I need to define a class of linked list,List, in a way such that object of class can be defined in two ways,
List obj1 = L1();//head=0
List obj2 = L2(given_arr[], size of array) // I would be given an array, whose elements are elements of list
so, I need to form a construter for both,
for obj1, Its easy.
List(){head=0};
But I am not abe to do so for second type of object.
I tried to form a program for this.
#include <iostream>
using namespace std;
class List {
class node {
public:
int val;
node* next;
};
public:
node* head;
int arr[];
List() { head = 0; }
List(int arr[], int size);
void addnode(int value) {
node* newnode = new node();
newnode->val = value;
newnode->next = NULL;
if (head == NULL) {
head = newnode;
} else {
node* temp = head; // head is not NULL
while (temp->next != NULL) {
temp = temp->next; // go to end of list
}
temp->next = newnode; // linking to newnode
}
}
void display() {
if (head == NULL) {
cout << "List is empty!" << endl;
} else {
node* temp = head;
while (temp != NULL) {
cout << temp->val << " ";
temp = temp->next;
}
cout << endl;
}
}
};
List::List(int arr[], int size) {
int i;
head->val = arr[0];
for (i = 0; i < size; i++) addnode(arr[i]);
}
int main() {
int barr[4] = {9, 89, 0, 43};
List* M = new List();
List* L = new List(barr[4], 4);
L->display();
return 0;
}
This program doesn't work. Please suggest a way to do so.
Make these changes to your main().
int main() {
int barr[] = {9, 89, 0, 43}; // No need to specify size if you're initializing
// List* M = new List(); // unused
// Your array is barr, barr[4] makes no sense. You also don't allocate the List,
// the list allocates
List L = List(barr, sizeof(barr) / sizeof(barr[0]);
L.display(); // -> to .
return 0;
}
This now compiles, but immediately segfaults. Simply running the program in the debugger shows a simple error. The line head->val = arr[0]; attempts to dereference a null pointer. Which takes us to the next thing. Use nullptr, not NULL or 0.
Your array constructor was over-complicated, you just need this:
List::List(int arr[], int size) {
for (int i = 0; i < size; i++) addnode(arr[i]);
}
Your addnode() function already handled an empty list. Fixing that, your code should run. I made a couple other small changes, mostly trimming cruft out. Here's your complete code:
#include <iostream>
using namespace std;
class List {
class node {
public:
int val;
node* next;
};
public:
node* head = nullptr;
List() = default;
List(int arr[], int size);
void addnode(int value) {
node* newnode = new node();
newnode->val = value;
newnode->next = NULL;
if (head == NULL) {
head = newnode;
} else {
node* temp = head; // head is not NULL
while (temp->next != NULL) {
temp = temp->next; // go to end of list
}
temp->next = newnode; // linking to newnode
}
}
void display() {
if (head == NULL) {
cout << "List is empty!" << endl;
} else {
node* temp = head;
while (temp != NULL) {
cout << temp->val << " ";
temp = temp->next;
}
cout << endl;
}
}
};
List::List(int arr[], int size) {
for (int i = 0; i < size; i++) addnode(arr[i]);
}
int main() {
int barr[] = {9, 89, 0, 43};
List L = List(barr, sizeof(barr) / sizeof(barr[0]));
L.display();
return 0;
}
Related
We are iterating through the linked list with the help of head, that is, we are updating our head as we move forward towards i th position. Please have a look at the fuction insertIthnode. I am inserting my Node at i th position are returning head - and it's still able to print the linked list. I don't know how? head is no longer pointing towards the first node then how is it still able to return a full linked list?
here's the code:
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int data) {
this->data = data;
next = NULL;
}
};
int length(Node *head) {
int x = 0;
Node *temp = head;
while (temp != NULL) {
x += 1;
temp = temp->next;
}
return x;
}
void printIthnode(Node *head, int i) {
int n = length(head);
if (i < 0 || i > n - 1) {
cout << -1 << endl;
return;
}
int count = 1;
while (count <= i) {
head = head->next;
count++;
}
if (head) {
cout << head->data << endl;
} else {
cout << "-1" << endl;
}
}
Node *takeinput() {
int data;
cin >> data;
Node *head = NULL;
Node *tail = NULL;
while (data != -1) {
Node *n = new Node(data);
if (head == NULL) {
head = n;
tail = n;
} else {
tail->next = n;
tail = n;
}
cin >> data;
}
return head;
}
void PrintLL(Node *head) {
Node *temp = head;
while (temp != NULL) {
cout << temp->data << " ";
temp = temp->next;
}
}
Node *insertIthnode(Node *head, int i, int data) {
if (i < 0) {
return head;
} else if (i == 0) {
Node *n = new Node(data);
n->next = head;
head = n;
return head;
}
int count = 1;
while (count <= i - 1 && head != NULL) {
head = head->next;
count++;
if (count == i - 1) {
Node *n = new Node(data);
n->next = head->next;
head->next = n;
return head;
}
return head;
}
}
int main() {
/*Node n1(1);
Node *head=&n1;
Node n2(2);
Node n3(3);
Node n4(4);
Node n5(5);
Node n6(6);
n1.next=&n2;
n2.next=&n3;
n3.next=&n4;
n4.next=&n5;
n5.next=&n6;
*/
Node *head = takeinput();
insertIthnode(head, 3, 7);
PrintLL(head);
}
In the main() function you are creating a head when you are taking input from the user with the help of the "takeInput()" function.
After that, you are calling the function "insertIthnode(head,3,7)" which is returning the head (since the return type is Node) but you are not receiving it in any variable so the head returned from the "insetIthnode" is lost.
Your original head remains the same as per of "takeInput()" function.
If you try to insert ith Node at Index 0 it won't print according to the inserted node.
The problem is that you consider the Node as the linked list. While this is valid, the whole point of the linked list is that you don't lose track of the head. You could use two approaches:
Don't iterate over the head. Instead, use a temporary reference to the head.
Implement a Linked List wrapper. You can keep a constant reference to the head while performing operations over the node.
You pass head by value. Any changes you do to the variable receiving the value of head inside the functions are made to the local variable inside the function only and will not be visible from the call site.
Take your PrintLL function as an example:
void PrintLL(Node *head) { // head is here a local variable
Node *temp = head;
while (temp != NULL) {
cout << temp->data << " ";
temp = temp->next;
}
}
This could be rewritten without the extra variable temp. The name head doesn't make it the same head you used to call the function with:
void PrintLL(Node* head) {
while (head != nullptr) {
cout << head->data << ' ';
head = head->next;
}
}
and it would not affect the head you passed in as a parameter.
Similarly:
void foo(int x) {
++x;
//
}
int main() {
int x = 10;
foo(x);
std::cout << x << '\n'; // prints 10
}
I am writing a code to rotate a singly linked list counter-clockwise by K number of nodes. I wrote the following code. For example for the input linked list, 1,2,3,4,5,6,7,8 the rotate function Node *rotate(Node *head, int k){} returns the head pointer to the linked list 5,6,7,8,1,2,3,4. I wrote the following code. In this code, If I call print(head) inside the rotate function then it gives correct output but once it returns the head pointer to main then it either throws SIGSEV or is producing 1,2,3,4.
#include <iostream>
using namespace std;
class List {
public:
int data;
List *next;
explicit List(int element) : data(element), next(nullptr){}
};
List *insert() {
int n, i, value;
List *temp = nullptr, *head = nullptr;
cin >> n;
for(i = 0; i < n; ++i) {
cin >> value;
if(i == 0) {
head = new List(value);
temp = head;
continue;
} else {
temp ->next = new List(value);
temp = temp->next;
}
}
return head;
}
void print(List *start) {
while(start != nullptr) {
cout << start ->data << " ";
start = start->next;
}
}
List* rotate(List* head, int k) {
List *traverse = head, *temp = head;
List *kth, *end;
int i = 0;
while(i < k - 1) {
traverse = traverse ->next;
++i;
}
kth = traverse;
while(traverse->next != nullptr) {
traverse = traverse->next;
}
end = traverse;
head = kth->next;
kth->next = nullptr;
end->next = temp;
print(head);
return head;
}
int main() {
int k;
List *head = insert();
cin >> k;
print(head);
cout << endl;
rotate(head, k);
print(head);
return 0;
}
PS: I am only allowed to change the rotate function.
I'm trying to implement my own version of a linked list for learning. I have the following code. The reverseList function works correctly and if I print it inside that function it is good.
However, when I leave the function and then call the print method I get the the first value and then nothing (null). I'm guessing when I get out of the function it brings me back to the original first ([99]) element which is now actually the last element. So my print method outputs the element sees null is the next and ends.
Or I was thinking the changes I was making in the function were somehow only in that function's scope even though I passed a pointer, but that doesn't make sense because if that's the case then I should have all the original data still.
struct ListNode
{
int value;
ListNode* next = NULL;
};
void insertRecList(ListNode* list, int value)
{
if(list->next == NULL)
{
ListNode* end = new ListNode;
end->value = value;
list->next = end;
}
else
insertRecList(list->next, value);
}
void printList(ListNode* list)
{
std::cout << list->value << std::endl;
while(list->next != NULL)
{
list = list->next;
std::cout << list->value << std::endl;
}
}
void reverseList(ListNode* list)
{
ListNode* next;
ListNode* prev = NULL;
ListNode* cur = list;
while(cur != NULL)
{
if(cur->next == NULL)
{
cur->next = prev;
break;
}
else
{
next = cur->next;
cur->next = prev;
prev = cur;
cur = next;
}
}
list = cur;
std::cout << cur->value << " list:" << list->value << std::endl;
}
void testLinkedList()
{
srand(time(NULL));
ListNode nodes;
nodes.value = 99;
int val;
for(int i = 0; i < 5; i++)
{
val = rand() % 30 + 1;
insertRecList(&nodes, i);
//insertList(&nodes, val);
}
printList(&nodes);
reverseList(&nodes);
printList(&nodes);
}
int main()
{
testLinkedList();
return 0;
}
Appreciative of any help you guys can give me,
Thanks!
Update:
By passing the ListNode *list to reverseList, you create a copy of your pointer which point to the same address with nodes. Inside the function, you assign list to the updated cur pointer but the copy will be destroyed at the end. list still points to the same address as before passing to reverseList but its next has changed.
I have modified your code a little bit:
#include <cstdlib>
#include <iostream>
struct ListNode
{
int value;
ListNode* next = nullptr;
};
void insertRecList(ListNode* list, int value)
{
if(list->next == nullptr)
{
ListNode* end = new ListNode;
end->value = value;
list->next = end;
}
else
insertRecList(list->next, value);
}
void printList(ListNode* list)
{
std::cout << list->value << std::endl;
while(list->next != nullptr)
{
list = list->next;
std::cout << list->value << std::endl;
}
}
void reverseList(ListNode** list)
{
ListNode* cur = *list;
ListNode* next = cur->next;
ListNode* prev = nullptr;
while(cur != nullptr)
{
next = cur->next;
cur->next = prev;
prev = cur;
cur = next;
}
*list = prev;
}
void cleanNodes(ListNode *list) {
// clean goes here
}
void testLinkedList()
{
srand(time(nullptr));
ListNode *nodes = new ListNode();
nodes->value = 99;
int val;
for(int i = 0; i < 5; i++)
{
val = rand() % 30 + 1;
insertRecList(nodes, i);
//insertList(&nodes, val);
}
printList(nodes);
reverseList(&nodes);
printList(nodes);
cleanNodes(nodes);
}
int main()
{
testLinkedList();
return 0;
}
Try to compile with: -std=gnu++11
You don't change nodes in reverseList you're just changing list you're just changing a pointer on your struct which is a temporary object so physically nodes steel the same and pointed on the same first element which now has next attribute pointing on Null so the result of printList is correct. You need to work with pointers e.g.
#include <iostream>
#include <cstdlib>
struct ListNode
{
int value;
ListNode* next = NULL;
~ListNode(){
if(this->next)
delete this->next;
}
};
void insertRecList(ListNode* list, int value)
{
if(list->next == NULL)
{
ListNode* end = new ListNode;
end->value = value;
list->next = end;
}
else
insertRecList(list->next, value);
}
void printList(ListNode* list)
{
std::cout << list->value << std::endl;
while(list->next != NULL)
{
list = list->next;
std::cout << list->value << std::endl;
}
}
ListNode * reverseList(ListNode* list)
{
ListNode* next;
ListNode* prev = NULL;
ListNode* cur = list;
while(cur != NULL)
{
if(cur->next == NULL)
{
cur->next = prev;
break;
}
else
{
next = cur->next;
cur->next = prev;
prev = cur;
cur = next;
}
}
std::cout << cur->value << " list:" << list->value << std::endl;
return cur;
}
void testLinkedList()
{
srand(time(NULL));
ListNode * nodes = new ListNode;
nodes->value = 99;
int val;
for(int i = 0; i < 5; i++)
{
val = rand() % 30 + 1;
insertRecList(nodes, i);
//insertList(&nodes, val);
}
printList(nodes);
nodes = reverseList(nodes);
printList(nodes);
delete nodes;
}
int main()
{
testLinkedList();
return 0;
}
Also, don't forget to delete object created dynamically
Reversing a linked list is not a fundamental operation. It does not belong among the basis operations of your class. It is easier (and safer) to implement it in terms of your other operations. Roughly:
Create an empty list.
While the first list is not empty, remove a node from the front of the first list and insert it into the front of the second list.
The second list is now the reverse of the original.
I'm trying to implement a stack using a doubly linked list. I know that the functions for my stack class (push, pop) should contain calls to member functions of my doubly linked list class, but I'm having trouble actually implementing that.
dlist.cpp:
#include <iostream>
#include <fstream>
#include <string>
#include "dlist.hpp"
using namespace std;
void dlist::appendNodeFront(int shares, float pps){
Node *n = new Node(shares, pps);
if(front == NULL){
front = n;
back = n;
}
else {
front->prev = n;
n->next = front;
front = n;
}
}
void dlist::appendNodeBack(int shares, float pps){
Node *n = new Node(shares, pps);
if(back == NULL){
front = n;
back = n;
}
else {
back->next = n;
n->prev = back;
back = n;
}
}
void dlist::display(){
Node *temp = front;
cout << "List contents: ";
while(temp != NULL){
cout << temp->value << " ";
temp = temp->next;
}
cout << endl;
}
void dlist::display_reverse(){
Node *temp = back;
cout << "List contents in reverse: ";
while(temp != NULL){
cout << temp->value << " ";
temp = temp->prev;
}
cout << endl;
}
void dlist::destroyList(){
Node *T = back;
while(T != NULL){
Node *T2 = T;
T = T->prev;
delete T2;
}
front = NULL;
back = NULL;
}
stack.cpp:
#include <iostream>
#include <fstream>
#include <string>
#include "stack.hpp"
using namespace std;
stack::stack(){
int i;
for(i = 0; i < 1500; i++){
shares[i] = 0;
pps[i] = 0;
}
first = 0;
}
void stack::push(int num, float price){
if(first ==(1500-1)){
cout << "Stack is full" << endl;
return;
}
first++;
shares[first] = num;
pps[first] = price;
return;
}
void stack::pop(int *num, float *price){
if(first == -1){
cout << "Stack is empty" << endl;
return;
}
num = &shares[first];
price = &pps[first];
cout << shares[first] << endl;
cout << pps[first] << endl;
shares[first] = 0;
pps[first] = 0;
first--;
return;
}
Should the push function in stack basically be a call to appendNodeFront() or appendNodeback()? Any help or advice is greatly appreciated!
You can create a stack class, then use linked list class as its container. In a linked list class there is virtually no limit to the number of items, so you add artificial limit to make it work like a stack. In a linked list, items can be added/removed anywhere in the list, you can limit add/remove the tail node only to make it work like stack. The example below demonstrate the usage.
Node that this is purely a programming exercise. Stack is relatively primitive compared to Doubly-linked list. Encapsulating a linked-list inside stack has no advantage. Also note, I declared all members as public for the sake of simplifying the problem, you may want to change some members to protected/private
#include <iostream>
#include <fstream>
#include <string>
using std::cout;
class Node
{
public:
Node *prev;
Node *next;
int shares;
float pps;
Node(int vshares, float vpps)
{
shares = vshares;
pps = vpps;
prev = next = nullptr;
}
};
class dlist
{
public:
Node *head;
Node *tail;
dlist()
{
head = tail = nullptr;
}
~dlist()
{
destroy();
}
void push_back(int shares, float pps)
{
Node *node = new Node(shares, pps);
if (head == NULL)
{
head = tail = node;
}
else
{
tail->next = node;
node->prev = tail;
tail = node;
}
}
void destroy()
{
Node *walk = head;
while (walk)
{
Node *node = walk;
walk = walk->next;
delete node;
}
head = tail = nullptr;
}
};
class stack
{
public:
int maxsize;
int count;
dlist list;
stack(int size)
{
count = 0;
maxsize = size;
}
void push(int num, float price)
{
if (count < maxsize)
{
list.push_back(num, price);
count++;
}
}
void pop()
{
Node *tail = list.tail;
if (!tail)
{
//already empty
return;
}
if (tail == list.head)
{
//only one element in the list
delete tail;
list.head = list.tail = nullptr;
count--;
}
else
{
Node *temp = list.tail->prev;
delete list.tail;
list.tail = temp;
list.tail->next = nullptr;
count--;
}
}
void display()
{
Node *walk = list.head;
while (walk)
{
cout << "(" << walk->shares << "," << walk->pps << ") ";
walk = walk->next;
}
cout << "\n";
}
};
int main()
{
stack s(3);
s.push(101, 0.25f);
s.push(102, 0.25f);
s.push(103, 0.25f);
s.push(104, 0.25f);
s.display();
s.pop();
s.display();
return 0;
}
My program should create a linked list and show it. My problem is when the addelemnt_end function ends, it doesn't update head and last.
I tried with debug and when my function is done, the info and next part from head and last are "unable to read memory".
struct node{
int info;
node *next;
};
node *head, *last;
void addelement_end(node *head, node *last, int element)
{if (head == NULL)
{ node *temp = new node;
temp->info = element;
temp->next = NULL;
last = temp;
head = temp;
}
else {node*temp = new node;
last->next = temp;
temp->info = element;
temp->next = NULL;
last = temp;
}
}
void show(node* head, node *last)
{
if (head==NULL)
cout << "Empty list";
else
while (head != NULL)
{
cout << head->info << " ";
head = head->next;
}
}
int _tmain(int argc, _TCHAR* argv[])
{
int x, n, i;
cout << "how many numbers";
cin >> n;
head = last = NULL;
for (i =1; i <= n; i++)
{
cin >> x;
addelement_end(head, last, x);
}
show(head, last);
return 0;
}
It's a very common error. Here is a similar illustration of the problem:
int change_a(int a) {
a = 42;
}
int main() {
int a = 10;
change_a(a);
printf("%d\n", a);
return 0;
}
This will print 10 because in the function change_a you are only modifying a copy of the value contained in the variable a.
The correct solution is passing a pointer (or using a reference since you are using C++).
int change_a(int *a) {
*a = 42;
}
int main() {
int a = 10;
change_a(&a);
printf("%d\n", a);
return 0;
}
But maybe you're going to tell me: "I'm already using a pointer!". Yes, but a pointer is just a variable. If you want to change where the pointer points, you need to pass a pointer to that pointer.
So, try this:
void addelement_end(node **head, node **last, int element)
{
if (*head == NULL)
{ node *temp = new node;
temp->info = element;
temp->next = NULL;
*last = temp;
*head = temp;
}
else {
node *temp = new node;
(*last)->next = temp;
temp->info = element;
temp->next = NULL;
*last = temp;
}
}