Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
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
Note:
Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.
class Solution(object):
def reverseKGroup(self, head, k):
count, node = 0, head
while node and count < k:
node = node.next
count += 1
if count < k: return head
new_head, prev = self.reverse(head, count)
head.next = self.reverseKGroup(new_head, k)
return prev
def reverse(self, head, count):
prev, cur, nxt = None, head, head
while count > 0:
nxt = cur.next
cur.next = prev
prev = cur
cur = nxt
count -= 1
return (cur, prev)
leetcode 206的扩展题，主要注意分组的时候新的head在哪，以及记得更新head.next
