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) {
        // write code here
        if(head == nullptr || head->next == nullptr)
            return head;

        if(head->next->next == nullptr){
            head->next = nullptr;
            head->next->next = head;
            return head->next;
        }

        ListNode *s1 = head;
        ListNode *s2 = s1->next;
        ListNode *s = s1->next->next;

        s1->next = nullptr;
        while(s2 != nullptr){
            s2->next = s1;
            s1 = s2;
            s2 = s;
            s = (s ? s->next : nullptr);
        }

        return s1;
    }
};