Leetcode #206. Reverse Linked List
Reverse a singly linked list.
一般，reverse list, reverse list by pair 等都需要三个指针，cur = head, prev, 和当前的下一个temp. 这里在head到达边界之后，还会再走一步，
所以对应新的表头的是prev.
1 class Solution(object): 2 def reverseList(self, head): 3 """ 4 :type head: ListNode 5 :rtype: ListNode 6 """ 7 8 prev = None 9 10 while head: 11 temp = head.next 12 head.next = prev 13 prev = head 14 head = temp 15 16 return prev
