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* newHead = nullptr;
        if (head != nullptr) {
            newHead = head;
            head = head->next;
        }else {
            return nullptr;
        }
        newHead->next = nullptr;
        while (head != nullptr) {
            ListNode* first = head->next;
            head->next = newHead;
            newHead = head;
            head = first;
        }
       return newHead;
    }
};