CS1102 Tut 4 Stacks, Queues, Recursion and The Big Oh

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CS1102 Tut 4 Stacks, Queues, Recursion and The
Big Oh

• Max Tan
• tanhuiyi_at_comp.nus.edu.sg
• COM1-01-09 Tel[phone]http//www.comp.nus.edu.s
  g/tanhuiyi

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First 15 minutes

• Any questions regarding tutorial/lecture
  materials?

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First 15 minutes

• No review, we will go through tutorial quickly
  and then quickly go through the midterm answers

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Question 1
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Question 1 solution

• 756

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Question 2

• Two stacks in a single array
• Just grow the stack from both ends of the array!

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Question 2
stack.pushStack1(o) pushStack1(Object o)
arraytop1 o top1
top1
top2
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Question 2
stack.pushStack2(o) pushStack2(Object o)
arraytop2 o top2--
top1
top2
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Question 2

• How to determine when to duplicate the array?

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Question 2
stack.pushStack2(o) pushStack2(Object o)
arraytop2 o top2--
top1
top2
When array is full, top 1 gt top 2!
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Question 2
Duplicate the array and copy the elements to the
end of the array
top1
top2
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Question 3

• Reversing a LinkedList recursively
• An easy way to think of recursion is imagine that
  the recursive function already works for
  sub-cases.
• For example, if you have to use recursion to
  solve a factorial function F(n), we can imagine
  it already works for F(n-1) and less.
• Since it works for F(n-1), to make it work for
  F(n), we simply multiply n to F(n-1) for F(n)!

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Question 3

• The idea is that if we can call recursive reverse
  for the rest of the list, our task is to make it
  work for just one node!

curr
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Question 3

• Set the current node to point to the previous
  node, and call recursiveReverse on the remaining
  nodes!

curr
recursiveReverse()
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Question 3

• But how does recursiveReverse know what the
  previous node is? It is a singly linkedlist, so
  you have to pass the reference to the previous
  node!

curr
recursiveReverse()
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Question 3

• public void recursiveReverse(ListNodeltIntegergt
  curr,
• ListNodeltIntegergt prev)
• ListNodeltIntegergt next curr.next
• //Update the new next pointer
• curr.next prev
• //Base case
• if(next null)
• this.head curr
• else
• recursiveReverse(next, curr)

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Question 4 - Overview

• Modified version of the Stack problem
• Pushing a stack onto another stack
• What happens if you pop and push one by one?

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Ordering is reversed!
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Question 4

• If you noticed, if you pop the elements and push
  it into another stack, the ordering is reversed
• The most obvious solution is to use a temporary
  Stack

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Question 4
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Question 4

• Attempt one
• public E pushRecursive(StackltEgt s)
• E element //base case if(s.empty() return
  null else element s.pop()
• push(element) pushRecursive(s) retur
  n element

What happens here if we push onto the stack first
before recursive calling pushRecursive? The stack
does not maintain its order!
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Question 4

• Attempt two
• public E pushRecursive(StackltEgt s)
• E element //base case if(s.empty() return
  null else element s.pop()
• pushRecursive(s)
• push(element) return element

What we actually want is to push the LAST element
of the stack first before pushing the FIRST
element so that we can maintain order!
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Question 4

• What about the efficiency?
• Not good because it requires the use of another
  stack, and requires the movement of all elements
  in the stack
• If we view it as a linked list

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Question 4
public E push(StackltEgt s) this.tail.next
s.head tail s.tail
tail2
tail1
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Of course we cant do this on a Java LinkedList as
the implementation of head and tail nodes are
hidden from us!
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head1
head2