给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。

示例 1：

输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]

示例 2：

输入：head = [5], left = 1, right = 1
输出：[5]

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseBetween(self, head: ListNode, left: int, right: int) -> ListNode:
        if left == right: 
            return head
        head = ListNode(0,head)
        position = 0
        begin_point, end_point = None, None #初始化剪切点
        pointer = head
        while(pointer):
            if position + 1 == left:
                begin_point = pointer #保存剪切点
            elif position == right:
                before_end_point = pointer 
                if (pointer.next):
                    end_point = pointer.next #保存剪切点
            position += 1
            pointer = pointer.next

        pointer = begin_point.next
        position = 0
        begin_point.next = before_end_point #左侧与反转链右侧
        reverse_list = end_point #右侧反转链左侧
        for p in range(right-left+1):
            if (pointer):
                temp = pointer.next
                pointer.next = reverse_list
                reverse_list = pointer
                pointer = temp
        return head.next

妈呀这题真的花了我好久的时间
不过最后的结果还是很满意的

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-linked-list-ii