题解 | #删除链表的节点#
删除链表的节点
https://www.nowcoder.com/practice/f9f78ca89ad643c99701a7142bd59f5d
/**
* struct ListNode {
* int val;
* struct ListNode *next;
* ListNode(int x) : val(x), next(nullptr) {}
* };
*/
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param head ListNode类
* @param val int整型
* @return ListNode类
*/
ListNode* deleteNode(ListNode* head, int val) {
// write code here
ListNode* start = new ListNode(-1); // 造一个不存实际数值的头结点，然后遍历查找
start->next = head;
head = start;
for (auto i = head; i; i = i->next)
{
auto p = i->next;
if (p->val == val)
{
i->next = p->next;
break;
}
}
return head->next;
}
};