题解 | #链表的中间结点# | C++

/**
 * 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* middleNode(ListNode* head) {
        ListNode *slow = head, *fast = head;
        while (fast!=nullptr) {
            if (fast->next == nullptr) break;;
            fast = fast->next->next;
            slow = slow->next;
        }
        return slow;
    }
};