题目
解题
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def reverseKGroup(head, k):
"""
翻转链表中每 k 个节点一组的节点。
:param head: 链表的头节点
:param k: 每组翻转的节点数量
:return: 翻转后的链表头节点
"""
def getKthNode(current, k):
"""
获取链表中第 k 个节点。
:param current: 当前节点
:param k: 节点数量
:return: 第 k 个节点或 None（如果不足 k 个节点）
"""
while current and k > 1:
current = current.next
k -= 1
return current
def reverseLinkedList(head, k):
"""
翻转链表的一部分。
:param head: 部分链表的头节点
:param k: 节点数量
:return: 翻转后的链表头节点
"""
previous, current = None, head
while k:
next_node = current.next
current.next = previous
previous = current
current = next_node
k -= 1
return previous
# 找到第 k 个节点
kth_node = getKthNode(head, k)
if not kth_node:
return head
next_group_head = kth_node.next
kth_node.next = None
# 翻转当前 k 个节点
new_head = reverseLinkedList(head, k)
# 递归处理剩余链表，并连接
head.next = reverseKGroup(next_group_head, k)
return new_head
def listToListNode(arr):
"""
将 Python 列表转换为链表。
:param arr: Python 列表
:return: 链表的头节点
"""
dummy = ListNode()
current = dummy
for val in arr:
current.next = ListNode(val)
current = current.next
return dummy.next
def listNodeToList(head):
"""
将链表转换为 Python 列表。
:param head: 链表的头节点
:return: Python 列表
"""
result = []
while head:
result.append(head.val)
head = head.next
return result
def testReverseKGroup():
"""
测试 reverseKGroup 函数。
"""
test_cases = [
([1, 2, 3, 4, 5], 2),
([1, 2, 3, 4, 5], 3),
([1, 2, 3, 4, 5], 1),
([1], 1),
([1, 2], 2)
]
for i, (input_list, k) in enumerate(test_cases):
head = listToListNode(input_list)
new_head = reverseKGroup(head, k)
output_list = listNodeToList(new_head)
print(output_list)
# 运行测试代码
testReverseKGroup()