從開始洗掉節點時，雙回圈鏈表在末尾留下尾隨0

有人可以說明為什么這個deleteB函式在洗掉后最后會留下一個尾隨 0。我出于好奇嘗試了這種方法。

我試圖訪問 head 的前一個指標并使其成為下一個指標指向head->next而不是head.

之后，我將更改前一個指標 ofhead->next并使其指向head->prev而不是head.

最后我改變free了.temphead

也許方法是錯誤的......如果我錯了，請指導我。

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

typedef struct Node
{
    int data;
    struct Node *prev;
    struct Node *next;
} Node;

void deleteB(Node **head)
{
    if (*head != NULL)
    {
        if ((*head)->next == *head)
        {
            *head = NULL;
            return;
        }
        Node *temp = *head;
        (*head)->prev->next = (*head)->next;
        (*head)->next->prev = (*head)->prev;
        *head = (*head)->next;
        free(temp);
        // Node *curr = *head;
        // while (curr->next != *head)
        // {
        //  curr = curr->next;
        // }
        // curr->next = (*head)->next;
        // (*head)->next->prev = curr;
        // *head = (*head)->next;
        // free(temp);
    }
}

void prepend(Node **head, int value)
{
    Node *newNode = (Node *)malloc(sizeof(Node));
    newNode->prev = NULL;
    newNode->next = NULL;
    newNode->data = value;
    if (*head == NULL)
    {
        *head = newNode;
        (*head)->next = *head;
        (*head)->prev = *head;
        return;
    }
    Node *temp = *head;
    while (temp->next != *head)
    {
        temp = temp->next;
    }
    temp->next = newNode;
    newNode->prev = temp;
    newNode->next = *head;
    *head = newNode;
}

void append(Node **head, int value)
{
    Node *newNode = (Node *)malloc(sizeof(Node));
    newNode->prev = NULL;
    newNode->next = NULL;
    newNode->data = value;
    if (*head == NULL)
    {
        *head = newNode;
        (*head)->next = *head;
        (*head)->prev = *head;
        return;
    }
    Node *temp = *head;
    while (temp->next != *head)
    {
        temp = temp->next;
    }
    temp->next = newNode;
    newNode->prev = temp;
    newNode->next = *head;
}

void display(Node *head)
{
    printf("\nPrinting the list: ");
    Node *temp = head;
    do
    {
        printf("-->%d", temp->data);
        temp = temp->next;
    } while (temp != head);
}

int main()
{
    Node *head = NULL;
    append(&head, 1);
    append(&head, 2);
    append(&head, 3);
    append(&head, 4);
    // insertAtN(&head, 9, 1);
    deleteB(&head);
    display(head);
    printf("\n");
    return 0;
}

uj5u.com熱心網友回復：

在（非空，回圈）雙向鏈表中插入節點時，需要設定 4 個指標。兩個指向遠離新節點，兩個指向它。您錯過了后兩者之一：

(*head)->prev = newNode;

其他一些評論：

• prepend具有與 相同的代碼append，只是在其后多了一條陳述句。所以避免代碼重復，讓prependcall append.

• 創建一個單獨的函式，僅用于構建新節點

• 不要將prevandnext成員初始化為NULL，因為這絕不是它們在回圈串列中應該具有的值。為什么不用自我參考來初始化它們......那么至少有可能這將是最終值，并且您永遠不必檢查它們的值是否為NULL.

• 在append(and ) 中，當您有從頭節點到該尾節點的直接prepend參考時，您沒有充分的理由temp沿著串列的所有節點向前走以找到最后一個節點！( )prev

• 不要強制轉換回傳的值malloc

• display將空串列傳遞給它時將具有未定義的行為。所以加了一個后衛。

更新代碼：

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

typedef struct Node
{
    int data;
    struct Node *prev;
    struct Node *next;
} Node;

void deleteB(Node **head)
{
    if (*head != NULL)
    {
        if ((*head)->next == *head)
        {
            *head = NULL;
            return;
        }
        Node *temp = *head;
        (*head)->prev->next = (*head)->next;
        (*head)->next->prev = (*head)->prev;
        *head = (*head)->next;
        free(temp);
    }
}

// Separate function, just to construct a node
Node *newNode(int value) 
{
    Node *node = malloc(sizeof(Node)); // Don't cast
    node->prev = node; // Circular list never has NULL here, so don't put it
    node->next = node;
    node->data = value;
    return node;
}

void append(Node **head, int value)
{
    Node *node = newNode(value);
    if (*head == NULL)
    {
        *head = node;
        return;
    }
    // We can find the tail with one step
    Node *tail = (*head)->prev;
    tail->next = node;
    node->prev = tail;
    node->next = *head;
    (*head)->prev = node; // This was missing
}

void prepend(Node **head, int value)
{
    append(head, value); // Code reuse
    *head = (*head)->prev;
}

void display(Node *head)
{
    printf("Printing the list: ");
    if (head == NULL) // Guard!
    { 
       return;
    }
    Node *temp = head;
    do
    {
        printf("-->%d", temp->data);
        temp = temp->next;
    } while (temp != head);
    printf("\n");
}

int main()
{
    Node *head = NULL;
    append(&head, 1);
    append(&head, 2);
    append(&head, 3);
    append(&head, 4);
    deleteB(&head);
    display(head);
    return 0;
}

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