I'm trying to delete a node in a doubly linked list, but the problem here is that when I have the list: 7-6-5-4-3-2-1, and try to delete 3, I get 7-6-5-4-3, I can't figure out the error. Please help!
void RemoveItem(int itm)
{
if (Head->item == itm) {
Head = Head -> Next;
cout<<"\nItem is at 1st node & removed\n";flag=1;
}
else if (Head->Next == NULL) {
cout<<" \n This is 1-node list & item not in it\n";
}
else {
node *current, *del = Head;
while(del->Next != NULL && del->item != itm) {
del = del->Next;
}
current = del;
current->Next = del->Next;
del->Next->previous = current;
delete(del);
del->Next = NULL;
flag = 1;
}
}
To delete a node you iterate until you reach the node of deletion del then set:
pseudocode:
del.prev.next = del.next
and
del.next.prev = del.prev
This makes no sense:
current = del;
current->Next = del->Next;
because it is the same as
del->Next = del->Next;
Try something like this
// some global int flag = 0
// some global node* Head pointing to head of list
void RemoveItem(int itm) {
if (Head->item == itm) {
node* del = Head;
Head = Head -> Next;
delete(del);
cout<<"\nItem is at 1st node & removed\n";
flag=1;
return;
}
if (Head->Next == NULL) {
cout<<" \n This is 1-node list & item not in it\n";
return;
}
node* current = Head;
node* del = Head;
while(del->Next != NULL && del->item != itm) {
del = del->Next;
}
// reached with del->Next == NULL || del->item == itm
current = del->previous;
if (del->Next == NULL) {
// at end of list
if (del->item != item) {
// item not found
// todo: signal error
return;
}
current->Next = NULL;
} else {
del->Next->previous = current;
current->Next = del->Next;
}
delete(del);
flag = 1;
}
Note: I fixed many bugs and tried to keep it within your logic. I would have written it different.
Related
void deletenode(string key) {
if (last == NULL) {
cout << "your circular linked list is an empty one" << endl;
}
else {
node* p = last->next;
node* prev = last;
do {
if (p->title == key) {
node* temp = p;
prev->next = p->next;
delete(temp);
}
else {
p = p->next;
prev = prev->next;
}
} while (p != last->next);
}}
I was trying to delete a node with key value. For instance, if node p->title is my key then I want to delete that node. However, I implemented it with other values but the code doesn't seem to work or delete a node with key value from my circular linked list. What is the mistake in the function?
circular linked list value "cat", "dog", "rat", "horse", the key to be deleted was "dog". When I traverse throughout the linked list after the deletion it still printed everything including "dog", which means deletion didn't work.
Anytime you write a "delete from the linked list" function, you have to account for the possibility that you are deleting from the "head" or whatever pointer you are referencing with the list. The common pattern is for the function to return the new head of the list if it changed, else return the current head.
Node* deletenode(Node* head, const string& key) {
// empty list
if (head == nullptr) {
return nullptr;
}
// single item list
if (head->next == nullptr || head->next == head) {
if (head->title == key) {
delete head;
head = nullptr;
}
return head;
}
// two or more item list, possibly circular
Node* prev = head;
Node* current = head->next;
Node* first = current;
while (current && current->title != key) {
prev = current;
current = current->next;
if (current == first) {
break;
}
}
if (current == nullptr || current->title != key) {
return head; // not found
}
prev->next = current->next;
if (current == head) {
head = current->next;
}
delete current;
return head;
}
I don't see the full code so I can't make a comment I tried to implement the function, hope it helps you with the comments.
void deleteNodeWithKey(node* head, string key)
{
node *curr = head;
node *last , *temp;
//Search for last node
while (curr->next != head)
{
curr = curr->next;
}
last = curr;
//If head is the desired key, make head's next new head
//and connect last node to new head
if (head->key == key)
{
temp = head->next;
delete head;
head = temp;
last->next = head;
return;
}
temp = head->next;
//Search for node with the given key
node *prev = head;
while (temp != head)
{
if (temp->key == key)
{
prev->next = temp->next;
delete temp;
return;
}
temp = temp->next;
prev = prev->next;
}
//If function gets here, key was not found
}
I made some changes to your code
void deletenode(string key) {
if (last == NULL) {
cout << "your circular linked list is an empty one" << endl;
}
else {
node* prev = last;
// If head is to be deleted
if (last->title == key) {
while (prev->next != last)
prev = (prev)->next;
prev->next = last->next;
free(last);
last = prev->next;
return;
}
node* p = last->next;
do {
if (p->next->title == key) {
node* temp = p->next;
p->next = temp->next;
delete(temp);
}
else {
p = p->next;
prev = prev->next;
}
} while (p != last->next);
}
}
void RemoveDuplicates(Slist& l)
{
if (l.head == NULL) {
return;
}
Node* cur = l.head;
while (cur != NULL && cur->next != NULL) {
Node* prev = cur;
Node* temp = cur->next;
while (temp != NULL) {
if (temp->data == cur->data) {
prev->next = temp->next;
cur->next = prev->next;
temp = prev->next;
}
else {
prev = prev->next;
temp = temp->next;
}
}
cur = cur->next;
}
}
Hi, I want to remove duplicates from linked list (0 is NULL)
input: 1->2->2->4->2->6->0
outPut: 1->2->4->6->0
Result after I run my program is:
1->2->6
Where am I wrong? Please help me
Here is my solution for your problem:
bool alreadyExist(Node head)
{
Node cur = head;
while(cur.next != nullptr)
{
if(cur.next->data == head.data) {
return true;
}
cur = *cur.next;
}
return false;
}
void RemoveDuplicates(Slist& l)
{
if (l.head == nullptr) {
return;
}
Node* head = l.head;
Node* curPtr = l.head->next;
while(curPtr != nullptr)
{
if(alreadyExist(*curPtr) == false)
{
head->next = curPtr;
head->next->prev = head;
head = head->next;
curPtr = curPtr->next;
}
else
{
Node* backup = curPtr;
curPtr = curPtr->next;
// delete duplicate elements from the heap,
// if the nodes were allocated with new, malloc or something else
// to avoid memory leak. Remove this, if no memory was allocated
delete backup;
}
}
}
Important: The destructor of the Node-object is NOT allowed to delete the linked object behind the next and prev pointer.
It results, for your input-example, in the output 1->4->2->6->0. Its not totally exact the order, you want as output, but each number exist only one time within the output. It only add the last time of a duplicate number.
I don't really know, if you use C or C++, but because I prefer C++, I replaced the NULL with nullptr in the code. The delete can be removed, if the objects are not on the HEAP create with malloc or new.
Here using the function returnReverseLinkedList I am returning the reversed linked list of the given linked list. But the problem with this approach is that i lose the original linked list. So I make another fucntion called createReversedLinkedList to make a copy of the original linked list and reverse the copy and maintain possession of both.
unfortunately createReversedLinkedList is giving Runtime error.
obviously my end goal is to check if the given linked list is palindrome or not. This issue is just a stepping stone.
Could someone tell me why?
//Check if a linked list is a palindrome
#include <iostream>
using namespace std;
class node
{
public:
int data;
node *next;
node(int data)
{
this->data = data;
this->next = NULL;
}
};
node *returnReverseLinkedList(node *head)
{
// Will Lose original Linked List
if (head == NULL)
return NULL;
else if (head != NULL && head->next == NULL)
return head;
node *prev = NULL;
node *curr = head;
node *tempNext = head->next;
while (tempNext != NULL)
{
curr->next = prev;
prev = curr;
curr = tempNext;
tempNext = tempNext->next;
}
curr->next = prev;
return curr;
}
node *createReversedLinkedList(node *head)
{
if (head == NULL)
return NULL;
else if (head != NULL && head->next == NULL)
return NULL;
else
{
node *temp = head;
node *newHead = NULL;
node *newTail = NULL;
while (temp != NULL)
{
node *newNode = new node(temp->data);
if (newHead == NULL)
{
newHead = newNode;
newTail = newNode;
}
else
{
newTail->next = newNode;
newTail = newNode;
}
}
return returnReverseLinkedList(newHead);
}
}
bool check_palindrome(node *head)
{
node *original = head;
node *reverse = returnReverseLinkedList(head);
while (original->next != NULL || reverse->next != NULL)
{
if (original->data != reverse->data)
return false;
cout << "debug 2" << endl;
original = original->next;
reverse = reverse->next;
}
return true;
}
// #include "solution.h"
node *takeinput()
{
int data;
cin >> data;
node *head = NULL, *tail = NULL;
while (data != -1)
{
node *newnode = new node(data);
if (head == NULL)
{
head = newnode;
tail = newnode;
}
else
{
tail->next = newnode;
tail = newnode;
}
cin >> data;
}
return head;
}
void print(node *head)
{
node *temp = head;
while (temp != NULL)
{
cout << temp->data << " ";
temp = temp->next;
}
cout << endl;
}
int main()
{
node *head = takeinput();
node *revese2 = createReversedLinkedList(head);
print(revese2);
// bool ans = check_palindrome(head);
// if (ans)
// cout << "true";
// else
// cout << "false";
// return 0;
}
As asked by the OP, building a reversed linked is simply done by building as you would a stack (e.g LIFO) rather than duplicating the same original forward chain. For example:
node *createReversedLinkedList(const node *head)
{
node *newHead = NULL;
for (; head; head = head->next)
{
node *p = new node(head->data)
p->next = newHead;
newHead = p;
}
return newHead;
}
Note we're not hanging our copied nodes on the tail of the new list; they're hanging on the head of the new list, and becoming the new head with each addition. That's it. There is no need to craft an identical list, then reverse it; you can reverse it while building the copy to begin with.
A note on the remainder of your code. You have a dreadful memory leak, even if you fix the reversal generation as I've shown above. In your check_palindrome function, you never free the dynamic reversed copy (and in fact, you can't because you discard the original pointer referring to its head after the first traversal:
bool check_palindrome(node *head)
{
node *original = head;
node *reverse = returnReverseLinkedList(head); // only reference to reversed copy
while (original->next != NULL || reverse->next != NULL)
{
if (original->data != reverse->data)
return false; // completely leaked entire reversed copy
original = original->next;
reverse = reverse->next; // lost original list head
}
return true;
}
The most obvious method for combating that dreadful leak is to remember the original list and use a different pointer to iterate, and don't leave the function until the copy is freed.
bool check_palindrome(const node *head)
{
bool result = true;
node *reverse = returnReverseLinkedList(head);
for (node *p = reverse; p; p = p->next, head = head->next)
{
if (p->data != head->data)
{
result = false;
break;
}
}
while (reverse)
{
node *tmp = reverse;
reverse = reverse->next;
delete tmp;
}
return result;
}
I have been stuck on this function to remove node from a list, if there are two names in the list they are both gone. If Anne and John are in the list and I want to delete Anne, then my list is empty, John is gone.
What am I missing to keep the connection in the list if I delete a node init?
bool ContactList::remove(string key)
{
NodePtr prev = NULL;
for(NodePtr temp = head; temp != NULL; temp = temp->link)
{
if(temp->data.key == key)
{
if(prev == NULL)
{
head = temp->link;
delete temp;
return true;
}
else
{
prev = temp->link;
delete temp;
return true;
}
}
}
return false;
}
You aren't keeping prev up to date in every iteration of your loop. You want something like:
prev = temp;
at the bottom of your for loop.
Try using this function
bool ContactList::remove(string key)
{
NodePtr prev = NULL;
for(NodePtr temp = head; temp != NULL; temp = temp->link)
{
if(temp->data.key == key)
{
if(prev == NULL)
{
head = temp->link;
delete temp;
return true;
}
else
{
prev->link = temp->link; // change.
delete temp;
return true;
}
}
prev = temp; // change.
}
return false;
}
This is my code for an insertion-sort, sorting via the cstring member of the node. The current code only inserts before the head. The code encapsulated in comments is the sorted insertion I am trying to make work. I'm thinking I have to use a predecessor and successor pointer, but maybe it's the comparison that is confusing me. Any help would be appreciated. Thanks!
#include <iostream>
#include "groclist.h"
void insert_item(Grocery_Item_Ptr &head, int quantity, const char name[])
{
bool exists = false;
bool done = false;
Grocery_Item_Ptr temp = NULL;
Grocery_Item_Ptr current = NULL;
Grocery_Item_Ptr pred = NULL;
Grocery_Item_Ptr succ = NULL;
if (head == NULL) {
head = new Grocery_Item;
head->quantity = quantity;
strncpy(head->name, name, MAX_ITEM_NAME_LEN);
head->next = NULL;
return;
}
else {
current = head;
while (current != NULL) {
if (strncmp(current->name, name, MAX_ITEM_NAME_LEN) == 0) {
current->quantity += quantity;
exists = true;
}
current = current->next;
}
if (exists) {
current = NULL;
return;
}
else {
current = head;
}
if (!exists) {
temp = new Grocery_Item;
temp->quantity = quantity;
strncpy(temp->name, name, MAX_ITEM_NAME_LEN);
/*
while (!done || current != NULL) {
if (strncmp(current->name, name, MAX_ITEM_NAME_LEN) < 0) {
pred = current;
succ = current->next;
current->next = temp;
temp->next = succ;
done = true;
}
if (!done) {
current = current->next;
}
}
*/
temp->next = head;
head = temp;
temp = NULL;
}
}
return;
}
One thing you are missing is keeping a reference to the predecessor while searching. This is necessary in order to keep the chain intact.
Here is a draft that should work (currently untested!):
while (!done || current != NULL)
{
//If we found the place to insert
if (strncmp(current->name, name, MAX_ITEM_NAME_LEN) < 0)
{
//If the place to insert was at head
if(pred == NULL)
{
//The new node becomes the head
head = temp;
}
else
{
//Set the previous nodes next to point at this node.
pred->next = temp;
}
//Always set the node to be inserted's next
//pointing to the node we should be inserted before
temp->next = current;
done = true;
}
if (!done)
{
//No match, keep looking but keep an updated pred pointer
pred = current;
current = current->next;
}
}
It's just pseudocode, but maybe it helps:
if(head == NULl)
{
//make newnode as head
}
if(head.name == new name)
{
//update quantity
}
if(head.name <new name)
{
//insert newnode before head
//make newnode as head
}
if (new name > head.name)
{
current = head;
succ = current.next;
while (succ && new name <succ.name)
{
curent = succ;
succ = succ.next
}
if(succ = NULL)
current->next = newnode
else
if new name = succ->name
update quantity
else
curent->next = newnode
newnode->next = succ;
}
I appreciate your input guys, if only for having me think about it in different ways. However my result is quite different, and I must truly thank my whiteboard.
if (strncmp(name, head->name, MAX_ITEM_NAME_LEN) < 0) {
temp->next = head;
head = temp;
temp = NULL;
}
else {
pred = head;
current = head->next;
do {
if (strncmp(name, current->name, MAX_ITEM_NAME_LEN) < 0) {
pred->next = temp;
temp->next = current;
done = true;
}
else if (current->next == NULL) {
current->next = temp;
done = true;
}
else {
pred = current;
current = current->next;
}
if (done) {
pred = NULL;
current = NULL;
}
} while (!done && current != NULL);
}