题解 | #反转链表#

/**
 * 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 == nullptr || head->next == nullptr) {
            return head;
        }
        ListNode *newHead = ReverseList(head->next);
        ListNode *tail = head, *cur = newHead;
        while (cur->next) {
            cur = cur->next;
        }
        cur->next = tail;
        tail->next = nullptr;
        return newHead;
    }
};