在Python中反转链表的内部节点的程序

假设我们有一个链表，还有两个值i和j，我们必须从第i个节点到第j个节点反转链表。最后返回更新后的列表。

因此，如果输入是[1,2,3,4,5,6,7,8,9]，i = 2，j = 6，则输出将为[1, 2, 7, 6, 5, 4, 3, 8, 9，]。

为了解决这个问题，我们将遵循以下步骤：

prev_head：创建一个带有与null相同值的链表节点，并指向节点
prev：prev_head，curr：node
迭代从0到i的所有值，做：
- prev：curr，curr：curr的下一个
rev_before：prev，rev_end：curr
迭代0到（j-i）的所有值，做：
- tmp：curr的下一个
- curr.next：prev
- prev，curr：curr，tmp
rev_before.next：prev，rev_end.next：curr
返回prev_head的下一个

让我们看以下实现以更好地理解：

样例

class ListNode:
   def __init__(self, data, next = None):
      self.val = data
      self.next = next

def make_list(elements):
   head = ListNode(elements[0])
   for element in elements[1:]:
      ptr = head
      while ptr.next:
      ptr = ptr.next
      ptr.next = ListNode(element)
   return head

def print_list(head):
   ptr = head
   print('[', end = "")
   while ptr:
      print(ptr.val, end = ", ")
      ptr = ptr.next
   print(']')

class Solution:
   def solve(self, node, i, j):
      prev_head = ListNode(None, node)
      prev, curr = prev_head, node

      for _ in range(i):
         prev, curr = curr, curr.next
   
      rev_before, rev_end = prev, curr

      for _ in range(j - i + 1):
         tmp = curr.next
         curr.next = prev
         prev, curr = curr, tmp

      rev_before.next, rev_end.next = prev, curr

      return prev_head.next

ob = Solution()
head = make_list([1,2,3,4,5,6,7,8,9])
i = 2
j = 6
print_list(ob.solve(head, i, j))