C++에서 연결된 목록을 뒤집는 방법

Question

C++에서 연결된 목록을 역순으로 변환하는 방법을 쓰고 있습니다. void 반환 형식 대신 Node*을 사용하려고하지만 여러 오류가 있습니다.

내 방법 코드는 ..

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; 
    } 

내가 수신하고 컴파일 시간 오류 메시지 ..

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

Node는 node으로 동일하지 않습니다 당신은 return 문

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; 
} 

누락

Answer 2

노드가 아니라 노드 여야합니다. 더하기 당신은 Node * 타입을 반환해야합니다.

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. 

}