昨天寫的遞歸有點(diǎn)傻惠勒，嗯姚建，沒注意到k值，改了一下烹棉，代碼如下：
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseKGroup(self, head, k):
"""
:type head: ListNode
:type k: int
:rtype: ListNode
"""
p = head
stack = []
m = k
while p and m:
stack.append(p)
m -= 1
p = p.next
if m > 0:
return head
newhead = ListNode(0)
q = newhead
while stack:
q.next = stack.pop()
q = q.next
q.next = self.reverseKGroup(p, k)
return newhead.next
第二種連接的辦法，真的很6
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseKGroup(self, head, k):
dummy = jump = ListNode(0)
dummy.next = l = r = head
while True:
count = 0
while r and count < k: # use r to locate the range
r = r.next
count += 1
if count == k: # if size k satisfied, reverse the inner linked list
pre, cur = r, l
for _ in range(k):
cur.next, cur, pre = pre, cur.next, cur # standard reversing
jump.next, jump, l = pre, l, r # connect two k-groups
else:
return dummy.next
