25. Reverse Nodes in k-Group
才发现没写过这个
def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode
]:
# check length
cur = head
for i in range(k):
if cur:
cur = cur.next
else:
return head
cur, prev = head, self.reverseKGroup(cur, k)
for i in range(k):
nxt = cur.next
cur.next = prev
prev = cur
cur = nxt
return prev
用递回挺好搞得
时隔13个月再次写出hard