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) return nullptr;
        auto *prev = new ListNode(head->val);

        ListNode *node = head->next;
        while(node!=nullptr){
                auto *next = new ListNode(node->val);
                next->next = prev;
                prev = next;
                node = node->next;
        }
        cout << endl;

        return prev;
    }
};