206. 反转链表

解法1

class Solution
{
public:ListNode *reverseList(ListNode *head){if(!head || !head->next)return head;ListNode *p;p = reverseList(head->next);head->next->next = head;head->next = nullptr;return p;}
};

解法2

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* reverseList(ListNode* head) {if(!head || !head->next) return head;ListNode *p = head, *res =nullptr;while(p){ListNode *curr = p->next;p->next = res;res = p;p = curr;}return res;}
};