题解 | #链表内指定区间反转#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @param m int整型 
# @param n int整型 
# @return ListNode类
#
class Solution:
    def reverseBetween(self , head: ListNode, m: int, n: int) -> ListNode:
        # 当只反转一个时，就不需要任何操作了
        if m == n:
            return head
        
		# 遍历链表
        a = []
        while head != None:
            a.append(head.val)
            head = head.next
    
		# 获取反转后的值排列，这步是关键
        a = a[:m-1]+a[m-1:n][::-1]+a[n:]

		# 创建新的链表
        newL = ListNode(a[0])
        res = newL
        for j in a[1:]:
            newL.next = ListNode(j)
            newL = newL.next

        # 记得添加这个
        newL.next = None
        return res