[leetcode][25]Reverse Nodes in k-Group
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.
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
# 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
"""
current = head
k_container = []
pre_k_node = None
while current:
next_node = current.next
k_container.append(current)
if len(k_container) == k:
reverse_k_nodes = k_container[::-1]
if pre_k_node is None:
for i in range(len(reverse_k_nodes)-1):
reverse_k_nodes[i].next = reverse_k_nodes[i+1]
head = reverse_k_nodes[0]
pre_k_node = reverse_k_nodes[-1]
else:
pre_k_node.next = reverse_k_nodes[0]
for i in range(len(reverse_k_nodes)-1):
reverse_k_nodes[i].next = reverse_k_nodes[i+1]
pre_k_node = reverse_k_nodes[-1]
k_container = []
current = next_node
if len(k_container) == 0:
if pre_k_node is None:
return None
else:
pre_k_node.next = None
else:
if pre_k_node is None:
return k_container[0]
else:
pre_k_node.next = k_container[0]
return head
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