题解 | #删除链表的节点#
删除链表的节点
http://www.nowcoder.com/practice/f9f78ca89ad643c99701a7142bd59f5d
简单的链表删除
/**
* 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* h = new ListNode(-1),*he = h;
h->next = head;
while(h->next){
if(h->next->val == val){
ListNode* tmp = h->next;
h->next = h->next->next;
free(tmp);
}
h = h->next;
}
return he->next;
}
};