Data Structure Lecture 3

Linear Data Structures

Stack

Topics Covered in This Lecture
• Applications of stack
– Reversing the string
– Balancing symbols
– Postfix expression evaluation
– Translating infix expression to postfix expression

Applications of Stack
• Reversing the string

– push each character on to a stack as it is read.

– When the line is finished, we then pop characters off the
stack, and they will come off in the reverse order.

Reversing the string
void ReverseRead(void)
{ //using static Stack class of Lecture#3
Stack<char> stack;//The Stack ‘stack’ is created
char item;
cin>>item;
while (!stack.isFull()&&item!='$')//type in $ from keyboard to stop input
{ stack.push(item); // push each character onto the stack
cin>>item;
}
while(!stack.isEmpty() )
{ item=stack.pop ();//Pop an element from stack
cout<<item;
}
}

• Balancing Symbols
Compilers use a program that checks whether every symbol (like braces,
parenthesis, brackets, etc) in a program is balanced.
The simple algorithm for this purpose is:
1. Make an empty stack.
2. Read the characters until end of file.
3. If the character is an open any thing, push it onto the stack.
4. If it is a close any thing, then
if the stack is empty report an error.
Otherwise Pop the Stack.
If the popped symbol is not the corresponding opening symbol, then
report an error.
5. At the end of the file, if the stack is not empty report an error.

Polish Notation

a–b/c+d*e

Precedence?

1. b/c
2. d*e
3. a – a1 /a1 = b/c /
4. t2 + a2 / t2 = a – b/c / /a2 = d*e/

Infix, Suffix, Prefix
Infix = a * b + c
((a*b) +c)
Priority:
1. a * b
2. a1 + c / a1 = a * b /
Prefix =
* a b , +a1 c
+*abc
Suffix =
ab* , a1c+
ab*c+

infix = a- b * c / d + e / f
suffix =a – bc* / d + e / f
a – bc*d/ + e / f
a – bc*d/ + ef/
abc*d/- + ef/
abc*d/-ef/+
prefix =a - *bc / d + e / f
a - /*bcd + e / f
a - /*bcd + /ef
-a/*bcd + /ef
+-a/*bcd/ef

Infix:
a+b*c–d/f+e
Suffix:
abc*+df/-e+
Prefix:
+-+a*bc/dfe

Postfix Expression Evaluation

– When a number is seen, it is pushed onto the stack

– When an operator is seen, then pop two elements from
stack and push the result onto the stack.
Now we evaluate the following postfix expression.
6523+8*+3+* 3

1. The first four are placed on the stack. The resulting stack is 2
5
6
stack

• evaluating the following postfix
expression.
3
6523+8*+3+* 2
5
6
stack
2. Next a + is read, so 3 and 2 are popped from the stack and their sum 5 is
pushed. 5
5
6
stack

expression.
6523+8*+3+* 5
5
6
stack
8
3. Next 8 is read and pushed.
5
5
6
stack

expression.
8
6523+8*+3+* 5
5
6
stack
4. Next a * is seen so 8 and 5 are popped as 8 * 5 = 40 is pushed
40
5
6
stack

expression.
6523+8*+3+* 40
5
6
stack
5. Next a + is read so 40 and 5 are popped and 40 + 5 = 45 is pushed.

45
6
stack

expression.
6523+8*+3+*
45
6
stack
6. Now 3 is pushed
3
45
6
stack

expression.
6523+8*+3+* 3
45
6
stack
7. Next symbol is + so pops 3 and 45 and pushes 45 + 3 = 48, so push 48 in stack.

48
6
stack

expression.
6523+8*+3+* 48
6
stack
7. Finally a * is seen and 48 and 6 are popped, the result 6 * 48 = 288 is pushed.

8. As there is no input, so pop the stack and we get the result.

288
stack

• Translating infix expressions to postfix expression
– When an operand is read, it is immediately placed onto the output.
– When an operator or left parenthesis comes then save it in the stack
initially stack is empty.
– If we see a right parenthesis, then we pop the stack, writing symbols
until we encounter a (corresponding) left parenthesis, which is
popped but not output.
– If we see any other symbol (‘+’, ‘*’, ‘(‘, etc) then we pop entries form
the stack until we find an entry of lower priority. One exception is
that we never remove a ‘(‘ from the stack except when processing a
‘)’.

– When the popping is done, we push the operand onto the stack.

– Finally, if we read the end of input, we pop the stack until it is empty,
writing symbols onto the output.

• Translating infix expressions to postfix expression
Convert the following infix expression to postfix expression.
a+b*c+(d*e+f)*g

1. First the symbol a is read, so it is passed through to the output a

2. Then + is read and pushed onto the stack. output
+
stack
3. Next b is read and passed through to the output. ab
output *
4. Next a * is read. The top entry on the operator stack has lower +
precedence than *, so nothing is output and * is put on the .
stack

Converting the following infix expression to postfix
expression.
a+b*c+(d*e+f)*g
5. Next, c is read and output. abc
output
6. The next symbol is a +. Checking the stack, we find that priority of stack top
symbol * is higher than + . So we pop a * and place it on the output, Pop the
other +, which is not of lower but equal priority, and then push +.

* abc*+
+
+ output
stack stack

expression.
a+b*c+(d*e+f)*g
7. The next symbol read is an ‘(‘, which, being of highest precedence, is placed on
the stack.
(
+
stack
8. Then d is read and output. abc*+d
output

expression.
a+b*c+(d*e+f)*g
9. We continue by reading a *. Since open parenthesis do not get removed except
when a closed parenthesis is being processed, there is no output and we push *
in stack
*
(
+
stack
10. Next, e is read and output.
abc*+de
output

expression.
a+b*c+(d*e+f)*g
11. The next symbol read is a +, since priority of stack top value is higher so we pop
* and push +.

+ abc*+de*
( output
+
stack
12. Now we read f and output f.
abc*+de*f
output

expression.
a+b*c+(d*e+f)*g
13. Now we read a ‘)’, so the stack is emptied back to the ‘(‘, we output a +.

abc*+de*f+
+
output
stack
14. We read a * next; it is pushed onto the stack.
*
+
stack
15. Now, g is read and output. abc*+de*f+g
output

expression.
*
a+b*c+(d*e+f)*g
+
stack
16. The input is now empty, so pop output symbols from the stack until it is empty.

abc*+de*f+g*+
output
stack

Assignment 1
Group Size: 2, Due Date : Feb 27, 2012

• Implement Applications of stack
1. Balancing of symbols
2. Decimal to Binary Conversion
3. Infix to Postfix conversion
4. Postfix expression evaluation
5. Infix to Prefix conversion
6. Prefix expression evaluation
1. Each application will be implemented as a separate
function. Function will use stack, which is already
implemented in a header file and is available for reuse.
2. Difference between CGPAs of both group members
should not be more than ONE

Data Structure Lecture 3

More Related Content

What's hot (20)

Viewers also liked (7)

Similar to Data Structure Lecture 3 (17)

More from Teksify (12)

Data Structure Lecture 3