Question

我试图做的是交换单链表的第一个和最后一个元素。到目前为止，我有以下代码，我在其中创建一个列表并添加一些数字。我的问题在于swapElements1函数。

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

struct node
{
    int number;
    struct node *next;
};

void addNodeSingle(struct node **head, int num, int thesi) //Function to insert new node at the beginning or the end of the list, depending on the value of "thesi"
{
    if (*head == NULL)
    {
        struct node *current;
        current = (struct node*) malloc (1*sizeof(struct node));
        current -> number = num;
        current -> next = NULL;
        *head = current;
    }

    else
    {
        if (thesi == 0)
        {
            struct node *current;
            current = (struct node*) malloc (1*sizeof(struct node));
            current -> number = num;
            current -> next = *head;
            *head = current;
        }

        else
        {
            struct node *current, *temp;
            current = (struct node*) malloc (1*sizeof(struct node));
            current -> number = num;
            temp = *head;
            while (temp -> next != NULL)
                temp = temp -> next;

            temp -> next = current;
            current -> next = NULL;
        }
    }
}

void displayList(struct node **head) //Function to display the list
{
    struct node *current;
    if(*head == NULL)
        printf("I lista einai adeia!\n");

    else
    {
    current= *head ;
        while(current != NULL)
        {
            printf("%d ",current -> number);
            current = current -> next;
        }
    }
}

void swapElements1(struct node **head) //(not working)Function to swap first and last element of the list
{
    struct node *current, *temp;
    current = temp = *head;

    while(current != NULL)
    {
        temp = current;
        current = current -> next;
    }

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

int main()
{
    struct node *head;
    head = NULL;

    addNodeSingle(&head,5,1);
    addNodeSingle(&head,6,1);
    addNodeSingle(&head,2,0);
    addNodeSingle(&head,7,0);
    addNodeSingle(&head,8,0);

    printf("List is: ");
    displayList(&head);
    swapElements1(&head);
    printf("\nNew list is: ");
    displayList(&head);
}

我得到的输出是：

列表是：8 7 2 5 6

新名单是：6

我需要的是：

列表是：8 7 2 5 6

新名单是：6 7 2 5 8

这里是demo

Answer 1

这显然是错误的：

*head = (*head)->next;
*head = temp;

只是用值temp覆盖前一个值。第一个声明甚至可能不存在。

您基本上需要两次掉期（技术上一次加分配和终止）

指向两个节点的指针
两个节点next指针

后者在技术上并不需要，但需要直接分配，并且新尾部需要将其next设置为null终止新名单。

下面显示了一个完整的示例，自由地评论了希望揭示正在发生的事情的算法。

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

struct node
{
    int data;
    struct node *next;
};

void swapFirstAndLast(struct node **head)
{
    // don't bother unless we have a list of at least two nodes
    if (!*head || !(*head)->next)
        return;

    // start with the head's next pointer (the second node in the list)
    struct node **pp = &(*head)->next;

    // walk the pointer-to-pointer down the list, each time grasping
    //  the next node's "next" pointer address until we reach a node
    //  whose 'next' is NULL. When that happens, `pp` will hold the
    //  address of the pointer pointing to the last node in the list
    while (*pp && (*pp)->next)
        pp = &(*pp)->next;

    // swap the pointer held in *head with *pp
    struct node *tmp = *head;
    *head = *pp;
    *pp = tmp;

    // save new head's next pointer to be the old head's next
    (*head)->next = (*pp)->next;

    // and finally, terminate the list.
    (*pp)->next = NULL;
}

void print_list(const struct node *head)
{
    while (head)
    {
        printf("%d ", head->data);
        head = head->next;
    }
    fputc('\n', stdout);
}

int main()
{
    struct node *head = NULL, **pp = &head;
    for (int i=1; i<=5; ++i)
    {
        *pp = malloc(sizeof **pp);
        (*pp)->data = i;
        pp = &(*pp)->next;
    }
    *pp = NULL;

    print_list(head);

    swapFirstAndLast(&head);

    print_list(head);
}

<强>输出

1 2 3 4 5 
5 2 3 4 1

我已经为你清理了清单（毫无疑问，你已经编码了这样的算法）。这方面的关键是如何使用指针指针来操纵链表的指针;不只是一堆临时指针。我强烈建议您在调试器中单步执行交换功能，观察整个过程中每个步骤发生的情况。

Answer 2

WhozCraig的答案已经很完美，但也许您可能希望通过一些调整来查看自己的代码。基本上，我所做的是在一次迭代之前停止搜索最后一个元素，因此temp将停在5并且当前为6.此外，你必须使用指针进行四次更改，例如WhozCraig。这是您的代码，略有修改：

void swapElements1(struct node **head) //(not working)Function to swap first and last element of the list
{
    struct node *current, *temp;
    current = temp = *head;

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

这适用于我的机器。我将while循环条件从current更改为current-＆gt; next并进行了正确的“指针更改”。这些指针的变化很难解释，通常我先将它们写在纸上然后用代码写出来。

C - 交换单链表中的第一个和最后一个元素

2 个答案: