doubleLinkedList複製構造函數

Question

我無法實現雙向鏈表的複製構造函數。我會在下面的for循環中設置尾部和先前的指針嗎？以下作品適用於單鏈表：

sortedList::sortedList(const sortedList& original) 
: size(original.size) 
{ 
if (original.head == NULL) 
    head = NULL; // original list is empty 

else 
{ 
    // copy first node 
    head = new listNode; 
    tail = new listNode; 
    assert(head != NULL); // check allocation 
    assert(tail != NULL); // check allocation 
    head->item = original.head->item; 
    tail->item = original.tail->item; 
    // copy rest of list 
    listNode *newPtr = head; // new list pointer 


    // newPtr points to last node in new list 
    // origPtr points to nodes in original list 
    for (listNode *origPtr = original.head->next; origPtr != NULL; origPtr = origPtr->next) 
    { 
     newPtr->next = new listNode; 
     assert(newPtr->next != NULL); 

     newPtr = newPtr->next; 

     newPtr->item = origPtr->item; 
     /* 
     newPtr = new listNode; 
     newPtr->item = origPtr->item; 
     newPtr->prev = origPtr->prev; 
     tail->next = origPtr; 
     tail = origPtr;*/ 
    } 

    newPtr->next = NULL; 
} 

} 

Answer 1

您需要爲循環內的每個元素設置prev指針。完成後，更新head->prev和tail->next。

這可能幫助：

for (listNode *origPtr = original.head->next; origPtr != original.head; origPtr = origPtr->next) 
    { 
     newPtr->next = new listNode; 
     assert(newPtr->next != NULL); 

     listNode * prev = newPtr; 
     newPtr = newPtr->next; 

     newPtr->item = origPtr->item; 
     newPtr->prev = prev; 
    } 

    tail->next = head; 
    head->prev = tail;