倒计时144t~冲鸭~~~~~

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

 

两种方法：

1.迭代：

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* reverseList(ListNode* head) {
12         if(head == 0)
13             return NULL;
14         ListNode *p = head->next;
15         head->next = 0;
16         while(p!=0)
17         {
18             ListNode *p1 = p->next;
19             p->next = head;
20             head = p;
21             p = p1;
22         }
23         return head;
24     }
25 };

24ms,9mb

2.递归：

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* reverseList(ListNode* head) {
12      if(head == 0 || head->next == 0)
13          return head;
14         ListNode * pNode = reverseList(head->next);
15         head->next->next = head;
16         head->next = 0;
17         return pNode;
18     }
19 };

16ms,9.3mb