Question

您好我已经编写了一个链表类的成员函数来复制偶数节点并删除奇数节点。

我的测试用例中的所有内容都是成功的，直到我尝试删除仅包含1个元素的列表头部。

我的测试程序指出无论出于何种原因，列表的长度大于零，这是不可能的，因为我明确地将headPtr的值设置为NULL。

void RemOddDupEven(Node*& headPtr)
{
Node *cur = headPtr;   // current node is set to head
Node *pred = 0;        // predecessor is NULL

if(headPtr == 0)       // check for empty list.
return;

// ensures will only run 1 time if there is 1 item in list.

while(cur != 0 && headPtr -> link != 0) // ensure there is a next link and more than 1 node in list.
{

if(cur -> data % 2 == 0)            // If the value is even
{
    Node *newNode = new Node;         // Create a new node
    newNode -> data = cur -> data;    // Set new Nodes data field
    newNode -> link = 0;              // set newNode link field to NULL

    if(cur == headPtr)                // if the current node is the head of the list
    {
     newNode -> link = headPtr;       // link field updated to head
     headPtr = newNode;               // newNode becomes the new Head of the list
    }
      else                            // current node is not the head of the list
      {
        pred -> link = newNode;       // update pred node to point to newNode
        newNode -> link = cur;        // update newNode to point to current
      }

    pred = cur;                       // update the pred node
    cur = cur -> link;                // update the current node
  }


     if(cur -> data % 2 == 1)              // check if this is odd
     {
     Node* nextNode = 0;                // Declare Next Node and set equal to

     if(cur -> link == 0)               // if there is no next Node then we are at the end of the list
     {
       delete cur;                      // delete the current Node
       cur = nextNode;
     }
      else{                             // else there is a next node defined
        nextNode = cur -> link;         // set the nextNode to point to next in list
        delete cur;                     // delete the current Node
        cur = nextNode;                 // assign the current Node to the next Node
      }

    if(pred)                            // if the pred is defined
     pred -> link = cur;                // previous node point to current node
      else
        headPtr = cur;                  // else we're at the head of the list
    }
  }    // end while
 }     // end method

这是我的列表功能代码

我检查列表长度的功能如下

int listLength(Node* headPtr){ // pass by value
   int length = 0;

   while(headPtr !=0){
     length++;
     headPtr=headPtr->link;
   }

   return length;
}

Answer 1

while(cur != 0 && headPtr -> link != 0)

如果headPtr是列表中唯一的节点，那么headPtr - ＆gt;事实上，link确实指向null，因此检查这不是解决该情况的最佳方法。您可以使用简单的else if()来应用逻辑，如此

 if(headPtr == 0)       // check for empty list.
    return;

    else if(headPtr -> link == NULL){
         // Do some stuff to delete only this node
    } 

    else { // handle as usual
    // ensures will only run 1 time if there is 1 item in list.

    while(cur != 0 && headPtr -> link != NULL) // ensure there is a next link and more than 1 node in list.
    {
      // The rest of your code
    }
   }

当列表中唯一的节点时删除链表的头部。单身联系

1 个答案: