链表：将节点从一个列表移动到另一个列表

Question

有两个链表source={3,2,1}和dest ={4,5,6,7}，链表的头指针分别在3和4。源中的头节点被删除，数据 3 被移动到目标列表，并在目标列表中作为新的头节点。

所以在第一轮 source ={2,1} dest ={3,4,5,6,7} 之后，源中的头现在指向 2，而目标中的头指向 3。最后我必须使 source = NULL and Dest = {1,2,3,4,5,6,7} head => 1。我可以通过每次调用下面的移动节点函数来做到这一点。但是当我在一个循环中运行时，它会一直循环。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

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

Answer 2

答案与@zavg 的答案有点不同。稍微调整一下就可以正常工作了。

正如我所说，源指针也应该更改。所以我将粘贴代码。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;
        }