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) {
        ListNode *first, *sec, *temp;
        first = head;
        
        if (head == nullptr || head->next == nullptr) return head;
        sec = head->next;
        head->next = nullptr;
        while (sec) {
            temp = sec;
            sec = sec->next;
            temp->next = first;
            first = temp;
        }
        return first;
        // write code here
    }
};