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25. Reverse Nodes in k-Group.py
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25. Reverse Nodes in k-Group.py
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'''
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.
'''
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
######## RECURSIVE ###########
def reverseKGroup(self, head, k):
if head==None:
return head
i=0
res = None
cur1 = head
while(i<k and cur1):
cur1 = cur1.next
i+=1
if i<k:
return head
i=0
#res = None
cur = head
while i<k and cur:
temp = cur.next
cur.next = res
res = cur
cur = temp
i += 1
if cur:
head.next = self.reverseKGroup(cur,k)
return res
######## ITERATIVE #############
def reverseKGroup_iter(self, head,k):
jump = ListNode(0)
dummy = jump
l = head
r = l
#dummy.next = r
while True:
count = 0
while count<k and r:
r = r.next
count += 1
if count==k:
pre , cur = r, l
for _ in range(k):
temp = cur.next
cur.next = pre
pre = cur
cur = temp
jump.next = pre
jump = l
l=r
else:
return dummy.next