Reverse Nodes in k-Group

题目：

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,

Given this linked list:

1->2->3->4->5

For k = 2, you should return:

2->1->4->3->5

For k = 3, you should return:

3->2->1->4->5

解题思路：
# Definition for singly-linked list.

# class ListNode:

# def __init__(self, x):

# self.val = x

# self.next = None

class Solution:

# @param {ListNode} head

# @param {integer} k

# @return {ListNode}

def reverseKGroup(self, head, k):

def reverseList(head,tail):

sentry = ListNode(-1)

sentry.next = tail.next

sentry_tail = tail.next

while head!=sentry_tail:

tmp = head

head = head.next

tmp.next = sentry.next

sentry.next = tmp

return sentry.next

sentry = ListNode(-1)

sentry.next, length = head, 0

tmp = sentry

while tmp.next:

tmp = tmp.next

length += 1

if length == k:

tail=tmp

head_new = tmp.next

if length<k:

return sentry.next

elif length==k:

return reverseList(head,tail)

else:

sentry.next = reverseList(head,tail)

head.next = self.reverseKGroup(head_new, k)

return sentry.next