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?

解決思路：使用原地改變鏈表的指針進行反轉。時間復雜度為O(n)，空間復雜度為O(1). 流程圖如下：

 1 class Solution(object):
 2     def reverseList(self, head):
 3         """
 4         :type head: ListNode
 5         :rtype: ListNode
 6         """
 7         pre = None   # 定義前驅節點
 8         cur = head   #  定義後驅節點
 9         while cur:     # 當cur為空時，循環結束
10             tem = cur.next       #
 
 使用臨時變量保存下一個節點
11             cur.next = pre       # 指向前驅節點
12             pre = cur            # 前驅節點復制到當前節點
13             cur = tem            # 臨時變量進行復制
14         return pre

【LeetCode每天一題】Reverse Linked List(鏈表反轉)