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
        ListNode *pre = head;
        ListNode *anext = pre->next;
        ListNode *nnext = anext->next;
        if (pre == nullptr || anext == nullptr)
            return head;
        while (anext != nullptr) {
            anext->next = pre;
            pre = anext;
            anext = nnext;
            nnext = nnext->next;
        }
        head->next = nullptr; # 千万别忘了将原头节点的next指针设为空。
        return pre;
    }
};