在C++中反轉鏈接列表的方法

Question

我正在寫一個方法來在C++中反轉鏈接列表。我試圖使用Node*而不是void返回類型，但面臨一些錯誤。

我的方法代碼..

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; 
    } 

的編譯時錯誤消息我收到了..

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 

Answer 1

Node不一樣node和你缺少一個return聲明

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; 
} 

Answer 2

它應該是節點而不是節點。再加上你必須返回Node *類型的東西。

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. 

}