为什么我的 LinkedList 在我的列表末尾读取一个额外的节点？

Question

我正在尝试为我的 LinkedList class 编写函数 moveLasttoFront，但它似乎在列表末尾的 nullptr 之前读取另一个节点.

例如，假设我有一个链表 {0, 1, 2, 8, 3, 4}。我想运行指针遍历列表，直到它等于 nullptr 并停止。但是它没有在 4 之后停止，而是再次迭代，然后停止（我认为）。这是我的代码：

  void moveLasttoFront ( ) {
    Node * last = head;
    Node * secLast = nullptr;

    while ( (*last).next != nullptr ) {
        secLast = last;
        last = (*last).next;
    }

    (*last).next = head;
    (*secLast).next = nullptr;
    head = (*last).next;
 }

例如，

在上面的列表中使用时，secLast 最终指向节点 4，而 last 最终指向 0。我不明白这个“0”是从哪里来的，因为在 4 之后不应该有另一个节点？ secLast 应指向 3，last 应指向 4。

如有任何帮助，我们将不胜感激。

谢谢！

P.S。对不起，如果这太过分了 vague/I 用错了。这是我第一次在这里提问:)

Answer 1

我用 C 做了这个例子。

Main.c

#include "linked_list.h"

Node* moveLastToFront(Node* head) {
    Node* secondLast;
    Node* last = head;
    while (last->next != NULL) {
        secondLast = last;
        last = last->next;
    }

    (*secondLast).next = NULL;
    (*last).next = head;
    head = last;

    return head;
}

int main(void) {
    Node* head = NULL;
    for (int i = 1; i < 10; i++)
        head = addnode(head, i);

    listprint(head);
    printf("\n");
    head = moveLastToFront(head);
    listprint(head);

    return 0;
}

linked_list.h

#ifndef LINKED_LIST_H
#define LINKED_LIST_H

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

Node *addnode(Node* node, int data);
void listprint(Node* node);
void freelist(Node* node);

#endif // LINKED_LIST_H

linked_list.c

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

Node* addnode(Node* node, int data) {
    if (node == NULL) {
        node = malloc(sizeof (Node));

        node->data = data;
        node->next = NULL;
    }
    else
        node->next = addnode(node->next, data);

    return node;
}

void freelist(Node* node) {
    if (node != NULL) {
        freelist(node->next);
        free(node);
    }
}

void listprint(Node* node) {
    static int i = 0;
    if (node != NULL) {
        printf("Node [%d] points to node [%d] which contains: %d\n", i, i + 1, node->data);
        i++;
        listprint(node->next);
    }
    else
        i = 0; // Reset i in case function is called again with another list
}

输出：

Node [0] points to node [1] which contains: 1
Node [1] points to node [2] which contains: 2
Node [2] points to node [3] which contains: 3
Node [3] points to node [4] which contains: 4
Node [4] points to node [5] which contains: 5
Node [5] points to node [6] which contains: 6
Node [6] points to node [7] which contains: 7
Node [7] points to node [8] which contains: 8
Node [8] points to node [9] which contains: 9

Node [0] points to node [1] which contains: 9
Node [1] points to node [2] which contains: 1
Node [2] points to node [3] which contains: 2
Node [3] points to node [4] which contains: 3
Node [4] points to node [5] which contains: 4
Node [5] points to node [6] which contains: 5
Node [6] points to node [7] which contains: 6
Node [7] points to node [8] which contains: 7
Node [8] points to node [9] which contains: 8

如有任何疑问，请随时提出！

为什么我的 LinkedList 在我的列表末尾读取一个额外的节点？

Why is my LinkedList reading an extra node at the end of my list?

c++

nodes

singly-linked-list

Main.c

linked_list.h

linked_list.c