翻转链表

题目：给出一个链表1->2->3->null，这个翻转后的链表为3->2->1->null

代码：

/**
* Definition of ListNode
*
* class ListNode {
* public:
* int val;
* ListNode *next;
*
* ListNode(int val) {
* this->val = val;
* this->next = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param head: The first node of linked list.
* @return: The new head of reversed linked list.
*/
ListNode *reverse(ListNode *head) {
// write your code here
if (head == NULL)
return NULL;
ListNode *pre = NULL;
ListNode *tmp = head->next;
while (head)
{
tmp = head->next;
head->next = pre;
pre = head;
head = tmp;
}
return pre;
}
};

截图：