237. Delete Node in a Linked List
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
Supposed the linked list is 1 -> 2 -> 3 -> 4
and you are given the third node with value 3
, the linked list should become 1 -> 2 -> 4
after calling your function.
题意:
删除链表指定节点。前提是只传删除节点给函数。但不包括删除尾节点。
思路:
由于没有链表前节点的存在,所以删除链表时无法改变前节点的指向。但是链表值是int型的,所有可以把当前节点的val和下个节点的val交换。然后删除下个节点即可。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
void deleteNode(struct ListNode* node)
{
if ( node->next == NULL )
{
return;
}
int tmp = 0;
tmp = node->val;
node->val = node->next->val;
node->next->val = tmp;
struct ListNode *list = node->next;
node->next = node->next->next;
free(list);
}
亿速云「云服务器」,即开即用、新一代英特尔至强铂金CPU、三副本存储NVMe SSD云盘,价格低至29元/月。点击查看>>
免责声明:本站发布的内容(图片、视频和文字)以原创、转载和分享为主,文章观点不代表本网站立场,如果涉及侵权请联系站长邮箱:is@yisu.com进行举报,并提供相关证据,一经查实,将立刻删除涉嫌侵权内容。