Simply Chained List - delete the first half of the list

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Good afternoon, you guys. I hope I’m publishing it right, because I’m new to the forum and new to programming. So, on a question that I am, more specifically in "Simple Chained Lists", is that the function below should eliminate the first half of the chained list. I would have to go through the entire list to get half off the list, in case it’s even.

Am I doing it right?

If you can help me with the code below, you are most welcome!!

Function below:

// (c) Remove a primeira metade da Lista. Caso a lista possua um número ímpar de elementos,
// considere que a primeira metade possui mais elementos
// (Ex: se a lista possuir 5 elementos, a primeira metade possui os 3 primeiros elementos).
// Se a lista tiver vazia, nada acontece.
void removePrimeiraMetade(Lista *L) {
    // IMPLEMENTE ESTA FUNÇÃO
    //Estou pensando no código
    if (listaEstavazia(L)){
      return;
    }
    else{
      No *p = L->inicio;
      No *anterior = NULL;
      while (p != NULL){
        if (L->tamanho % 2 == 0){
          anterior->prox = no->prox;
          L->tamanho--;
        }
      }
    }
}

Struct _no, _produto and _lista (it’s not in order, but it’s here):

typedef struct _produto {
    int num_serie; // numero de série do produto
    char nome[64];
    double preço;
} Produto;

// struct que define um nó curcular duplamente encadeado
typedef struct _no {
    Produto *prod;
    struct _no *prox;
} No;

// struct que define uma Lista Ligada Simples
typedef struct _lista {
    No *inicio;
    No *fim;
    int tamanho; // numero de nós da lista
} Lista;

Code update above:

The code is basically like this, right?

if (listaEstaVazia(L)){
    return;
}
else {
    int tamanho = L->tamanho;
    if (tamanho % 2 == 0){
        for (int i=0; i<(tamanho / 2); i++){
            L->inicio = L->inicio->prox;
            L->tamanho--;
        }
    }
    else {
        double impar = round(L->tamanho / 2);
        for (int i=0; i<impar; i++){
            L->inicio = L->inicio->prox;
            L->tamanho--;
        }
    }
}
c list chain-list

  • Take the list size and divide by 2, call this value loops. Then run a for that goes up to the amount of loops, for each cycle you remove the first item from the list.

  • What if the list size is odd? For example, I have a list of size 5, the "half" is 3, and I must pick them up. It’s basically smeelhante when I make in even list.

  • In this case, you decide whether to round up or down. If you are using a variable of type int, it will ignore the decimal places, which gives the effect of "rounding down".

  • Got it. In this case, then, I must declare a new float variable?

  • You can just divide the L->tamanho for 2, make a for loop that traverses all nodes from the L->start to the node in question, deleting them and then changing the L->start to the final node

1 answer

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// (c) Remove the first half of the List. If the list has an odd number of elements, // consider that the first half has more elements // (Ex: if the list has 5 elements, the first half has the first 3 elements). // If the list is empty, nothing happens.

Pay attention to the data...

You don’t have to do any of this module stuff and count if it’s even and stuff.

Understand that:

  • if the list is empty will do nothing
  • If the list only has one guy then it doesn’t have half and it won’t do anything
  • from there will erase at least one: then:
  • delete 1
  • delete half of the others

Examples:

  • if they were 2: go delete one and 1/2 = 0 remains: delete 1, ok
  • if they were 3: delete one and 2/2 = 1: delete 2, ok
  • if they were 4: delete one and 3/2 = 1: delete 2, ok
  • if they were 5, the example: delete 1, 4/2 = 2: delete 3, ok

You don’t need to test anything else.

post a complete program that can be compiled

I’d have to go through the entire list to get half off the list, if you are even

And why? Just set how many you will remove, as I said. Position yourself at the beginning and delete the first one and then half of the others in a loop. The entire division, as you know, will truncate values.

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