题解 | #链表中的节点每k个一组翻转#
链表中的节点每k个一组翻转
https://www.nowcoder.com/practice/b49c3dc907814e9bbfa8437c251b028e
# 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 k int整型
# @return ListNode类
#
class Solution:
def reverseKGroup(self , head: ListNode, k: int) -> ListNode:
# write code here
n, cur = 0, head
while cur:
n += 1
cur = cur.next
dummy = ListNode(0)
dummy.next = head
p0 = dummy
prev, cur = None, p0.next
while n >= k:
n -= k
for _ in range(k):
nxt = cur.next
cur.next = prev
prev = cur
cur = nxt
nxt = p0.next
p0.next.next = cur
p0.next = prev
p0 = nxt
return dummy.next
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