elif l.val < r.val:
s.append(l)
l = l.next
elif l.val > r.val:
s.append(r)
r = r.next
# Append the rest of l
while l != None:
s.append(l)
l = l.next
# Append the rest of r
while r != None:
s.append(r)
r = r.next
for i in range(1, len(s)):
first = s[i - 1]
second = s[i]
first.next = second
return s[0]
from LINK import makeLL, printLL
a = [1, 3, 5, 7, 9]
b = [2, 4, 6, 8, 10]
headA = makeLL(a)
headB = makeLL(b)
printLL(A(headA, headB))
return head
# 2pass n using dummy node
def B(head, n):
# Use dummy to make it easy to return
dummy = node(0)
dummy.next = head
length = 0
curr = head
while curr:
length += 1
curr = curr.next
nodeToRem = length - n
curr = dummy
while nodeToRem > 0:
nodeToRem -= 1
curr = curr.next
curr.next = curr.next.next
return dummy.next
from LINK import makeLL, printLL
a = [1, 2, 3, 4, 5]
head = makeLL(a)
printLL(A(head, 2))
# Reverse right part of LL
revLeft = []
while left:
revLeft.append(left.pop())
# Reorder based on constraint
ret = []
while revLeft or right:
if revLeft:
ret.append(revLeft.pop())
if right:
ret.append(right.pop())
# Assign next values
for i in range(1, len(ret)):
ret[i - 1].next = ret[i]
ret[-1].next = None
# Return head
return head
# a=[1,2,3,4] # [1,4,2,3]
# a=[1,2,3,4,5] # [1,5,2,4,3]
a = [1, 2, 3, 4, 5, 6] # [1,6,2,5,3,4]
from LINK import makeLL, printLL
head = makeLL(a)
head = A(head)
printLL(head)
if not ret: return ret[0] = node(ret[0]) # kntime for i in range(1,len(ret)): ret[i] = node(ret[i]) ret[i-1].next = ret[i] return ret[0] import heapq as hq def B(lists): h = [] for i in lists: head = i while head: h.append(head.val) head = head.next hq.heapify(h) return h from LINK import makeLL, printLL a=[1,4,5] b=[1,3,4] c=[2,6] headA=makeLL(a) headB=makeLL(b) headC=makeLL(c) printLL(A([headA,headB,headC]))
# n via recursion WIP def B(head): head = B_helper(head,None) return head def B_helper(node,prev): if node == None: return prev tmp = node nextN = node.next node.next = prev return B_helper(nextN, tmp) # n using prev variable def C(head): prev = None curr = head while curr: tmp = curr.next curr.next = prev prev = curr curr = tmp return prev # because curr is None, and prev is the new head (formerly tail) from LINK import makeLL, printLL a=[1,2,3,4,5,6,7] head=makeLL(a) printLL(A(head))