正在链表拷贝正确的这种方法吗？ [英] Is this method of making copy of a linked list correct?

问题描述

void InsertAtTail(struct node** headref,int val)
{
    struct node *current,*newnode;
    current=*headref;
    newnode=malloc(sizeof(struct node));

    if(current==NULL)
    {
        newnode->data=val;
        newnode->next=NULL;
        *headref=newnode;
        current=*headref;
    }

    else
    {

        while(current->next!=NULL)
        {
            current=current->next;
        }

        newnode->data=val;
        newnode->next=NULL;
        current->next=newnode;
    }
}

struct node* CopyList(struct node* headref)
{
    struct node* newlist=NULL;
    struct node* current;
    current=headref;


    if(current==NULL)
    {
        newlist=current;
    }

    else
    {
        while(current!=NULL)
        {
            InsertAtTail(&newlist, current->data);
            current=current->next;
        }

    }
    return (newlist);
}

我所经历的斯坦福大学的CS101笔记，发现code制作链表的副本。但所使用的指针的尾节点为好。我写了这个code，而不使用（尾指针）。我是新来的链接列表。请告诉我，如果我可以继续这种方式为好。当我打印的原件及复印件的地址，两人都不同。我正在使用X code C。

I was going through Stanford's CS101 notes and found code for making a copy of linked list. But it used a pointer to the tail node as well. I have written this code without using that(tail pointer). I am new to linked list. Please tell me if I can proceed this way as well. When I printed the original and copy's addresses, both were different too. I am using c in Xcode.

推荐答案

作品不正，虽短：

void InsertAtTail(struct node** ref,int val)
{
    while (*ref != NULL) {
        ref = &(*ref)->next;
    }
    struct node *newnode = malloc(sizeof(struct node));
    newnode->data=val;
    newnode->next=NULL;
    *ref = newnode;
}

和列表复制应该改写。N²/ 2

And the list copying should be rewritten: N²/2.

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