0

我有以下短代码，它是解决链表反转问题的解决方案。

void backwardslist(atom** head) {
atom* first;
atom* second;

if (*head == NULL) return; //if list is empty

first = *head;
second = first->next; // intuitive

if (second == NULL) return; 

backwardslist(&second); // recursive call with 2nd one as head, after we got variables
                     first and second

first->next->next = first; // when we get to the end, we rearrange it
first->next = NULL; // so last one is pointing to first, first is pointing to NULL

*head = second; // I dont understand this part, so the head is changing from the last, 
                   to the second element as the recursion goes to the beginning or am i 
                   missing something?

}

不是第二个=（指向递归中两个指针中的第二个的指针）吗？

所以第一次，我明白，它应该指向最后一个，

但是随着递归的建立，它不断地改变*头到秒。

正在使用的第二个自动取款机中有什么？

感谢你们

4

2 回答 2

0

void recursiveReverse(struct node** head_ref) { struct node* first; struct node* rest;

/* empty list */
if (*head_ref == NULL)
   return;  

/* suppose first = {1, 2, 3}, rest = {2, 3} */
first = *head_ref; 
rest  = first->next;

/* List has only one node */
if (rest == NULL)
   return;  

/* reverse the rest list and put the first element at the end */
recursiveReverse(&rest);
first->next->next  = first; 

/* tricky step -- see the diagram */
first->next  = NULL;         

/* fix the head pointer */
*head_ref = rest;

}

于 2014-04-19T21:25:18.157 回答
0

想到的一个简单答案是递归调用您的函数，直到到达终点。然后返回最后一个节点。当递归函数返回时，将返回节点的next指针设置为头节点。

  1) A->B->C->D
    2) A->B->C->D->C
    3)  A->B->C->D->C->B
    4)  A->B->C->D->C->B->A
    5)           D->C->B->A->Null
于 2013-08-25T22:48:00.830 回答