在 C++ 中反转链表的方法答案

【问题标题】：Method to Reverse a Linked list in c++在 C++ 中反转链表的方法
【发布时间】：2014-06-24 05:35:56
【问题描述】：

我正在编写一种在 C++ 中反转链表的方法。我正在尝试使用 Node* 而不是 void 返回类型，但遇到了许多错误。

我的方法代码..

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

我收到的编译时错误消息..

    solution.cc: In function 'Node* Reverse(Node*)':
    solution.cc:24:22: error: cannot convert 'Node*' to 'Reverse(Node*)::node*' in initialization
  node* current = head;
                  ^
    solution.cc:28:24: error: invalid use of incomplete type 'struct Reverse(Node*)::node'
     next  = current->next; 
                    ^
    solution.cc:23:14: error: forward declaration of 'struct Reverse(Node*)::node'
   struct node* prev   = NULL;
          ^
    solution.cc:29:16: error: invalid use of incomplete type 'struct Reverse(Node*)::node'
     current->next = prev;  
            ^
    solution.cc:23:14: error: forward declaration of 'struct Reverse(Node*)::node'
   struct node* prev   = NULL;
          ^
    solution.cc:33:10: error: cannot convert 'Reverse(Node*)::node*' to 'Node*' in assignment
 head = prev;
      ^
    solution.cc:34:1: error: no return statement in function returning non-void [-Werror=return-type]
   }
   ^
    cc1plus: some warnings being treated as errors

【问题讨论】：

node和Node的定义是什么？
1 问题是你在这个方法之前一定没有完全声明struct node，你可能只是像struct node一样放了一个前向声明；请附上完整的声明。您还说您的函数将返回类型 o Node* 但您不返回任何内容。
您无需在任何地方输入struct。

标签： c++ pointers return-type singly-linked-list

【解决方案1】：

Node 与 node 不同，您缺少 return 语句

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

【讨论】：

【解决方案2】：

它应该是节点而不是节点。另外，您还必须返回 Node* 类型的内容。

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

   return head;          //return head as it will return the whole list.

 }

【讨论】：

为什么会出现无故内存泄漏？