为什么在cpp中反转链表需要临时节点？

Question

这是工作代码：

Node* Reverse(Node *head)
{
    Node *prev = NULL;
    Node *next;
    Node *current = head;
    while (current != NULL)
    {
        next = current -> next;
        current -> next = prev;
        prev = current;
        current = next;
    }
    head = prev;

    return head;
}

当我删除临时变量 current 时，这不起作用。

Node* Reverse(Node *head)
{
    Node *prev = NULL;
    // Node *current = head;
    Node *next;
    while (head != NULL)
    {
        next = head -> next;
        head -> next = prev;
        prev = head;
        head = next;
    }
    return head;
}

我知道我们需要 next 因为我们必须在失去访问权限之前跟踪它。但是我想不出为什么临时变量 current 是必要的？任何帮助将不胜感激。

Answer 1

您不需要 current 变量。（尽管它的名字 head 会令人困惑）

代码不同，因为您没有在第二个代码段中将 prev 分配给 head。

Node* Reverse(Node *head)
{
    Node *prev = NULL;
    // Node *current = head;
    Node *next;
    while (head != NULL)
    {
        next = head -> next;
        head -> next = prev;
        prev = head;
        head = next;
    }
    
    // or head=prev; return head;
    return prev; // <--- 
}

为什么在cpp中反转链表需要临时节点？

Why is temporary Node necessary in Reversing a linked list in cpp?

c++

linked-list