在链表中移动指针

Question

问题:给定一个有三个指针的链表:第一个指向第一个节点，第二个指向第三个节点，第三个指向最后一个节点。返回指向同一列表的单个指针，使第5个指针是第一个，第一个指针是最后一个。

这是课堂上给我们的一个问题，我理解这个问题有困难，但这是我尝试的解决方案。

//List is the first pointer
//p is the second pointer (pointing to the third node)
//q is the last pointer (pointing to the last node)
 R = p -> next //R, a name to a pointer i gave that is between p and q
 p -> next = R -> next // don't even know what this means but wrote it down anyways

在此之后我卡住了，任何帮助都很感激，但我会感激完整的解决方案。

我将进一步欣赏使用STL

Answer 1

可以这样做:

list *tmp, *tmp2, *pf, p3, pl;    //pf: first, p3: 3rd, pl: last;
tmp = p3->next->next;    //pointer of the 4th element to the 5th;
p3->next->next = tmp->next;    //4th element now pointing to the 6th (since 5th moves to the beggining);
pl->next = pf;    //make the first the last;
tmp2 = pf->next;    //save the pointer to the 2nd;
pf->next = NULL;    //pf is now last (-> pf->next has to be NULL);
tmp->next = tmp2;    //old 5th now pointing to the 2nd (as it should be the first);
pf = tmp;    //make the 5th the first -> this is what you want to return;

这里基本上做了什么:你把第5个元素(它的指针)拿出来，因此你必须把4和6连接起来。现在把第一个放在最后。这很容易，因为last->next是NULL。你现在要做的最后一件事就是把5号变成1号。你需要让它指向2。就是这样。据我所知，您应该将其包装在一个函数中，该函数首先返回