LeetCode 92. Reverse Linked List II--Python 解法--反转部分链表--笔试算法题

题目地址：Reverse Linked List II - LeetCode

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

这道题目是我之前做一家公司笔试时候遇到的算法题，反转部分链表，难度比反转全部链表大。

当时做题的时候不会告诉你错误的样例，现在重新写，在特殊情况的判断中依然会有出错，以后需要重新再做。

Python解法如下：

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:def helper(reverse_head, count):if reverse_head is None:return reverse_headcurr = reverse_headprev = Nonefor _ in range(0, count):next_node = curr.nextcurr.next = prevprev = currcurr = next_nodereverse_head.next = currreturn previf n - m == 0:return headreverse_before_head = headif m == 1:reverse_head = helper(reverse_before_head, n - m + 1)return reverse_headfor _ in range(1, m - 1):reverse_before_head = reverse_before_head.nextreverse_head = helper(reverse_before_head.next, n - m + 1)reverse_before_head.next = reverse_headreturn head

时间复杂度为O(n)，空间复杂度为O(1)。

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