[LeetCode]题解（python）：092 Reverse Linked List II

题目来源

---

https://leetcode.com/problems/reverse-linked-list-ii/

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.

---

题意分析

---

Input:一个链表

Output:翻转后的链表

Conditions:给定翻转的起始位置和终止位置，翻转链表

---

题目思路

---

关键是理清变换的思路，注意在链表的操作中，经常设置一个空的头节点。

代码思路如图（注意哪些在变化，至于旋转了，程序员你懂的）

---

AC代码（Python）

 # Definition for singly-linked list.
 # class ListNode(object):
 #     def __init__(self, x):
 #         self.val = x
 #         self.next = None

 class Solution(object):
     def reverseBetween(self, head, m, n):
         """
         :type head: ListNode
         :type m: int
         :type n: int
         :rtype: ListNode
         """
         if head == None or head.next == None:
             return head
         newHead = ListNode(0)
         newHead.next = head
         head1 = newHead
         for i in range(m - 1):
             head1 = head1.next
         p = head1.next
         print p.val
         for i in range(n - m):
             tmp = head1.next
             head1.next = p.next
             p.next = p.next.next
             head1.next.next = tmp
             print p.val, head1.val
         return newHead.next