题解 | #链表的中间结点#
链表的中间结点
https://www.nowcoder.com/practice/d0e727d0d9fb4a9b9ff2df99f9bfdd00
/** * 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 */ //采用快慢指针,快的指针到头,慢的指针刚好一半 ListNode* middleNode(ListNode* head) { ListNode* slow = head, *fast = head; while(fast && fast->next){ slow = slow->next; fast = fast->next->next; } return slow; } };