如何初始化此数据结构 [英] How to initialize this data structure

问题描述

这是一个双向链接列表.除了下一个和上一个指针之外，每个元素还具有一个子指针，该子指针可以指向也可以不指向单独的双向链接列表.这些子列表可以自己拥有子列表.

像这样的结构:

It is a doubly linked list. In addition to next and previous pointers, each element has a child pointer, which may or may not point to a separate doubly linked list. These child lists could have their child lists on their own.

Structure like this:

struct nodeT
{
   struct nodeT *prev;
   struct nodeT *next;
   struct nodeT *child;
   int value;
}

谁能给我一个提示，我如何初始化这样的结构?有了给定的数据.
谢谢大家〜

好吧，我可以想到的一种方法就是输入此结构的扁平化版本，我可以从扁平化版本转换为目标版本，但是我正在寻找更好的解决方案.

我自己实现了一些代码，不知道它是否有效.
希望你们能给我一些帮助！

Could anyone give me a clue how may I initialize one structure like this? With given data.
Thank you guys~

Well, one way I can think is to just input a flatten version of this structure, I can transform from flatten version to target one, but I am looking for a better solution.

I implement some code myself, no idea it works or not.
I hope you guys can give me some help!

typedef struct nodeT{
struct nodeT *next;
struct nodeT *prev;
struct nodeT *child;
int value;
}node;

//flatten fn:
/*void Flatten(node *head,node **tail)
{
	node *currentNode=head;
	while(currentNode)
	{
		if(node->child)
		{
			Append(currentNode->child,**tail)
		}
		node=node->next;
	}
}
void Append(node *child,**tail)
{
	child->prev=*tail;
	(*tail)->next=child;
	node *curNode=child;
	while(curNode)
	{
		*tail=curNode;
		curNode=curNode->next;
	}
}
*/
void Initialize(node **currentNode)
{
	int data,tag_1,tag_2;
	printf("Input data of current node:\n");scanf("%d",&data);
	(*currentNode)->value=data;
	node **temp_1;node **temp_2;
	printf("Does current node have a child?(1 for yes, 0 for no)\n");scanf("%d",&tag_1);
	printf("Does current node have next node?(1 for yes, 0 for no)\n");scanf("%d",&tag_2);
	if(tag_1==1)
	{
		(*currentNode)->child=(node *)malloc(sizeof(node));
		temp_1=&((*currentNode)->child);	
		Initialize(temp_1);
	}
	if(tag_2==1)
	{
		(*currentNode)->next=(node *)malloc(sizeof(node));
		temp_2=&((*currentNode)->next);
		(*currentNode)=(*temp_1)->prev;
		Initialize(temp_2);
	}
}

推荐答案

请在对问题的评论中看到我的问题.

如果可以，将其设置为C ++，而不是C，然后使用一组构造函数对其进行初始化.另外，请遵循封装原则( http://en.wikipedia.org/wiki/Encapsulation_％28object-oriented_programming％29 [ ^ ])，将这些成员隐藏在访问修饰符private下(类默认为private，但struct默认为public)，将其替换为仅显示所需内容的函数.

它为您提供线索吗?

—SA

Please see my question in my comment to the question.

If you can, make it C++, not C and initialize it using a set of constructors. Also, follow the encapsulation principles (http://en.wikipedia.org/wiki/Encapsulation_%28object-oriented_programming%29[^]), hide these members under access modifier private (private is default for classes, but default for struct is public), replace it with functions which expose only what is needed.

Does it provide you a clue?

—SA

我首先将节点中的每个属性都清除为0.然后，当您导航插入您的节点时，您将始终有一个基准来进行测试下一个节点，您已经找到了树的底部.

您可以简化函数中的几件事，然后可能会更加清楚如何初始化节点.当前编写函数Initialize的方式，您必须像这样调用它:

I would start by clearing out each property in your node to 0. Then you will always have a baseline to test against when you navigate to insert your next node and you have found the bottom of your tree.

You can simplify a few things in your function, then it might be a bit clearer how to initialize your node. The way you currently have the function Initialize written, you have to call it like this:

Node elt;

Initialize(&(&elt));

或

or

Node elt;
Node *pElt = &elt;
Initialize(&pElt);

从要初始化的结构开始.您没有为实际节点分配任何新内存，因此可以删除*之一，并使用单个指针而不是指向指针的指针:

Start with the struct that you will pass into be initialized. You are not allocating any new memory for the actual node, so you can remove one of the *, and use a single pointer rather than a pointer to a pointer:

void Initialize(node *currentNode)

然后，您可以像这样调用该函数:

You can then call the function like this:

Node elt;
Initialize(&elt);

另外，您不必将函数中的每次访问都重新引用到currentNode两次；
第一次:(* currentNode)
第二次:->

您可以将所有

Also, you don''t have to dereference every access in your function to currentNode twice;
first time: (*currentNode)
second time: ->

You can replace all of the

(*currentNode)->

替换为

currentNode->

除此之外，我不确定您将如何初始化这些项目.我可以告诉你，如果用户对第一个问题的回答为否"，那么是否有一个孩子，但对下一个节点为是"，则不会初始化temp_1，并且您将尝试访问未初始化的指针.

另外，我看不到任何初始化您的上一个数据成员的代码.

最后一个技巧，就像对Initialize(¤tNode)的调用一样，如果将temp_1和temp_2的定义更改为使用单个指针而不是像这样的指针，则可以使用指针简化调用: br/>

Beyond that, I am not sure exactly how you need the items to be initialized. I can tell you that if the user answers no for the first question of is there a child, but yes for a next node, temp_1 will not be initialized, and you will attempt to access an uninitialized pointer.

Also, I do not see any code that initializes your prev data member.

One final tip, just like with the call to Initialize(¤tNode), you can simplify your calls with the pointers, if you change the definitions for temp_1 and temp_2 to use a single pointer rather than a pointer to a pointer like this:

node *temp_2;

...

currentNode->next = (node *)malloc(sizeof(node));
temp_2            = currentNode->next;
currentNode       = temp_1->prev;

Initialize(temp_2);

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