leetcode 【 Reverse Linked List II 】 python 实现
题目：Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
代码：oj测试通过 Runtime: 65 ms
1 # Definition for singly-linked list.
2 # class ListNode:
3 # def __init__(self, x):
4 # self.val = x
5 # self.next = None
6
7 class Solution:
8 # @param head, a ListNode
9 # @param m, an integer
10 # @param n, an integer
11 # @return a ListNode
12 def reverseBetween(self, head, m, n):
13 if head is None or head.next is None:
14 return head
15 dummyhead = ListNode(0)
16 dummyhead.next = head
17
18 p = dummyhead
19
20 for i in range(m-1):
21 p = p.next
22
23 curr = p.next
24 for i in range(n-m):
25 tmp = curr.next
26 curr.next = tmp.next
27 tmp.next = p.next
28 p.next = tmp
29
30 return dummyhead.next
思路：
不妨拿出四本书，摞成一摞（自上而下为 A B C D），要让这四本书的位置完全颠倒过来（即自上而下为 D C B A）：
盯住书A，每次操作把A下面的那本书放到最上面
初始位置：自上而下为 A B C B
第一次操作后：自上而下为 B A C D
第二次操作后：自上而下为 C B A D
第三次操作后：自上而下为 D C B A
小白觉得四本书的例子，貌似可以更有助于解释代码，欢迎大侠拍砖指导。
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