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) {
        //处理空链表和单节点链表
        if (head == nullptr || head->next == nullptr) {
            return head;
        }

        ListNode* prev = nullptr;
        ListNode* current = head;
        ListNode* next = nullptr;
        while (current != nullptr) {
            next = current->next;//保存下一个
            current->next = prev;//反转当前节点
            prev = current;
            current = next;
        }
        return prev;
    }
};