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) { if(!head || !head->next) return head; ListNode* Pre = nullptr,*Phead = head,*temp = nullptr; while (Phead) { temp = Phead->next; Phead->next = Pre; Pre = Phead; Phead = temp; } return Pre; } };