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*  dummyNode = new ListNode(-1);
        dummyNode->next = head;

        ListNode* curr = head;
        ListNode* prev = nullptr;

        while(curr != nullptr ){
            ListNode* next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
        }
        return dummyNode->next = prev;
    }
};