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
need a helper function to reverse k nodes and then put the reversed list together. this should
be O(n) time.
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseNodes(self, startNode, endNode):
dummyNode = ListNode(None)
dummyNode.next = startNode
while dummyNode.next != endNode:
tmpNode = startNode.next
startNode.next = tmpNode.next
tmpNode.next = dummyNode.next
dummyNode.next = tmpNode
return endNode, startNode
def reverseKGroup(self, head, k):
"""
:type head: ListNode
:type k: int
:rtype: ListNode
"""
if not head:
return head
dummyNode = ListNode(None)
dummyNode.next = head
startNode = head
while startNode.next:
endNode = startNode
for count in range(k - 1):
endNode = endNode.next
if not endNode.next:
return dummyNode
newstart, newend = self.reverseNodes(startNode, endNode)
startNode.next = newstart
startNode = newend
return dummyNode.next