C - 交换双向链表中的第一个和最后一个元素

Question

我正在尝试交换双向链表的第一个和最后一个元素。到目前为止，我有以下代码，我在其中创建了一个列表并向其中添加了一些数字。但是两次输出都是相同的列表。

#include <stdio.h>
#include <stdlib.h>

struct node2 {
    int number;
    struct node2 *next, *prev;
};

void addNodeDouble(struct node2 **head, struct node2 **tail, int num, int thesi) {
    if (*head == NULL) {
        struct node2 * current;
        current = (struct node2 *)malloc(1 * sizeof(struct node2));
        current->number = num;
        current->prev = NULL;
        current->next = NULL;
        *head = current;
        *tail = current;
    } else {
        if (thesi == 1) {
            struct node2 *current, *temp;
            current = (struct node2 *)malloc(1 * sizeof(struct node2));
            current->number = num;
            temp = *head;
            while (temp->next != NULL)
                temp = temp->next;

            temp->next = current;
            current->prev = *tail;
            current->next = NULL;
            (*tail)->next = current;
            *tail = current;
        } else  {
            struct node2 *current;
            current = (struct node2 *)malloc(1 * sizeof(struct node2));
            current->number = num;
            current->next = *head;
            (*head)->prev = current;
            *head = current;
        }
    }
}

void ReversedisplayList(struct node2 **head, struct node2 **tail) {
    struct node2 *current;
    if (*head == NULL)
        printf("I lista einai adeia!\n");
    else {
        current = *tail;
        while (current != NULL) {
            printf("%d ", current->number);
            current = current->prev;
        }
    }
}

void swapElements2(struct node2 **head, struct node2 **tail) {
    struct node2 *current, *temp;

    temp = (*tail)->prev;
    current = *tail;

    temp->next = *head;
    current->next = (*head)->next;
    (*head)->next = NULL;
    *head = current;
}

int main() {
    struct node2 *head, *tail;
    head = tail = NULL;

    addNodeDouble(&head, &tail, 4, 1);
    addNodeDouble(&head, &tail, 8, 1);
    addNodeDouble(&head, &tail, 3, 0);
    addNodeDouble(&head, &tail, 1, 1);
    addNodeDouble(&head, &tail, 7, 0);

    printf("\n\nDoubly linked list (reversed): ");
    ReversedisplayList(&head, &tail);

    swapElements2(&head, &tail);
    printf("\nChanged list: ");
    ReversedisplayList(&head, &tail);
}

我得到：

Doubly linked list (reversed): 1 8 4 3 7 
Changed list: 1 8 4 3 7

但我想要：

Changed list: 7 8 4 3 1

Answer 1

您忘记更改 (*head) -> prev 和 (*tail) -> prev。

(*head)->prev = temp;
(*tail)->prev = NULL;

Answer 2

要交换第一个元素和头元素和尾元素，您必须完成以下过程。首先我们必须在某个临时变量中获取 tail 的前一个节点和 head 的下一个节点，并交换 head 和 tail 的 next 和 prev 指针。

void swapElements2(struct node2 **head, struct node2 **tail) {
    struct node2 *ttail, *thead;

    ttail = (*tail) -> prev;
    thead = (*head) -> next;

    (*head) -> next = NULL;
    (*tail) -> prev = NULL;

    (*head) -> prev = ttail;
    (*tail) -> next = thead;

    ttail -> next = (*head);
    thead -> prev = (*tail);

    (*tail)  = ttail -> next;
    (*head)  = thead -> next;
}

C - 交换双向链表中的第一个和最后一个元素

C - Swap first and last element in doubly linked list

c

swap

linked-list

doubly-linked-list