题解 | #反转链表#
反转链表
https://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca
/**
* 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
* @return ListNode类
*/
ListNode* ReverseList(ListNode* head) {
// write code here
if(!head)
return head;
ListNode* pre = head;
ListNode* cur = head->next;
pre->next = nullptr;
while(cur)
{
head = cur;
cur = cur->next;
head ->next = pre;
pre = head;
}
return head;
}
};