value, depending on how you sort them, to point at the other one. For example, if you are sorting from smallest to biggest, set the smallest node's next pointer to point to the bigger one and return the pointer of the smallest one.
If you want to add one of the nodes into a list of nodes, then you must loop through the list until you find one that is larger or smaller than the node you want to insert. I recommend using a while loop.
If you want to merge two lists, then you must make a new loop that inserts each element of one list into the other list.
There is no need to make a new node for any of this, just use the ones you have and change the next pointers.
Hope this helps.

How I can get back in singly linked list?

I have a singly linked list. From each node in the list, I can get to the next element. Is it possible instead to get to the previous element? Note that I cannot use a doubly linked list because of memory constraints.
If you must use a singly linked list, then one way to approach this is to traverse the list again until you find the prior item. You will know when you are at the prior item when its Next field is equal to the current item.
Pseudo C code:
Node* GetPreviousNode(Node* currentNode)
{
Node* iteratorNode = GetHeadNode();
while (iteratorNode != NULL)
{
if (iteratorNode->Next == currentNode)
return iteratorNode;
iteratorNode = iteratorNode->Next;
}
return NULL;
}
You can't, you have to use a double-linked-list where you have a pointer to the previous and next elements. A single linked list simply points to the next element and doesn't keep track off previous elements.
Try this:
http://en.literateprograms.org/Doubly_linked_list_(Java)
There a 2 different methods for traversing a singly linked list both ways.
Method 1: The link field (of the structure) is actually a 'difference' field. The link is computed
as the address of the next item minus the address of the previous item, therefore ...
next = address of the previous item + link of the current item.
prev = address of the next item - link of the current item.
If you can do arithmetic with pointers, you will be in business. If not, you will have to use a union (in C++) or use method 2. See: Algorithms + Data Structures = Programs by Niklaus Wirth - Prentice-Hall
1976 - page 19.
Method 2: The link field is the 'exclusive or' (^) of the address of the previous item with the address
of the next item. Exclusive or is an inverse of itself, therefore ...
next = address of the previous item ^ link of the current item.
prev = address of the next item ^ link of the current item.
If you have a node that you are sure of, is before the given node in a singly linked list, then you can make a function:-
struct node* prev(struct node* before, struct student_detail* current)
{
struct student_detail* temp = before;
while (temp->next != NULL)
{
if(temp->next == current)
{
return temp;
}
temp = temp-> next;
}
return NULL;
}
and use it as in
example_node = prev(before, example_node);
If you have head, you can always use that as the prev node.
You can theoretically go back, it generally beats the whole concept behind a singly linked list and it's more trouble than necessary but if you have to, here's how to implement it in C, assuming of course, you already declared your Node struct:-
Node* go_back(Node* head, Node* current_node){
//Traverse the linked list until we discover the next node
//in our traversal points to our current node
Node* previous_node = head;
while(previous_node->next != NULL){
if(previous_node->next == current_node)
return previous_node->next; //found previous node
previous_node = previous_node->next;
}
return previous_node //we found nothing, give up
}

Sorting Nodes (Linked-list) C++

I'm trying to sort a linked-list. I have a Node called head and it's pointing to the next node, etc.
But when I try to sort the Nodes by the value they carry, I get the sort working because I see it printing out the stuff in the if-statement, but I'm not getting back the linked-list. Where did I go wrong?
Node* head;
void sortlist(){
Node * runner = head;
Node * runner2;
for(runner = head; runner->next != NULL; runner = runner->next){
for(runner2 = runner->next; runner2->next != NULL; runner2 = runner2->next){
if(runner->freq < runner2->freq){
cout<< runner->freq<< " is LT "<<runner2->freq<< endl;
Node * temp = runner;
runner = runner2;
runner2 = temp;
}
}
}
head = runner;
}
I'm only getting back the first Node.
In order to swap two elements in a linked list, consider what you need to change. For example, to get from
Head -> First -> Second -> (NULL)
to
Head -> Second -> First -> (NULL)
you need to update: Head.next, First.next and Second.next. You don't change any of those things when trying to swap nodes, so it can't possibly do what you expect.
Just swapping the values (ie, swap(runner->freq, runner2->freq)) would be much simpler.
you will stop when runner->next == NULL;, which is supposed to be the last element. And then you set head = runner;, which means the head will always be the last element after this routine. Furthermore, I do not trust this swapping.
It seems you vaguely want to do an insertion sort. If you want to do a simple sorting on linked lists, I suggest you to use selection sort: You create another empty list l2, and each time you remove the minimum element from your first list, and add it as the head of l2. The code for the second list is simple:
void prepend(Node* node, Node** list){
//null checks if you want
node->next = *list;
*list=node->next;
}