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Linked List and Pointers in C


Linked List and Pointers in C

By : CLeOpatra
Date : November 19 2020, 03:01 PM
will be helpful for those in need The way the code is posted does not work, because it indeed misses deletion of the initial node in the head pointer. However, you omitted some code which apparently fixes the problem. Here is the code from the original delete:
code :
//found a match, update the link
if(current == head) {
    //change first to point to next link
    head = head->next;
} else {
    //bypass the current link
    previous->next = current->next;
}
struct node* delete(int key) {
    struct node** ptrCurrent = &head;
    while (*ptrCurrent) {
        if ((*ptrCurrent)->key == key) {
            struct node* tmp = *ptrCurrent;
            *ptrCurrent = (*ptrCurrent)->next;
            free(tmp);
            break;
        }
        ptrCurrent = &((*ptrCurrent)->next);
    }
}


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Pointers to pointers: partitioning a linked list

Pointers to pointers: partitioning a linked list


By : ROtrex
Date : March 29 2020, 07:55 AM
help you fix your problem I'm trying to become more facile with pointers. So, for fun, I took the following C++ function that partitions a linked list around a value
code :
void partitionList(lnode<int> *& head, int val) {
    ...
    head = front; // <= head is modified
}
void partitionList(lnode<int> ** head, int val) {
    ...
    head = &front; // try it with *head = front
}
void partitionList(lnode<int> *& head, int val) {
   ...
   lnode<int> * curr = head;
void partitionList(lnode<int> ** head, int val) {
    ...
    lnode<int> * curr = *head;
Trying to bubble sort a linked list by only rearranging pointers. My code has a workaround for a 2 node list that I want

Trying to bubble sort a linked list by only rearranging pointers. My code has a workaround for a 2 node list that I want


By : Shashank P
Date : March 29 2020, 07:55 AM
wish help you to fix your issue You can play a classic trick: take a pointer head and make it subsequently point to the next member of the previous node:
code :
Node *sortList(Node *head) {
    // make sure the list is not empty 
    if (head != NULL) {
        // bubble sort
        for (;;) {
            int swapped = 0;
            Node **linkp = &head;
            Node *currNode = head;
            Node *nextNode;

            while ((nextNode = currNode->next) != NULL) {
                if (strcmp(currNode->word, nextNode->word) > 0) {
                    currNode->next = nextNode->next;
                    nextNode->next = currNode;
                    *linkp = currNode = nextNode;
                    swapped++;
                }
                link = &currNode->next;
                currNode = nextNode;
            }
            if (swapped == 0)
                break;
        }
    }
    return head;
}
Converting linked list from raw pointers to smart pointers

Converting linked list from raw pointers to smart pointers


By : Howard Chang
Date : October 22 2020, 02:01 PM
hop of those help? When constructing C++ class instances, you must use new and delete, instead of malloc and free. malloc and free are C library functions, that know absolutely nothing about C++ class constructors, destructors, and everything else that goes with what a C++ class is all about.
The shown code is attempting to construct an instance of the node class by using malloc. that won't work. new must be used to construct it:
code :
std::shared_ptr<node> link = new node;
Pointers to pointers - linked list mess

Pointers to pointers - linked list mess


By : Abigail Winograd
Date : March 29 2020, 07:55 AM
I wish this help you I'm writing a simple C program to manage a linked list defined as follow: , This code has (at least) two bugs:
code :
void insert(List l, int value)
{
  List p;

  // Find end of list or highest item less than value
  for(p = l; p->next != NULL && p->next->value < value; p = p->next);

  if (p->value >= value) {
    // Over-write p with new value, and insert p as a new one after.
    // This saves having to modify l itself.
    int tmpval = p->value;
    p->value = value;
    p->next = new_node(tmpval, p->next);
  } else {
    // Insert new item after p
    p->next = new_node(value, p->next);
  }
}
void print_list(List l){
  List aux;
  for(aux = l; aux != NULL; aux = aux->next)
    printf("Valor: %d\n", aux->value);
}
typedef struct node{
  int value;
  struct node *next;
} List
my_func(List *l, ...)
Linked List: Finding intersection of 2 linked lists by swapping their pointers after reaching the end of one of them

Linked List: Finding intersection of 2 linked lists by swapping their pointers after reaching the end of one of them


By : diabloReigns80
Date : March 29 2020, 07:55 AM
Does that help They both rely on getting the pointers such that they are the same distance from the intersection, then walking them forward simultaneously until they meet.
Your first approach explicitly calculates the number of times you have to walk the pointer on the longer list forward, while your second approach does this implicitly by making both pointers take (m+n) steps.
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