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) {
    //    if(head == nullptr || head -> next == nullptr) return head;
    //    ListNode* prev = ReverseList(head->next);
    //    head->next -> next= head;
    //    head-> next = nullptr;
    //    return prev;
    ListNode* prev = nullptr;
    ListNode* next;
    while(head != nullptr)
    {
        next = head ->next;
        head ->next = prev;
        prev = head;
        head = next;

    }
    return prev;
    }
};

画图即可