反转链表 - C ++ [英] Reversing Linked List - C++

问题描述

我写了一个函数，应该颠倒一个列表。

I wrote a function that should reverse a list.

到目前为止，我只能反转两个项目，但没有更多。我检查和双检查，仍然找不到问题。我甚至使用调试器来查看每个指针的值。运行调试器时，我收到了以下消息：

So far, I can reverse only two items, but no more. I checked and double checked and still can't find the problem. I even used the Debugger to see the value of each pointer. When running the debugger, I received the message:

程序中出现访问冲突（分段错误）。

An access violation (segmentation fault) raised in your program.

这是我第一次使用链接列表的作业，所以我还在学习。

This is my first assignment with linked lists so I am still learning.

我在Dev-C ++中写道：

Here is the code I wrote in Dev-C++:

List::ListNode *List::Reverse_List(ListNode *head)
{
    ListNode *cur = head;
    ListNode *forward = NULL;
    ListNode *previous = NULL;

    while (cur != NULL)
    {
        head = cur; //set the head to last node
        forward = head->next;  //save the next pointer in forward
        cur->next = previous;  //change next to previous
        previous = cur;
        cur = forward;

        cout << "cur= " << cur->item << endl; //this is just to display the current value of cur

        return head;
    }
}

推荐答案

代码关闭，它早退了。

Your code is close, it is returning early.

List::ListNode *List::Reverse_List(ListNode *head) 
{
    ListNode *cur = head;
    ListNode *forward = NULL;
    ListNode *previous = NULL;

    while (cur != NULL) {
        //There is no need to use head here, cur will suffice
        //head = cur; //set the head to last node
        forward = cur->next; //save the next pointer in forward

        cur->next = previous; //change next to previous
        previous = cur;
        cur = forward;

        cout << "cur= " << cur->item << endl; //this is just to display the current value of cur

        //don't return here you have only adjusted one node
        //return head;
    }

    //at this point cur is NULL, but previous still holds the correct node
    return previous;
}

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