What I am trying to do is check if node after value is specific number and if is, then delete previoues node.
Something like:
1,2,3,4,5,4
if next node number is 4 then delete this node.
1,2,3,4,5,4 -> 1,2,4,4
node* temp = head;
while (head != NULL) {
if (head->next->number == 4) {
delete temp;
}
head = head->next;
}
Struggling at this moment as compiler crashes.
You delete the head before you advance it:
node *temp = NULL;
while (head->next != NULL) {
if (head->next->number == 4) {
temp = head;
}
if ( temp == NULL ){
head = head->next;
}
else{
head = head->next->next;
delete temp;
temp= NULL:
}
}
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);
}
}
I am trying to delete a node from a linked list using this function:
void del_node(int del_data)
{
node* temp = NULL;
node* trail = NULL;
node* del_ptr = NULL;
temp = head;
trail = head;
while (temp != NULL && temp->data != del_data)
{
trail = temp;
temp = temp->next;
}
if (temp != NULL) {
del_ptr = temp;
temp = temp->next;
trail->next = temp;
delete(del_ptr);
}
}
It seems like it deletes it fine until i print the linked list using this:
void print()
{
node* temp = NULL;
temp = head;
while (temp != NULL)
{
cout << temp->data << " ";
temp = temp->next;
}
cout << endl;
}
and it starts outputting seemingly random numbers, can anybody help me with this, really confused as this code comes from a tutorial.
Your algorithm doesn't manage the head pointer correctly whatsoever. Any changes that ultimately should modify the head pointer don't, and that's a huge problem. A pointer to pointer algorithm not only solves this problem, it also delivers a considerably more succinct solution:
void del_node(int del_data)
{
struct node **pp = &head;
while (*pp && (*pp)->data != del_data)
pp = &(*pp)->next;
if (*pp)
{
node *tmp = *pp;
*pp = tmp->next;
delete tmp;
}
}
This will work for any list condition including:
An empty list. i.e. head is null.
A single-node list. If the value matches head->data it will properly delete and reset the node pointer.
A multi-node list. The first matching node will be removed, and it will properly fix up the head node pointer if that was the matching location.
All of the above, in cases where there is no matching node, the list remains unchanged.
Fulfilling all of that in such a short algorithm + implementation is beneficial.
I'll comment on your code inline:
void del_node(int del_data)
{
node* temp = NULL;
node* trail = NULL;
node* del_ptr = NULL;
temp = head;
trail = head;
// This is fine, but recommend you use nullptr instead of NULL.
// This will find the first instance of data matches del_data,
// But if you are trying to delete all instances of del_data,
// You'll need to do this a little differently.
while (temp != NULL && temp->data != del_data)
{
trail = temp;
temp = temp->next;
}
// This if is fine, but see previous comment about using nullptr
// instead of NULL.
if (temp != NULL) {
del_ptr = temp;
temp = temp->next;
// Problematic: What if trail is null?
trail->next = temp;
delete(del_ptr);
}
}
Your code isn't bad. I wouldn't have written exactly like this, but I'm going to replace your if-statement:
if (temp != nullptr) {
// If trail is nullptr, then we're deleting from the head
if (trail == nullptr) {
head = temp->next;
}
else {
trail->next = temp->next;
}
delete(temp);
}
There's no need for the temporary. Just point around temp as you see in the if-else block and then delete temp.
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 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;
}