Transcript Lecture Notes for Stack

Part-B1
Stacks
Abstract Data Types (ADTs)
An abstract data
type (ADT) is an
abstraction of a
data structure
An ADT specifies:



Data stored
Operations on the
data
Error conditions
associated with
operations
Example: ADT modeling a
students record

The data stored are
 Student name, id No., as1,
as2,as3, exam

The operations supported are
 int averageAs(as1,as2,as3)
 Int finalMark(as1, as2,as3, exam) )

Error conditions:
 Calculate the final mark for
absent student
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The Stack ADT (§4.2)
The Stack ADT stores
arbitrary objects
Insertions and deletions
follow the last-in first-out
scheme
Think of a spring-loaded
plate dispenser
Main stack operations:


push(object): inserts an
element
object pop(): removes and
returns the last inserted
element
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Auxiliary stack
operations:



object top(): returns the
last inserted element
without removing it
integer size(): returns the
number of elements
stored
boolean isEmpty():
indicates whether no
elements are stored
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Stack Interface in Java
Java interface
corresponding to
our Stack ADT
Requires the
definition of class
EmptyStackException
public interface Stack {
public int size();
public boolean isEmpty();
public Object top()
throws EmptyStackException;
public void push(Object o);
public Object pop()
throws EmptyStackException;
}
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Exceptions
Attempting the
execution of an
operation of ADT may
sometimes cause an
error condition, called
an exception
Exceptions are said to
be “thrown” by an
operation that cannot
be executed
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In the Stack ADT,
operations pop and
top cannot be
performed if the
stack is empty
Attempting the
execution of pop or
top on an empty
stack throws an
EmptyStackException
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Applications of Stacks
Direct applications


Undo sequence in a text editor
Chain of method calls in the Java Virtual
Machine
Indirect applications


Auxiliary data structure for algorithms
Component of other data structures
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Array-based Stack (Implementation)
A simple way of
implementing the
Stack ADT uses an
array
We add elements
from left to right
A variable t keeps
track of the index of
the top element
Algorithm size()
return t + 1
Algorithm pop()
if isEmpty() then
throw EmptyStackException
else
tt1
return S[t + 1]
…
S
0 1 2
t
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Array-based Stack (cont.)
The array storing the
stack elements may
become full
A push operation will
then throw a
FullStackException


Algorithm push(o)
if t = S.length  1 then
throw FullStackException
else
tt+1
Limitation of the arrayS[t]  o
based implementation
Not intrinsic to the
Stack ADT
…
S
0 1 2
t
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Array-based Stack (Cont.)
A Stack might be
empty
top returns the
element at the top
of the Stack, but
does not remove the
top element. When
the Stack is empty,
an error occurs.
Algorithm isEmpty()
if t<0 then return true
else return false
Algorithm top()
if isEmpty() then
throw EmptyStackException
return S[t ]
…
S
0 1 2
t
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Performance and Limitations for
array-based Stack
Performance



Let n be the number of elements in the stack
The space used is O(n)
Each operation runs in time O(1)
Limitations


The maximum size of the stack must be defined a
priori and cannot be changed
Trying to push a new element into a full stack
causes an implementation-specific exception
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Array-based Stack in Java
public class ArrayStack implements Stack{
// holds the stack elements
private Object S[ ];
// index to top element
private int top = -1;
// constructor
public ArrayStack(int capacity) {
S = new Object[capacity]);
}
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Array-based Stack in Java
public Object pop() throws EmptyStackException{
if isEmpty()
throw new EmptyStackException
(“Empty stack: cannot pop”);
Object temp = S[top];
// facilitates garbage collection
S[top] = null;
top = top – 1;
return temp;
}
public int size() {
return (top + 1);
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Array-based Stack in Java
public boolean isEmpty() {
return (top < 0);
}
public void push(Object obj) throws FullStackException {
if (size() == capacity)
throw new FullStackException("Stack overflow.");
S[++top] = obj;
}
public Object top() throws EmptyStackException {
if (isEmpty())
throw new EmptyStackException("Stack is empty.");
return S[top];
}
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Parentheses Matching
An expression, i.e.,[(2+3)*x+5]*2.
Each “(”, “{”, or “[” must be paired with
a matching “)”, “}”, or “[”





correct: ( )(( )){([( )])}
correct: ((( )(( )){([( )])}
incorrect: )(( )){([( )])}
incorrect: ({[ ])}
incorrect: (
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Parentheses Matching Algorithm
Algorithm ParenMatch(X,n):
Input: An array X of n tokens, each of which is either a grouping symbol, a
variable, an arithmetic operator, or a number
Output: true if and only if all the grouping symbols in X match
Let S be an empty stack
for i=0 to n-1 do
if X[i] is an opening grouping symbol then
S.push(X[i])
else if X[i] is a closing grouping symbol then
if S.isEmpty() then
return false {nothing to match with}
if S.pop() does not match the type of X[i] then
return false {wrong type}
if S.isEmpty() then
return true {every symbol matched}
else
return false {some symbols were never matched}
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Parentheses Matching Example 1
Input: () (() [()])
i
X[i]
Operation
Stack
0
(
Push (
1
)
Pop (
Test if ( and X[i] match?
(
YES
2
(
Push (
(
3
(
Push (
((
4
)
Pop (
Test if ( and X[i] match?
(
5
[
Output
Push [
YES
([
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Parentheses Matching Example 1
Input: () (() [()])
i
X[i] Operation
Stack Output
6
(
Push (
([(
7
)
Pop (
Test if ( and X[i] match?
([
8
9
]
)
YES
Pop [
Test if [ and X[i] match?
YES
Pop (
Test if ( and X[i] match?
YES
Test if stack is Empty?
YES
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(
TRUE
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Parentheses Matching Example 2
Input: ( () [] ]()
i
X[i]
Operation
Stack
0
(
Push (
(
1
(
Push (
((
2
)
Pop (
Test if ( and X[i] match?
(
YES
3
[
Push [
([
4
]
Pop [
Test if [ and X[i] match?
(
YES
Pop (
Test if ( and X[i] match ?
NO
5
]
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Output
FASLE
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Computing Spans (not in book)
7
We show how to use a stack 6
as an auxiliary data structure 5
in an algorithm
4
Given an array X, the span
3
S[i] of X[i] is the maximum
2
number of consecutive
elements X[j] immediately
1
preceding X[i] and such that 0
X[j]  X[j+1]
Spans have applications to
financial analysis

E.g., stock at 52-week high
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X
S
1
6
1
3
1
2
3
4
2
5
3
4
2
1
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Quadratic Algorithm
Algorithm spans1(X, n)
Input array X of n integers
Output array S of spans of X
#
S  new array of n integers
n
for i  0 to n  1 do
n
s1
n
while s  i  X[i  s]  X[i-s+1] 1 + 2 + …+ (n  1)
ss+1
1 + 2 + …+ (n  1)
S[i]  s
n
return S
1
X[]= 1,2,3, …, n-1, n.
Algorithm spans1 runs in O(n2) time
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Computing Spans with a Stack
We push the n elements in the
stack
We sacn the elements in the
stack in the reverse order.


We find the last element x[j]
with x[j]<=x[i] and x[j-1]>x[i]
and set s[i]=i-j+1.
If no such a j exists, s[i]=i+1.
Compute the s[i] for the
remaining i’s as follows:



for (i=n-2; i>=0; i--)
if (s[i+1]>1 & s[i]==0) then
s[i]=s[i+1]-1
7
6
5
4
3
2
1
0
0 1 2 3 4 5 6 7
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Example:
i= 0, 1, 2, 3, 4, 5, 6, 7
X[i]= 6, 3, 4, 1, 2, 3, 5, 4
S[i]= 1, 1, 2, 1, 2, 3, 4, 1.
4
5
3
2
1
4
3
6
Stack
5
3
2
1
4
3
6
Stack
4
3
6
Stack
6
Stack
Stack
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Linear Algorithm
Each index of the array
 Is pushed into the
stack exactly one
 Is popped from the
stack at most once
The statements in the
two while loops are
executed at most n
times
Algorithm spans2 runs in
O(n) time
for (i=n-2; i>=0; i--)
if (s[i+1]>1 &
s[i]==0) then
s[i]=s[i+1]-1
Algorithm spans2(X, n)
S  new array of n integers
A  new empty stack
for i  0 to n  1 doA.push(i)
i=n-1 ; j=n-1;
while (i<=0) do
while (A.isEmpty() 
X[A.top()]  X[j] ) do
j  A.pop()
if A.isEmpty() then
S[i]  i + 1
else
S[i]  i  j +1
i=j-1;
return S
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#
n
1
n
1
n
n
n
1
n
n
1
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Summary of Stack
Understand the definition of Stack (basic)
Applications
1.
2.
Parentheses (moderate)
Computing Spans (not required)
 Implementation of Stack is not required.
 You should be able to use stack to write
programs to solve problems
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Week 3: Tutorial Arrangement
Week 3 Tutorial (Feb. 6, 2006,
6:30p.m. – 7:20p.m) will be in
MMW2478 (CS lab, second floor, Lift 18,
on the left hand side after you go
through the tunnel.)
Tunnel connecting city U and Festival Walk
Lift 18
City Uni.
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Remarks
Emphasize the concept of ADT.
More examples about ADT
Delete the Span application examples.
Add Queue part (perhaps to week 3)
Teach slowly.
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