r = r.next
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))
d[curr] = True
curr = curr.next
return False
# LC SOLN
# Using slow and fast pointers to check for cycle
# n time but O(1) space
def B(head):
if not head or not head.next:
return False
slow = head
fast = head.next
while slow != fast:
if not fast or not fast.next:
return False
slow = slow.next
fast = fast.next.next
return True
from LINK import makeLL, printLL, getTail
a = [3, 2, 0, -4]
head = makeLL(a)
tail = getTail(head)
# Make cycle between -4 -> 2
tail.next = head.next
print B(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]))