struct Node{
string val;
Node* next;
};
Node* makeList ()
{
string current;
Node* n;
Node* head= NULL;
Node* temp = n;
while(cin>>current && !cin.fail())
{
n = new Node;
n->val = current;
temp ->next = n;
temp = temp -> next;
}
n->next = NULL;
return n;
}
I am trying to learn about linked lists, and this function makeList() is supposed to create and return a linked list using input from a list of strings. To be honest, I'm kind of lost. Any help would be greatly appreciated.
First of all, you are returning the last node of the linked list.. I think you should return the head and assign it to the first Node.
Secondly you are using cin.fail() for a string which I think should not be done. cin.fail() will work if there is a data mismatch and for string I think it is rare.
The function would look somewhat like:
Node* makeList ()
{
string current;
Node* n;
Node* head= NULL;
Node* temp = n;
while(cin>>current && !cin.fail())
{
if(current == "0")
break;
n = new Node;
n->val = current;
temp ->next = n;
temp = temp -> next;
if(!head)
head = n;
}
n->next = NULL;
return head;
}
First of all, since your temp represents the last element I would put it to NULL at the beginning (nullptr is more in the spirit of C++, so I'll use it in the text that comes).
After that in a while loop ,when you are adding a new element, you should write n->next=nullptr,since the pointer next of the new element(if you're always adding it to the back of the list) will always point to nullptr. In your implementation your new element n is always pointing to itself.Later on in your while loop you need to check if head==nullptr, if that's true than you should assign head to the new element that you made head=n. If head is not equal to nullptr then you need to add your element n to the back temp->next=n. An at the and of the loop you should assign the n element as last-temp=n (that has to be outside of the else block since it is done in both of the above mentioned cases).
I'm afraid answers above all got some bugs...
Node *make_link_list_from_input(){
string value;
Node *head = nullptr;
Node *current = nullptr;
Node *last = nullptr;
while (cin >> value){
current = new Node();
if(head== nullptr){
head = current;
}
if(last!= nullptr){
last->next=current;
}
last=current;
}
if(last != nullptr) {
last->next = nullptr;
}
return head;
}
Related
This is a basic function that takes an iterator position and deletes the node in this position but it gives me a runtime error. what am i doing wrong?
iterate erase(iterate position)
{
iterate i;
Node<T>* temp = head;
if (head == NULL) {
cout << "empty list" << endl;
}
else if (position.pointer == head) {
head = temp->next;
temp->next->previous = NULL;
delete position.pointer;
}
else {
while (temp != NULL) {
if (temp == position.pointer->previous) {
temp->next = position.pointer->next;
temp->next->previous = temp;
i.pointer = temp->next;
delete position.pointer;
return i;
}
}
}
Your function is lacking adequate return statements. There are multiple flows that can cause the function to exit, but only one of them has a return statement. So the return value will largely be indeterminate, causing undefined behavior for any caller that tries to use the return value.
In any case, your while loop iterates forever, because you are not updating temp on each iteration of the loop. You also have a NULL pointer dereference if position is pointing at the last node in the list, as you are not checking the new temp->next for NULL before accessing temp->next->previous.
But, you really don't need the while loop at all. The thing about a double-linked list is that, given any node in the list, you have direct access to the nodes that are surrounding it on both sides. So there is no need to iterate the list hunting for nodes.
Try something more like this instead:
iterate erase(iterate position)
{
Node<T> *temp = position.pointer;
if (!temp) return end();
Node<T> *next = temp->next;
Node<T> *previous = temp->previous;
if (next) next->previous = previous;
if (previous) previous->next = next;
if (temp == head) head = next;
//if (temp == tail) tail = previous;
delete temp;
iterate i;
i.pointer = next;
return i;
}
Alternatively:
iterate erase(iterate position)
{
Node<T> *temp = position.pointer;
if (!temp) return end();
Node<T> *dummy; // <-- only if no tail ...
Node<T> **previous = (temp->next) ? &(temp->next->previous) : &dummy/*&tail*/;
Node<T> **next = (temp->previous) ? &(temp->previous->next) : &head;
*previous = temp->previous;
*next = temp->next;
delete temp;
iterate i;
i.pointer = *next;
return i;
}
i need to make a function to sum 2 piecewise linear functions (both decreasing or increasing) and insert them into a third list in an ascending order based on the x-axis coordinates of each point. So i have created multiple functions, all seem to check out except this one yet i do not know what's the problem. It's not inputting anything at all.
struct coords has double x,y;
dList has : coords pt;
node has : node *head, *tail;
node *prev, *next;
dList insert(dList L, coords point) {
node *temp;
temp = new node;
if (temp == NULL) {
cout << "error";
exit(1);
}
temp->next = NULL;
temp->prev = NULL;
temp->pt = point;
if (L.head == NULL || L.tail == NULL) {
L.head = temp;
L.tail = temp;
return L;
}
if (L.head->pt.x > temp->pt.x) {
temp->next = L.head;
L.head->prev = temp;
L.head = temp;
return L;
}
if (L.tail->pt.x < temp->pt.x) {
temp->prev = L.tail;
L.tail->next = temp;
L.tail = temp;
return L;
}
node *cur;
cur = L.head->next;
while (cur->pt.x < temp->pt.x)
cur = cur->next;
temp->next = cur->next;
temp->prev = cur;
cur->next->prev = temp;
cur->next = temp;
return L;
}
The case in which the node to be inserted is in the middle is the problem. You should be looking one node ahead instead of looking at the current one. Try working it out on paper and you will see how that makes a difference:
node * cur;
// also start at head here
cur=L.head;
while(cur->next->pt.x<temp->pt.x)
cur=cur->next;
temp->next=cur->next;
temp->prev=cur;
cur->next->prev=temp;
cur->next=temp;
You should also consider passing a dList L as a pointer to the function and also returning it as a pointer:
// this way you won't be making a copy of it, you may run into trouble if you don't have your copy constructor implemented
dList* insert(dList* L,coords point)
I hope that this helps you.
I am writing a simple function to insert at the end of a linked list on C++, but finally it only shows the first data. I can't figure what's wrong. This is the function:
node* Insert(node* head, int data)
{
if (head == NULL) {
head = new node();
head->data = data;
head->link = NULL;
return head;
}
else {
node* temp = head;
while (temp != NULL) {
temp = temp->link;
}
node* temp2 = new node();
temp2->data = data;
temp2->link = NULL;
(temp->link) = temp2;
return head;
}
}
Change the condition in while construct from:
while (temp!=NULL) {
temp=temp->link;
}
To
while (temp->link!=NULL) {
temp=temp->link;
}
In statement, temp->link = temp2, temp is a null pointer. You were dereferencing a NULL pointer.
To append a node at the back, temp pointer should point to the last node of the linked list. So, in the while loop, you need to just stop linked list traversal when you have reached the last node, i.e, the node whose link member points to nothing (has NULL). while (temp->link!=NULL) will stop at the last node as last node will have link member pointing to NULL.
You can simplify your logic by doing this:
void Insert(node **pnode, int data)
{
while (*pnode) {
pnode = &(*pnode)->link;
}
*pnode = new node(data, NULL);
}
assuming you have a node constructor that initializes data and link from arguments.
Instead of calling it as
head = Insert(head, 42);
you'd now do
Insert(&head, 42);
change while(temp!=NULL) to while(temp->link!=NULL)
node* Insert(node* head, int data)
{
if (head == NULL) {
head = new node();
}
else {
while (head->link != NULL) {
head = head->link;
}
head = head->link = new node();
}
head->data = data;
head->link = NULL;
return head;
}
I was writing a simple function to insert at the end of a linked list on C++, but finally it only shows the first data. I can't figure what's wrong. This is the function:
void InsertAtEnd (node* &firstNode, string name){
node* temp=firstNode;
while(temp!=NULL) temp=temp->next;
temp = new node;
temp->data=name;
temp->next=NULL;
if(firstNode==NULL) firstNode=temp;
}
What you wrote is:
if firstNode is null, it's replaced with the single node temp which
has no next node (and nobody's next is temp)
Else, if firstNode is not null, nothing happens, except that the temp
node is allocated and leaked.
Below is a more correct code:
void insertAtEnd(node* &first, string name) {
// create node
node* temp = new node;
temp->data = name;
temp->next = NULL;
if(!first) { // empty list becomes the new node
first = temp;
return;
} else { // find last and link the new node
node* last = first;
while(last->next) last=last->next;
last->next = temp;
}
}
Also, I would suggest adding a constructor to node:
struct node {
std::string data;
node* next;
node(const std::string & val, node* n = 0) : data(val), next(n) {}
node(node* n = 0) : next(n) {}
};
Which enables you to create the temp node like this:
node* temp = new node(name);
You've made two fundamental mistakes:
As you scroll through the list, you roll off the last element and start constructing in the void behind it. Finding the first NULL past the last element is useless. You must find the last element itself (one that has its 'next' equal NULL). Iterate over temp->next, not temp.
If you want to append the element at the end, you must overwrite the last pointer's NULL with its address. Instead, you write the new element at the beginning of the list.
void InsertAtEnd (node* &firstNode, string name)
{
node* newnode = new node;
newnode->data=name;
newnode->next=NULL;
if(firstNode == NULL)
{
firstNode=newnode;
}
else
{
node* last=firstNode;
while(last->next != NULL) last=last->next;
last->next = newnode;
}
}
Note, this gets a bit neater if you make sure never to feed NULL but have all lists always initialized with at least one element. Also, inserting at the beginning of list is much easier than appending at the end: newnode->next=firstNode; firstNode=newnode.
The last element in your list never has it's next pointer set to the new element in the list.
The problem is that you are replacing the head of the linked list with the new element, and in the process losing the reference to the actual list.
To insert at the end, you want to change the while condition to:
while(temp->next != null)
After the loop, temp will point to the last element in the list. Then create a new node:
node* newNode = new node;
newNode->data = name;
newNode->next = NULL;
Then change temps next to this new node:
temp->next = newNode;
You also do not need to pass firstNode as a reference, unless you want NULL to be treated as a linked list with length 0. In that case, you will need to significantly modify your method so it can handle the case where firstNode is NULL separately, as in that case you cannot evaluate firstNode->next without a segmentation fault.
If you don't want to use reference pointer, you could use pointer to pointer. My complete code goes like below:
void insertAtEnd(struct node **p,int new_data)
{
struct node *new_node=(struct node *)malloc(sizeof(struct node));
new_node->data=new_data;
new_node->next=NULL;
if((*p)==NULL)//if list is empty
{
*p=new_node;
return;
}
struct node* last=*p;//initailly points to the 1st node
while((last)->next != NULL)//traverse till the last node
last=last->next;
last->next=new_node;
}
void printlist(struct node *node)
{
while(node != NULL);
{
printf("%d->",node->data);
node=node->next;
}
}
int main()
{
struct node *root=NULL;
insertAtEnd(&root,1);
insertAtEnd(&root,2);
insertAtEnd(&root,3);
insertAtEnd(&root,4);
insertAtEnd(&root,5);
printlist(root);
return 0;
}
Understanding the need of the below two variables is key to understanding the problem:
struct node **p: Because we need to link it from the root node created in the main.
struct node* last: Because if not used, the original content will be changed with the contents of the next node inside the while loop. In the end only 2 elements will be printed, the last 2 nodes, which is not desired.
void addlast ( int a)
{
node* temp = new node;
temp->data = a;
temp->next = NULL;
temp->prev=NULL;
if(count == maxnum)
{
top = temp;
count++;
}
else
{
node* last = top;
while(last->next)
last=last->next;
last->next = temp;
}
}
#include <bits/stdc++.h>
using namespace std;
class Node
{
public:
int data;
Node *next;
};
void append(Node *first, int n)
{
Node *foo = new Node();
foo->data = n;
foo->next = NULL;
if (first == NULL)
{
first = foo;
}
else
{
Node *last = first;
while (last->next)
last = last->next;
last->next = foo;
}
}
void printList(Node *first)
{
while (first->next != NULL)
{
first = first->next;
cout << first->data << ' ';
}
}
int main()
{
Node *node = new Node();
append(node, 4);
append(node, 10);
append(node, 7);
printList(node);
return 0;
}
Output: 4 10 7
You can use this code:
void insertAtEnd(Node* firstNode, string name)
{
Node* newn = new Node; //create new node
while( firstNode->next != NULL ) //find the last element in yur list
firstNode = firstNode->next; //he is the one that points to NULL
firstNode->next = newn; //make it to point to the new element
newn->next = NULL; //make your new element to be the last (NULL)
newn->data = name; //assign data.
}
void InsertAtEnd (node* &firstNode, string name){
node* temp=firstNode;
while(temp && temp->next!=NULL) temp=temp->next;
node * temp1 = new node;
temp1->data=name;
temp1->next=NULL;
if(temp==NULL)
firstNode=temp1;
else
temp->next= temp1;
}
while loop will return at temp==null in your code instead you need to return last node pointer from while loop like this
while(temp && temp->next!=NULL) temp=temp->next;
and assign a new node to next pointer of the returned temp node will add the data to the tail of linked list.
#include <iostream>
using namespace std;
struct Node
{
int item; // storage for the node's item
Node* next; // pointer to the next node
};
Node* addNode(Node*& head, int data , int& count)
{
Node * q; // new node
q = new Node; // allocate memory for the new mode
q->item = data; // inserting data for the new node
q->next = head; // point to previous node ?? how would i do that? ( am i doing it correctly?)
count++; // keep track of number of node
head = q;
return q;
}
int main()
{
int a, count=0;
int data;
bool repeat;
Node *head= NULL;
//^^ assuming thats creating the first node ^^
do
{
cout << "please enter the data for the next node" <<endl;
cin >> data;
addNode(head, data, count);
cout << "do you wish to enter another node? (enter true or false)" << endl;
cin >>repeat;
}
while (repeat == true);
// assuming this is the print function
while(head != NULL)
{
cout << "output" << temp->item << endl;
cout << temp->next << endl;
}
system("pause");
return 0;
}
okey i tried adding a new element in the list how would i move the head around like a LIFO memory (stack) so the last element is on the very top..
any help would be appreciated ! The pointers and the nodes are messing with my brain lately ....
temp is an uninitialized pointer. So -
temp-> item = a; // temp is not initialized or pointing to a memory location
// that has Node object to use operator ->
First, temp needs to be allocated memory location using new.
temp = new Node;
temp -> item = a;
And now assign it head. Similarly allocate memory for the child nodes too in the while loop. And return all the resources acquired from child to head using delete before program termination.
You seem to have some misunderstandings here:
Your "head" is the start of the list. It's always the start.
You add append elements to a linked list by assigning them to the last node's next pointer.
Third, you're not allocating anything.
Node *head= new Node();
Node *temp = new Node();
cout<<"enter something into data"<<endl;
cin >> a ;
temp->item = a;
head->next = temp;
Now ... to add the next thing, you either need to keep track of the last node (tail), or traverse the list to find the last node.
Node *nextNode = new Node();
nextNode->item = 0.0;
Node *i;
for (i = head; i->next != null; i = i->next);
i->next = nextNode;
This is O(n) execution time. By keeping track of the tail you make it O(1):
Node *head= new Node();
Node *tail = head;
Node *temp = new Node();
cout<<"enter something into data"<<endl;
cin >> a ;
temp->item = a;
tail->next = temp;
tail = temp;
Node *nextNode = new Node();
nextNode->item = 0.0;
tail->next = nextNode;
tail = nextNode;
EDIT: As pointed out, if you want to prepend to the list, you would:
temp->next = head;
head = temp;
Since I'm not sure every answer completely answers it, here's a linked list implementation (written without testig:
// your (correct) structure
struct Node
{
float item; // storage for the node's item
Node* next; // pointer to the next node
};
Now we need two pointers somewhere to look after the list:
/* some pointers */
struct List
{
Node* head;
Node* tail;
};
Now we need to create some elements. As others have said, you can do that with new:
/* create some elements we want to link in */
Node* elem1 = new Node();
Node* elem2 = new Node();
Node* elem3 = new Node();
/* maybe even set their properties! */
elem1->item = 3.14;
elem2->item = 3.14;
elem3->item = 3.14;
Notice how I didn't try to add these elements to a list yet? That's because I've got a function in mind which looks like this:
void addtolist(List &list, Node* node)
{
/* if no head, initialise the list */
if ( list->head == NULL )
{
list->head = node;
list->tail = node;
}
else if ( list->head != NULL && list->tail != NULL )
{
/* access the tail element and set its
next to this ptr.
Move tail to this node */
list->tail->next = node;
list->tail = node;
}
else
{
/* probably raise an exception! */
}
}
You can call this by doing this:
List l;
addtolist(l, elem1); /* etc */
Deleting elements is somewhat more tricky, since you have to go to that element, remember its previous element, grab it's next element, join them up and delete the Node* you're on.
Now for traversing lists... your terminology HEAD|TAIL reminds me of Erlang and tail recursion, where the current element is referred to as the head and the remainder the tail. If I write:
Node* cur = l.head;
while ( cur != NULL )
{
// do something with cur.item ?
cur = cur->next;
}
You can see this happening. Replacing cur with head here would be harmless thanks to the List struct.
Finally, I've used a very C-like approach here, but there's scope for templates:
template<typename T>
struct node
{
T item; // storage for the node's item
Node<T>* next; // pointer to the next node
};
and encapsulating the List struct as a class:
template<typename T>
class List
{
protected:
Node<T>* head;
Node<T>* tail;
public:
void addtolist(Node<T>* node);
Node<T>* gethead();
Node<T>* gettail();
}
Which brings you a little bit closer to std::list.
Additionally note that you are doing an implicit cast from int to float on
temp-> item = a;
as a is an int, while temp->item is a double.
To solve your problem: You want to allocate a new structure before accessing temp, thus
temp = new Node();