Reverse a singly linked list.
Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL
Follow up:
A linked list can be reversed either iteratively or recursively. Could you implement both?
解决思路：使用原地改变链表的指针进行反转。时间复杂度为O(n)，空间复杂度为O(1). 流程图如下：
1 class Solution(object):
2 def reverseList(self, head):
3 """
4 :type head: ListNode
5 :rtype: ListNode
6 """
7 pre = None # 定义前驱节点
8 cur = head # 定义后驱节点
9 while cur: # 当cur为空时，循环结束
10 tem = cur.next # 使用临时变量保存下一个节点
11 cur.next = pre # 指向前驱节点
12 pre = cur # 前驱节点复制到当前节点
13 cur = tem # 临时变量进行复制
14 return pre