Question

我有以下代码来反转链表。我在while循环中感到困惑，所以如果有人可以提供有关其实际工作方式的直观解释，我一定会感激不尽。

 static void Reverse (struct node** headRef)
{
     struct node* result = NULL;
     struct node* current = *headref;
     struct node* next;

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

        current = next;
     }     
     *headRef = result;

}

Answer 1

好的，这是我试图让valya的答案更清晰（尽管我认为它已经相当不错了）：

说我们有这个清单：

// a->b->c->d->e->NULL

我们从第一个节点a开始，该节点包含指向next的指针b：

// a->b ...

第next = current->next;行将next设置为b（足够简单）。下一行current->next = result;执行此操作：

// NULL<-a  b ... (notice there is no longer a pointer from a to b)

然后我们result = current;将result设置为a（再次，足够简单）。最后，我们拥有所有重要的current = next;，将current设置为b。

因此，在while循环的下一次迭代中，next设置为b，result设置为a，current设置为{{ 1}}，我们重新开始：

然后我们再做一次：

next = current->next;

// NULL<-a<-b  c ...
current->next = result;

result = current;

一旦我们到达链表中的最后一项（本例中为next = current->next; // NULL<-a<-b<-c d ... current->next = result; result = current;），就会发生这种情况：

现在，因为next = current->next; // next becomes NULL // NULL<-a<-b<-c<-d<-e current->next = result; result = current; // result is now e current = next; // current is now NULL为NULL，while循环终止，我们留下：

current

正如您现在所看到的那样，*headRef = result;指向headRef，将e视为链接列表中的新第一项，e指向{ {1}}，e->next指向d等。

Answer 2

我用dot制作了一个图表，我认为它将以图形方式解释正在发生的事情：

thumbnail

Link to full-sized image

如果有人关心的话，这是（草率）点源：

digraph g {
    label = "Start"
    subgraph cluster_1
    {
        a1 -> b1 -> c1;
        current1 -> a1;
        result1
        a1 [label="a"]
        b1 [label="b"]
        c1 [label="c"]
        current1 [label="current"]
        result1 [label="result"]
    }
    label = "Once through loop"
    subgraph cluster_2
    {
        current2 -> b2;
        result2 -> a2;
        b2 -> c2;
        a2 [label="a"]
        b2 [label="b"]
        c2 [label="c"]
        current2 [label="current"]
        result2 [label="result"]
    }
    label = "Twice through loop"
    subgraph cluster_3
    {
        current3 -> c3;
        result3 -> b3;
        b3 -> a3;
        a3 [label="a"]
        b3 [label="b"]
        c3 [label="c"]
        current3 [label="current"]
        result3 [label="result"]
    }
    label = "Final"
    subgraph cluster_4
    {
        result4 -> c4 -> b4 -> a4;
        a4 [label="a"]
        b4 [label="b"]
        c4 [label="c"]
        current4 [label="current"]
        result4 [label="result"]
    }
    label = ""

}

Answer 3

查看this网站以获取直观表示看起来好像是一个很好的代码项目解释here（见技术3）。

Answer 4

列表看起来像：

=>(1) => => => => => => => => =>(10)

我们颠倒了列表的每一部分

<=(1) => => => => => => => => =>(10)
<=(1) <= => => => => => => => =>(10)
<=(1) <= <= => => => => => => =>(10)
...
<=(1) <= <= <= <= <= <= <= <= <=(10)

所以，现在开始到底，我们可以从不同的角度查看列表，看看：

=>(10) => => => => => => => => =>(1)

Answer 5

有关详细说明，请参阅here。以下是摘录：

public class List {
    private class Node {
        int data;
        Node next;
    }
   private Node head;

   public void reverse() {
      if (head == null) return;
      Node prev = null,curr = head,next = null;
      while (curr != null) {
         next = curr.next;
         curr.next = prev;
         prev = curr;
         curr = next;
      }
      head = prev;
   }
}

The list:
==========
1->2->3->4->5
------------------------------------------------------------------
Running of the code:
============================
At start:
**********
head = 1;
prev = null,curr = 1, next = null
1st iteration of while loop:
******************************
next = 2;
2->3->4->5
curr.next = null;
1->null
prev = 1;
1->null
curr = 2;
2->3->4->5
2nd iteration of while loop:
******************************
next = 3
3->4->5
curr.next = 1
2->1->null
prev = 2
2->1->null
curr = 3
3->4->5
3rd iteration of while loop:
******************************
next = 4
4->5
curr.next = 2
3->2->1->null
prev = 3
3->2->1->null
curr = 4
4->5
4th iteration of while loop:
******************************
next = 5
5->null
curr.next = 3
4->3->2->1->null
prev = 4
4->3->2->1->null
curr = 5
5->null
5th iteration of while loop:
******************************
next = null
null
curr.next = 4
5->4->3->2->1->null
prev = 5
5->4->3->2->1->null
curr = null
There is no 6th iteration and the while loop terminates since 
curr is null and the check is for curr != null.
last statement: 
==================
head = prev
This statement leaves the list reversed with referencing head with the reversed node prev 
to become the new head node.

可视化解释链接列表数据结构代码反转所需的指导？

5 个答案: