Top MCQs on Linked List Data Structure with Answers

Question 41

Consider the queues Q1 containing four elements and Q2 containing none (shown as the Initial State in the figure). The only operations allowed on these two queues are Enqueue(Q, element) and Dequeue(Q). The minimum number of Enqueue operations on Q1 required to place the elements of Q1 in Q2 in reverse order (shown as the Final State in the figure) without using any additional storage is___________. 

• 12

• 9

• 4

• 0

Question 42

Correct Program to find Middle of a Linked list ?

• class Node:
     def __init__(self, k):
         self.data = k
         self.next = None

  def printList(head):
     curr = head
     while curr != None:
         print(curr.data)
         curr = curr.next
     print()
  
  def printMiddle(ptr):
     if head == None:
         return
     count = 0 
     curr = head
     while curr :
         curr = curr.next
         count+=1
  
     curr = head
     for i in range (count//2):
         curr = curr.next
     print(curr.data)  

  head = Node(10)
  head.next = Node(10)
  head.next.next = Node(20)
  printList(head)
  printMiddle(head)

• class Node:
     def __init__(self, k):
         self.data = k
         self.next = None

  def printList(head):
     curr = head
     while curr != None:
         print(curr.data)
         curr = curr.next
     print()
  def printMiddle(ptr):
     if head == None:
         return
     count = 0 
     curr = head
     while curr :
         curr != curr.next
         count+=1
  
     curr = head
     for i in range (count//2):
         curr = curr.next
     print(curr.data)
  

  head = Node(10)
  head.next = Node(10)
  head.next.next = Node(20)
  printList(head)
  printMiddle(head)

• class Node:
     def __init__(self, k):
         self.data = k
         self.next = None

  def printList(head):
     curr = head
     while curr == None:
         print(curr.data)
         curr = curr.next
     print()

  
  
  def printMiddle(ptr):
     if head == None:
         return
     count = 0 
     curr = head
     while curr :
         curr = curr.next
         count+=1
  
     curr = head
     for i in range (count//2):
         curr = curr.next
     print(curr.data)
  
  

  head = Node(10)
  head.next = Node(10)
  head.next.next = Node(20)
  printList(head)
  printMiddle(head)

• class Node:
     def __init__(self, k):
         self.data = k
         self.next = None

  def printList(head):
     curr = head
     while curr != None:
         print(curr.data)
         curr = curr.next
     print()
  

  head = Node(10)
  head.next = Node(10)
  head.next.next = Node(20)

  
  printList(head)
  printMiddle(head)