单链表以k个节点为一组进行翻转
解题思路：
首先把前K个结点看成一个子链表，采用前面介绍的方法进行翻转，把翻转后的子链表链接到头结点后面，然后把接下来的K个结点看成另外一个单独的链表进行翻转，把翻转后的子链表链接到上一个已经完成翻转子链表的后面
图示：
代码实现
# -*-coding:utf-8-*-
"""
@Author : 图南
@Software: PyCharm
@Time : 2019/9/7 11:37
"""
class Node:
def __init__(self, data=None, next=None):
self.data = data
self.next = next
def con_link(n):
head = Node()
cur = head
for i in range(1, n + 1):
node = Node(i)
cur.next = node
cur = node
return head
def print_link(head):
if head is None or head.next is None:
return
cur = head.next
while cur:
print(cur.data, end=" ")
cur = cur.next
print()
def reverseKNode(head, k):
if head is None or head.next is None:
return
if k == 0 or k == 1:
return head
pre = head
begin = head.next
while begin:
end = begin
for i in range(k-1):
if end.next:
end = end.next
else:
return head
next= end.next
end.next = None
r_head, r_end = reverseLink(begin)
r_end.next = next
pre.next = r_head
pre = r_end
begin = next
return head
def reverseLink(head):
if head is None or head.next is None:
return
r_end = head
pre = head
r_head = pre.next
pre.next = None
while r_head.next:
next = r_head.next
r_head.next = pre
pre = r_head
r_head = next
r_head.next = pre
return r_head, r_end
if __name__ == '__main__':
n = int(input("请输入n："))
k = int(input("请输入k："))
head = con_link(n)
print_link(head)
head = reverseKNode(head, k)
print_link(head)
运行结果：
