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) {
ListNode *first, *sec, *temp;
first = head;
if (head == nullptr || head->next == nullptr) return head;
sec = head->next;
head->next = nullptr;
while (sec) {
temp = sec;
sec = sec->next;
temp->next = first;
first = temp;
}
return first;
// write code here
}
};
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