C中根据用户请求的元素个数创建链表

Question

所以我想创建一个双向链表的节点结构:

typedef struct Node { //node structure
    int value;
    struct Node *next;
    struct Node *previous;
} node;

起初，我创建了如下列表:

//allocate memory for the nodes
    node *first = malloc(sizeof(*first));
    node *second = malloc(sizeof(*second));
    node *third = malloc(sizeof(*third));
    node *fourth = malloc(sizeof(*fourth));
    node *fifth = malloc(sizeof(*fifth));

    { //define the nodes and link them together.(circular linked list)
        first->value = 1;
        first->next = second;
        first->previous = fifth;

        second->value = 2;
        second->next = third;
        second->previous = first;

        third->value = 3;
        third->next = fourth;
        third->previous = second;

        fourth->value = 4;
        fourth->next = fifth;
        fourth->previous = third;

        fifth->value = 5;
        fifth->next = first;
        fifth->previous = fourth;
    }

但后来我意识到我不知道列表是否应该由 5 个元素组成，想向用户询问列表的大小。 问题是，当我尝试实现一个名为 createList() 的函数并将列表大小作为参数传递时，我不知道应该如何创建这样的列表。我想首先使用 ->next 和 ->previous 值创建一个临时指针作为 NULL 但是我将如何遍历输入 n 并在 next 节点为 NULL 时创建新节点？

谢谢

Answer 1

你必须创建新节点 n(input size) 次并附加到链表的尾部，你可以使用任何循环构造来创建新节点，下面是示例代码。 已编辑:循环双向链表的条件，早些时候我注意到您要求循环 dll，我认为它只是 dll。

#include<stdio.h>
#include<stdlib.h>
typedef struct Node { //node structure
    int value;
    struct Node *next;
    struct Node *previous;
} node;

void createList(struct Node ** head, int size) {

    // temp variable to iterate list
    struct Node *temp = *head;
    int val;
    while (size--) {
        scanf("%d", &val);


        if (temp == NULL) {
            // create head node for empty linked list;
            struct Node *node = (struct Node*)malloc(sizeof( Node));
            node->previous = NULL;
            node->next = NULL;
            node->value = val;
            *head = node;
            temp = node;
        } else {
            // if head node exists append the numbers at the end of the linkedlist
            struct Node *node = (struct Node*)malloc(sizeof(struct Node));
            node->previous = temp;
            node->next = NULL;
            node->value = val;
            temp->next = node;
            temp = temp->next;
        }
    }

    // for circular linked list
    temp->next = *head;
    (*head)->previous = temp;
}

void printList (struct Node *head) {
    node *temp = head;
    while (head->next != temp) {
        printf("%d\n", head->value);
        head = head->next;
    }
    // print last node
    printf("%d\n", head->value);
}
int main () {

    struct Node *head = NULL;
    int size;
    scanf("%d", &size);
    createList(&head, size);
    printList(head);
    return 0;
}