链接列表：将节点从一个列表移动到另一个列表

Question

有2个列表source={3,2,1}和dest ={4,5,6,7}其中链接列表的头指针分别位于3和4中。删除源头节点，将数据3移动到目标列表，并将其作为目标列表中的新头节点。

所以在第一轮source ={2,1} dest ={3,4,5,6,7}之后，现在head的源代码指向2，head代表dest指向3.最后，我必须制作source = NULL and Dest = {1,2,3,4,5,6,7} head => 1。每次我都可以通过调用移动节点函数来实现。但是当我在一个循环中运行时，它会保持循环。这是错误的代码。请告诉我为什么会出现循环问题。

 typedef struct node{ 
int data; 
struct node* next; 
}Node; 

    void push(Node** headRef, int data){ 
Node* newNode = (Node*) malloc(sizeof(newNode)); 
newNode->data = data; 
newNode->next = *headRef; 
*headRef = newNode; 
    } 
    Node* pushtop(){ 
Node* head = NULL; 
int i; 
for(i = 1; i<=3; i++){ 
push(&head,i); 
} 
return head; 
    } 

    Node* pushbottom(){ 
Node* head = NULL; 
int i; 
for(i=7; i>=4; i--){ 
push(&head,i); 
} 
return head; 
    } 

    void moveNode(Node** source,Node** dest){ 
Node* ptr = *source; 
Node* current = NULL; 
while(ptr!=NULL){ // here the continuous looping occurs 
    current=ptr; 
    current->next = *dest 
    *dest = current;  
    *source = ptr->next; 
    ptr = ptr->next; 
    } 
    Node* test = *dest; 
    printf("\nthe then moved list is\n\n"); 
    while(test!=NULL){ 
     printf("%d\n",test->data); 
     test = test->next; 
     } 
     } 
    int main(){ 
Node* headA = pushtop(); 
Node* headB = pushbottom(); 
moveNode(&headA, &headB); 
    return 0; 
} 

请检查移动节点While循环部分。

Answer 1

答案与@zavg的答案有点不同。一点点调整使它工作正常。

正如我所说的源指针也应该改变。所以我会粘贴代码。感谢@zavg的帮助。

 while(ptr != NULL) { // here is the correct code. 
     current = ptr->next; 
     ptr->next = *dest; 
     *dest = ptr;  
     ptr = current; 
     *source = ptr; 
     } 

     printf("%p\n",*source); //it will print Nil. 

     Node* test = *dest; 
     printf("\nthe then moved list is\n\n"); 

     while(test!=NULL){ 
      printf("%d\n",test->data); 
      test = test->next; 
     } 

Answer 2

Node* ptr = NULL; 
Node* current = *source; 
while(current != NULL) { // here the continuous looping occurs 
    ptr = current->next; 
    current->next = dest; 
    dest = current;  
    current = ptr; 
}