So basically, what I'm trying to do is going through all the nodes and verify if node.value is <= cost. If it is I need to remove that node and in the end I want to store the nodes that weren't remove in a new struct. I'm not sure how exactly am I suppose to do this.
This struct can be an example:
struct node {
int value;
node * next;
}
I'm going through all the nodes and remove only the ones that doen't have the required.
node * verify_money(node * head, int cost)
{
node * iterator = head;
while(iterator != NULL){
if(iterator.value <= cost) {
/*remove this node*/
}
iterator = iterator.next;
}
return /*struct without nodes removed/*
}
I want to get the remaining nodes.
What you are asking for really depends on your requirements, which you did not make clear enough.
If you are expected to modify the input list that is being verified, you can do something like this:
node * verify_money(node * head, int cost)
{
node * iterator = head;
node * previous = NULL;
while (iterator) {
node *next = iterator->next;
if (iterator->value <= cost) {
if (previous) previous->next = next;
if (head == iterator) head = next;
delete iterator;
}
else
previous = iterator;
iterator = next;
}
return head;
}
If you are expected to return a new list without modifying the original list, you can do something like this instead:
node * verify_money(node * head, int cost)
{
node * new_head = NULL;
node ** new_node = &new_head;
node * iterator = head;
while (iterator) {
if (iterator->value > cost) {
*new_node = new node;
(*new_node)->cost = value;
(*new_node)->next = NULL;
new_node = &((*new_node)->next);
}
iterator = iterator->next;
}
return new_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;
}
#include <iostream>
#include <string.h>
namespace forward_circular_linked_list {
typedef struct Node {
std::string data;
Node *nextNode;
}Node;
class ForwardCircularLinkedList {
private:
Node *head;
public:
ForwardCircularLinkedList() : head(nullptr) {}
void AddItem(std::string data) {
Node * newNode = new Node();
newNode->data = data;
if(head == nullptr)
{
head = newNode;
newNode->nextNode = head;
} else{
Node * copyOfHead = head;
while(copyOfHead->nextNode != head)
{
copyOfHead = copyOfHead->nextNode;
}
copyOfHead->nextNode = newNode;// process last node
newNode->nextNode = head;
}
}
void print()
{
Node * copyOfHead = head;
do
{
std::cout<<copyOfHead->data;
copyOfHead = copyOfHead->nextNode;
}while(copyOfHead != head);
}
public:
static void Test() {
ForwardCircularLinkedList list;
list.AddItem("Hello");
list.AddItem(" ");
list.AddItem("World");
list.AddItem("!");
list.print();
}
};
}
here, a do-while is being used to print the elements of the list.
At the current setup, Can I use a while or for loop to print the list?
Note: I am considering do-while and while as different looping structures.
Yes, you can use do-while or a for loop.
But do-while is more natural because it checks the condition after the body of code.
You have a circular data structure and (presumably) you want to print each element once.
Doing only one round of circling.
do{...move circulator}while(compare with head) has the right logic.
CGAL implements "circulators" and does exactly that, it starts from "head" does something and increments the circulator until it is head once again.
See https://doc.cgal.org/latest/Circulator/classCirculator.html (scroll to Example).
Note the example also checks for emptyness at start, but probably you want.
(In my mind a circular buffer is never empty, but I accept other opinions.)
With while you have:
Node * copyOfHead = head;
do
{
std::cout<<copyOfHead->data;
copyOfHead = copyOfHead->nextNode;
}while(copyOfHead != head);
With for you can have
Node * copyOfHead = head;
for(;;){
std::cout<<copyOfHead->data;
copyOfHead = copyOfHead->nextNode;
if(copyOfHead == head) break;
}
or
for(Node * copyOfHead = head;;){
std::cout<<copyOfHead->data;
copyOfHead = copyOfHead->nextNode;
if(copyOfHead == head) break;
}
or
for(Node * copyOfHead = head; ; copyOfHead = copyOfHead->nextNode){
std::cout<<copyOfHead->data;
if(copyOfHead->nextNode == head) break;
}
or (exploiting that the body of the loop evaluates to bool:true)
for(
Node * copyOfHead = head;
std::cout<<copyOfHead->data;
copyOfHead = copyOfHead->nextNode
) if(copyOfHead->nextNode == head) break;
The main advantage of for is the initialization, but still it is not worth it.
You can of course do the step outside the loop but then you have repeated code, etc.
(NOT RECOMMENDED, it may even have a bug)
Node * copyOfHead = head;
std::cout<<copyOfHead->data;
copyOfHead = copyOfHead->nextNode;
for(; copyOfHead != head ;copyOfHead = copyOfHead->nextNode){
std::cout<<copyOfHead->data;
}
So, there you have, do-while is exactly what you want for this kind of data structure! and for (or while-only) is exactly what you don't want.
I'm trying to write an insert function for string values for a circular doubly linked list. I saw that creating a dummy node is beneficial in doing this so I can eliminate special cases like when the list is empty. The problem is I'm not finding alot of good information on dummy head nodes. I understand their purpose, but I don't understand how I create/implement it.
appreciate all the code examples guys, tried to figure it out on my own getting a little stuck though if someone can look at it.
#include <iostream>
#include <string>
using namespace std;
typedef string ListItemType;
struct node {
node * next;
node * prev;
ListItemType value;
};
node * head;
node * dummyHead = new node;
void insert(const ListItemType input, node * & within);
void main(){
insert("bob",dummyHead);
}
void insert( const ListItemType input, node * &ListHead){
node *newPtr = new node;
node *curr;
newPtr->value = input;
curr = ListHead->next; //point to first node;
while (curr != ListHead && input < curr->value){
curr = curr->next;
}
//insert the new node pointed to by the newPTr before
// the node pointed to by curr
newPtr->next = curr;
newPtr->prev = curr->prev;
curr->prev = newPtr;
newPtr->prev->next = newPtr;
}
For a circular doubly linked list, you can setup 1 sentinel node where both "next" and "prev" points to itself when list is empty. When list is not empty, sentinel->next points to first element and sentinel->prev points to last element. With this knowledge, your insert and remove function would look something like this.
This is very basic and your LinkedList and Node class maybe implemented differently. That is OK. The main thing is the insert() and remove() function implementation that shows how sentinel node(s) removes the need for checking for NULL values.
Hope this helps.
class DoublyLinkedList
{
Node *sentinel;
int size = 0;
public DoublyLinkedList() {
sentinel = new Node(null);
}
// Insert to the end of the list
public void insert(Node *node) {
// being the last node, point next to sentinel
node->next = sentinel;
// previous would be whatever sentinel->prev is pointing previously
node->prev = sentinel->prev;
// setup previous node->next to point to newly inserted node
node->prev->next = node;
// sentinel previous points to new current last node
sentinel->prev = node;
size++;
}
public Node* remove(int index) {
if(index<0 || index>=size) throw new NoSuchElementException();
Node *retval = sentinel->next;
while(index!=0) {
retval=retval->next;
index--;
}
retval->prev->next = retval->next;
retval->next->prev = retval->prev;
size--;
return retval;
}
}
class Node
{
friend class DoublyLinkedList;
string *value;
Node *next;
Node *prev;
public Node(string *value) {
this->value = value;
next = this;
prev = this;
}
public string* value() { return value; }
}
Why are you trying to use dummy node?
I hope you can handle it without a dummy node.
Eg:
void AddNode(Node node)
{
if(ptrHead == NULL)
{
ptrHead = node;
}else
{
Node* itr = ptrHead;
for(int i=1; i<listSize; i++)
{
itr = itr->next;
}
itr->next = node;
}
listSize++;
}
The above one is an example to handle the linked list without dummy node.
For a circular double linked list without a dummy node, the first node previous pointer points to the last node, and the last node next pointer points to the first node. The list itself has a head pointer to first node and optionally a tail pointer to last node and/or a count.
With a dummy node, the first node previous pointer points to the dummy node and the last node next pointer points to the dummy node. The dummy nodes pointers can point to the dummy node itself or be null.
The HP / Microsoft STL list function uses a dummy node as a list head node with the next pointer used as a head pointer to the first real node, and the previous pointer used as a tail pointer to the last real node.
#include <iostream>
#include <string>
using namespace std;
typedef string ElementType;
struct Node
{
Node(){}
Node(ElementType element, Node* prev = NULL, Node* next = NULL):element(element){}
ElementType element;
Node* prev;
Node* next;
};
class LinkList
{
public:
LinkList()
{
head = tail = dummyHead = new Node("Dummy Head", NULL, NULL);
dummyHead->next = dummyHead;
dummyHead->prev = dummyHead;
numberOfElement = 0;
}
void insert(ElementType element)
{
Node* temp = new Node(element, NULL, NULL);
if (0 == numberOfElement)
{
head = tail = temp;
head->prev = dummyHead;
dummyHead->next = head;
tail->next = dummyHead;
dummyHead->prev = tail;
}
else
{
tail->next = temp;
temp->prev = dummyHead->next;
temp->next = dummyHead;
dummyHead->next = temp;
tail = temp;
}
numberOfElement += 1;
}
int length() const { return numberOfElement; }
bool empty() const { return head == dummyHead; }
friend ostream& operator<< (ostream& out, const LinkList& List);
private:
Node* head;
Node* tail;
Node* dummyHead;
int numberOfElement;
};
ostream& operator<< (ostream& out, const LinkList& List)
{
Node* current = List.head;
while (current != List.dummyHead)
{
out<<current->element<<" ";
current = current->next;
}
out<<endl;
return out;
}
int main()
{
string arr[] = {"one", "two", "three", "four", "five"};
LinkList list;
int len = sizeof(arr) / sizeof(arr[0]);
for (int i = 0; i < len; ++i)
{
list.insert(arr[i]);
}
cout<<list<<endl;
}
I think this code can help you. When you want to implement some data structure, you must have a clear blueprint about it.
Do the following inside the constructor
ptrHead = new Node();
listSize = 1;
if you have tail also,
ptrHead->next = ptrTail;
The above code will create dummy node.
Make sure you implementation should not affected by this dummy node.
eg:
int getSize()
{
return listSize-1;
}
Here is my code to delete all the nodes having the value passed in the argument.
typedef struct nodetype
{
int data;
struct nodetype * next;
} node;
typedef node * list;
void Linklist::deleteNode(list * head, int value)
{
list current = *head;
list previous = *head;
while(current != NULL)
{
if(current->data != value)
{
previous = current;
current = current->next;
}
else if (current->data == value)
{
previous->next = current->next;
delete current;
current = previous->next;
}
}
}
But here if all the elements in the linklist is say 2, then it should delete all the elements in the linklist and finally head should also become NULL so that if I pass this head to count the number of nodes in the list it should say that the list is empty and other similar operations.
According to my current implementation the head is not becoming NULL for the above mentioned case.
Please suggest the modification so that head should become NULL if the linklist has all the nodes with the same value passed in the function argument.
I modified my code as follows and its working file now
void Linklist::deleteNode(list *head, int value)
{
list * current = head;
list * previous = head;
bool flag = false;
while(*current != NULL)
{
if((*current)->data != value)
{
*previous = *current;
*current = (*current)->next;
}
else if ((*current)->data == value)
{
flag = true;
(*previous)->next = (*current)->next;
delete *current;
*current = (*previous)->next;
}
}
if(!flag)
cout<<"Element not found in the linklist\n";
cout<<"Count is "<<Linklist::count(*head)<<endl;
}
I've been trying to figure out how to reverse the order of a doubly-linked list, but for some reason, in my function void reverse() runs while loop once and then crashes for some reason. To answer some questions ahead, I'm self-teaching myself with my brothers help. This isn't all of the code, but I have a display() function which prints all nodes chronologically from start_ptr and a switch which activates certain functions like
case 1 : add_end(); break;
case 2 : add_begin(); break;
case 3 : add_index(); break;
case 4 : del_end(); break;
case 5 : del_begin(); break;
case 6 : reverse(); break;
This is the geist of my code:
#include <iostream>
using namespace std;
struct node
{
char name[20];
char profession[20];
int age;
node *nxt;
node *prv;
};
node *start_ptr = NULL;
void pswap (node *pa, node *pb)
{
node temp = *pa;
*pa = *pb;
*pb = temp;
return;
}
void reverse()
{
if(start_ptr==NULL)
{
cout << "Can't do anything" << endl;
}
else if(start_ptr->nxt==NULL)
{
return;
}
else
{
node *current = start_ptr;
node *nextone = start_ptr;
nextone=nextone->nxt->nxt;
current=current->nxt;
start_ptr->prv=start_ptr->nxt;
start_ptr->nxt=NULL;
//nextone=nextone->nxt;
while(nextone->nxt!= NULL)
{
pswap(current->nxt, current->prv);
current=nextone;
nextone=nextone->nxt;
}
start_ptr=nextone;
}
}
Try this:
node *ptr = start_ptr;
while (ptr != NULL) {
node *tmp = ptr->nxt;
ptr->nxt = ptr->prv;
ptr->prv = tmp;
if (tmp == NULL) {
end_ptr = start_ptr;
start_ptr = ptr;
}
ptr = tmp;
}
EDIT: My first implementation, which was correct but not perfect.
Your implementation is pretty complicated. Can you try this instead:
node * reverse(Node * start_ptr)
{
Node *curr = start_ptr;
Node * prev = null;
Node * next = null;
while(curr)
{
next = curr->nxt;
curr->nxt = prev;
curr->prv = next;
prev = curr;
curr = next;
}
return start_ptr=prev;
}
Here is my updated solution:
node * reverse()
{
node *curr = start_ptr;
node * prev = NULL;
node * next = NULL;
while(curr)
{
next = curr->nxt;
curr->nxt = prev;
curr->prv = next;
prev = curr;
curr = next;
}
return start_ptr=prev;
}
The logic was correct. But the issue was that I was accepting in input argument start_ptr. Which means that I was returning the local copy of it. Now it should be working.
You can simplify your reverse() quite a bit. I'd do something like this:
void reverse()
{
if(start_ptr == NULL)
{
cout << "Can't do anything" << endl;
}
else
{
node *curr = start_ptr;
while(curr != NULL)
{
Node *next = curr->next;
curr->next = curr->prev;
curr->prev = next;
curr = next;
}
start_ptr = prev;
}
}
Explanation: The basic idea is simply to visit each Node and swap the links to previous and next. When we move curr to the next Node, we need to store the next node so we still have a pointer to it when we set curr.next to prev.
Simple solution. reverses in less than half a number of total iterations over the list
template<typename E> void DLinkedList<E>::reverse() {
int median = 0;
int listSize = size();
int counter = 0;
if (listSize == 1)
return;
DNode<E>* tempNode = new DNode<E>();
/**
* A temporary node for swapping a node and its reflection node
*/
DNode<E>* dummyNode = new DNode<E>();
DNode<E>* headCursor = head;
DNode<E>* tailCursor = tail;
for (int i = 0; i < listSize / 2; i++) {
cout << i << "\t";
headCursor = headCursor->next;
tailCursor = tailCursor->prev;
DNode<E>* curNode = headCursor;
DNode<E>* reflectionNode = tailCursor;
if (listSize % 2 == 0 && listSize / 2 - 1 == i) {
/**
* insert a dummy node for reflection
* for even sized lists
*/
curNode->next = dummyNode;
dummyNode->prev = curNode;
reflectionNode->prev = dummyNode;
dummyNode->next = reflectionNode;
}
/**
* swap the connections from previous and
* next nodes for current and reflection nodes
*/
curNode->prev->next = curNode->next->prev = reflectionNode;
reflectionNode->prev->next = reflectionNode->next->prev = curNode;
/**
* swapping of the nodes
*/
tempNode->prev = curNode->prev;
tempNode->next = curNode->next;
curNode->next = reflectionNode->next;
curNode->prev = reflectionNode->prev;
reflectionNode->prev = tempNode->prev;
reflectionNode->next = tempNode->next;
if (listSize % 2 == 0 && listSize / 2 - 1 == i) {
/**
* remove a dummy node for reflection
* for even sized lists
*/
reflectionNode->next = curNode;
curNode->prev = reflectionNode;
}
/**
* Reassign the cursors to position over the recently swapped nodes
*/
tailCursor = curNode;
headCursor = reflectionNode;
}
delete tempNode, dummyNode;
}
template<typename E> int DLinkedList<E>::size() {
int count = 0;
DNode<E>* iterator = head;
while (iterator->next != tail) {
count++;
iterator = iterator->next;
}
return count;
}
I suggest maintaining a link to the last node.
If not, find the last node.
Traverse the list using the "previous" links (or in your case, prv).
There is no need to actually change the links around. Traversing using the prv pointer will automatically visit the nodes in reverse order.
Look at
valuesnextone=nextone->nxt->nxt;
Here nextone->nxt can be null.
Apart from that, try to use pointers to pointers in the swap function.
Your pswap function is wrong
your should swap the pointer not try to create temporary objects and swap them.
Should be like that (there might be other mistake later)
void pswap (node *&pa, node *&pb)
{
node* temp = pa;
pa = pb;
pb = temp;
return;
}
A very simple and O(n) solution using two pointers:
start = head of the doubly LL
struct node *temp, *s;
s = start;
while(s != NULL){
temp = s->prev;
s->prev = s->next;
s->next = temp;
s = s->prev;
}
//if list has more than one node
if(current != NULL){
start = temp->prev;
}
My code for reversing doubly linked list,
Node* Reverse(Node* head)
{
// Complete this function
// Do not write the main method.
if(head != NULL) {
Node* curr = head;
Node* lastsetNode = curr;
while(curr != NULL) {
Node* frwdNode = curr->next;
Node* prevNode = curr->prev;
if(curr==head) {
curr->next = NULL;
curr->prev = frwdNode;
lastsetNode = curr;
}
else {
curr->next = lastsetNode;
curr->prev = frwdNode;
lastsetNode = curr;
}
curr = frwdNode;
}
head = lastsetNode;
}
return head;
}
I thought I'd add a recursive solution here.
node* reverse_and_get_new_head(node* head) {
if (head == nullptr) { return nullptr; }
// This can be avoided by ensuring the initial,
// outer call is with a non-empty list
std::swap(head->prev, head->next);
if (head->prev == nullptr) { return head; }
return reverse_and_get_new_head(head->prev);
}
void reverse() {
start_ptr = reverse_and_get_new_head(start_ptr);
}