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 stringvoid ReverseRead(void){ //using static Stack class of Lecture#3

Stack<char> stack;//The Stack ‘stack’ is createdchar 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 SymbolsCompilers 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, PrefixInfix = a * b + c((a*b) +c) Priority: 1. a * b2. a1 + c / a1 = a * b / Prefix = * a b , +a1 c+*abcSuffix = ab* , a1c+ab*c+

Infix, Suffix, Prefixinfix = 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, Suffix, Prefix

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.6 5 2 3 + 8 * + 3 + *

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

3

2

5

6

stack


• evaluating the following postfix expression.6 5 2 3 + 8 * + 3 + *

3

2

5

6

stack2. Next a + is read, so 3 and 2 are popped from the stack and their sum 5 is

pushed. 5

5

6

stack



5

5

6

stack3. Next 8 is read and pushed. 8

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

8

5

5

6

stack



5. Next a + is read so 40 and 5 are popped and 40 + 5 = 45 is pushed.

40

5

6

stack

45

6

stack



6. Now 3 is pushed

45

6

stack

3

45

6

stack



7. Next symbol is + so pops 3 and 45 and pushes 45 + 3 = 48, so push 48 in stack.

3

45

6

stack

48

6

stack



7. Finally a * is seen and 48 and 6 are popped, the result 6 * 48 = 288 is pushed.

288

stack

48

6

stack

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

Applications of 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.

Applications of Stack• 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

output2. Then + is read and pushed onto the stack.

+

stack

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 .

3. Next b is read and passed through to the output. ab

output *

+

stack

Applications of StackConverting the following infix expression to postfix

expression.a+b*c+(d*e+f)*g5. Next, c is read and 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 +.

+

stack

abc*+

output

*

+

stack

abc

output


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.

(

+

stack8. Then d is read and output. abc*+d

output


expression.a+b*c+(d*e+f)*g9. 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

*

(

+

stack10. Next, e is read and output.

abc*+de

output


expression.a+b*c+(d*e+f)*g11. 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



13. Now we read a ‘)’, so the stack is emptied back to the ‘(‘, we output a +.

14. We read a * next; it is pushed onto the stack.

abc*+de*f+

output+

stack

*

+

stack15. Now, g is read and output. abc*+de*f+g

output



16. The input is now empty, so pop output symbols from the stack until it is empty.

abc*+de*f+g*+

outputstack

*

+

stack

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

• Implement Applications of stack1. Balancing of symbols2. Decimal to Binary Conversion3. Infix to Postfix conversion4. Postfix expression evaluation5. Infix to Prefix conversion6. 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

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Data Structure Lecture 3