用 Python 编写一个按大小为 k 的组反转链表的程序

pythonserver side programmingprogramming更新于 2026/2/3 7:40:17

假设我们有一个单链表和另一个值 k，我们必须反转每 k 个连续的节点组。

因此，如果输入为 List = [1,2,3,4,5,6,7,8,9,10], k = 3，则输出为 [3, 2, 1, 6, 5, 4, 9, 8, 7, 10, ]

为了解决这个问题，我们将遵循以下步骤 −

tmp := 一个值为 0 的新节点
tmp 的下一个节点 := 节点
prev := null, curr := null
lp := temp, lc := curr
cnt := k
当 curr 不为空时，执行
- prev := null
- 当 cnt > 0 且 curr 不为空时，执行
  - following := next of curr
  - next of curr := prev
  - prev := curr, curr := following
  - cnt := cnt - 1
- next of lp := prev, next of lc := curr
- lp := lc, lc := curr
- cnt := k
返回 tmp 的下一个

让我们看看下面的实现以便更好地理解 −

示例

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, k):
      tmp = ListNode(0)
      tmp.next = node
      prev, curr = None, node
      lp, lc = tmp, curr
      cnt = k
      while curr:
         prev = None
         while cnt > 0 and curr:
            following = curr.next
            curr.next = prev
            prev, curr = curr, following
            cnt -= 1
         lp.next, lc.next = prev, curr
         lp, lc = lc, curr
         cnt = k
      return tmp.next
ob = Solution()
head = make_list([1,2,3,4,5,6,7,8,9,10])
print_list(ob.solve(head, 3))