将节点添加到链表的前面 [英] Adding node to the front of a linked list

问题描述

我真的很困惑，这是怎么回事。

我有函数

  void addToFront（int data）
 {
 Node * tmp = new Node（）; 
 tmp  - > data = data; 
 tmp  - >下一页
 head = tmp; 
} 
  

所以当我们执行 tmp-> next = head ，我们使 tmp 指针指向头指向（列表的当前第一个元素）？因为这是它的感觉，但不是只是让它指向 head ？然后，当我们 head = tmp 时，我们正在指向我们创建的新节点，正确？

解决方案

这是您的列表（假设）

  + + ----- + + ----- + ----- + + ----- + ------ + 
 | A | +  - | +  - | | 
 + ---- + ----- + + ----- + ----- + + ----- + ------ + 
 / 
  

头指向 A

$

 <$> 
 c $ c> + ---- + ---- + 
 | B | | 
 + ---- + ---- + 
 / 
  

code> tmp 现在指向 B 。在 tmp-> data 中插入数据，然后使 tmp - > next = head; 。现在我们有：

  + ---- + ---- + + ---- + -  + + ----- + ----- + + ----- + ------ + 
 | B | next | ----> | A | +  - | +  - | | 
 + ---- + ---- + + ---- + ----- + + ----- + ----- + + ----- + --- --- + 
 / 
  

然后你使 / code>指向 B 而不是 A 。

  + ---- + ---- + + ---- + ----- + + ----- + ----- + + ----- + ------ + 
 | B | next | ----> | A | +  - | +  - | | 
 + ---- + ---- + + ---- + ----- + + ----- + ----- + + ----- + --- --- + 
 / \ 
  

head & tmp 现在指向 B 。

I'm really confused as to what exactly is going on here..

I have the function

void addToFront(int data)  
{
  Node* tmp = new Node();
  tmp -> data = data;
  tmp -> next = head;
  head = tmp;
}

So when we do the line tmp-> next = head, we are making the tmp pointer point to what head is pointing to (the current first element of the list)? Because that's what it feels like, but wouldn't that just make it point to head? And then when we do head = tmp, we are making head point to the new node that we created, correct?

解决方案

This is how your list was (suppose)

 +----+-----+   +-----+-----+   +-----+------+
 |  A |     +-->|     |     +-->|     |      |
 +----+-----+   +-----+-----+   +-----+------+
 /

Head is pointing to location A.

You make a new node

 +----+----+
 | B  |    |
 +----+----+
 /

tmp now points to B. You insert data in tmp->data, and then make tmp -> next = head;. So now we have:

 +----+----+     +----+-----+   +-----+-----+   +-----+------+
 | B  |next|---->|  A |     +-->|     |     +-->|     |      |
 +----+----+     +----+-----+   +-----+-----+   +-----+------+
                  /

And then you makehead to point to B instead of A.

 +----+----+     +----+-----+   +-----+-----+   +-----+------+
 | B  |next|---->|  A |     +-->|     |     +-->|     |      |
 +----+----+     +----+-----+   +-----+-----+   +-----+------+
 /  \

head & tmp point to B now.

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