Transcript CS200 - Data Structures

CS2006 - Data Structures I
Chapter 7
Stacks III
Topics
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Applications
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Infix to postfix expression
Evaluate postfix expression
Infix Expressions
Binary operators appear between operands:
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W-X/Y–Z
(4+3)*2
(2+3)/(9-4)
Order of evaluation is determined by:
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precedence rules
parentheses
association (L to R or R to L)
Application: Algebraic Expressions
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Infix Expression Evaluation
 To evaluate an infix expression
Convert the infix expression into postfix
 Evaluate the postfix expression using stacks
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Postfix Expressions
Binary operators appear after both
operands:
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Order of operations is completely
determined by the expression
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X + Y in postfix form is: X Y +
no parentheses or precedence required
for example:
A * B - C becomes A B * C -
Postfix Expression
Infix
6-1
(4+3)*2
(2+3)/(9-4)
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Postfix
Postfix Expression Evaluation
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Assumptions:
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The string is syntactically correct postfix
expression
No unary operators are present
No exponentiation operators are present
Postfix Expression Evaluation
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Evaluation of postfix expressions makes use of an
operand stack.
Algorithm:
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Parse expression from left to right
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When an operand is encountered, push it onto stack
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when an operator is encountered, pop the last two operands
off the stack and apply the operation, and push the result onto
the stack
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when the expression is completely scanned, the final result will
be on the stack
2 3 + 9 4 -
Postfix Expression Evaluation
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Pseudocode:
for ( each character ch in the string )
{ if (ch is an operand )
Push value that operand ch represents onto stack
else
/ / ch is an operator named op
{
/ / evaluate and push the result
operand2 = top of stack
Pop the stack
operand1 = top of stack
Pop the stack
result = operand1 op operand2
Push Result onto stack
} / / end if
} / / end for
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Example
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Evaluate 2 3 4 5 + 2 * 1 + + 3 2 * + *
Converting Infix into Postfix
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Since postfix expressions are easily
evaluated, the easiest way to evaluate
infix is to convert from infix to postfix and
then evaluate.
This conversion uses an operator stack
since low precedence operators must be
saved and applied after high precedence
ones.
Converting Infix into Postfix
The operands always stay in the same
order with respect to one another
An operator will move only to “ the right”
with respect to the operands
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If in infix expression, the operand x precedes the
operator op, in the postfix, the operand x also
precedes the operator x
All parentheses are removed
Converting Infix into Postfix
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Converting Infix into Postfix
 Scan infix string from left to right
 Each time an operand is encountered, copy it
to output
 When a bracket is encountered, check its
orientation.
 Push
left brackets onto stack
 If it is a right bracket, pop all operators on the
stack and copy them to output until a matching
left bracket is encountered. Discard the left
bracket
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Converting Infix into Postfix
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Converting Infix into Postfix
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When an operator is encountered, check the top
item of the stack
If the priority is >= the current operator, pop the top
operator and copy it to output
 Continue until an operator of lower priority is encountered
or until the stack is empty
 Finally, push current operator onto the stack
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When the end of the expression is reached,
copy the remaining stack contents to output
in the order popped
Converting Infix into Postfix
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Converting Infix into Postfix
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Precedence Table
Operator
Name
Precedence when on
stack
Precedence when
on input
(
Opening
parentheses
always stack
0
7
)
Closing parentheses
never stacked
0
^
Exponentiation
6
5
*/
Multiply & divide
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3
+-
Add & Subtract
2
1
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Conversion Example
W - X /Y + Z
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'-' is pushed onto operator stack
'/' has higher precedence than '-' on stack - it gets
pushed
'+' has lower precedence than '/' on stack - pop '/' and
output - then '-' is on top of stack - since subtraction
associates left to right - pop '-' off stack and output - push
'+' onto stack
when end of expression is reached pop remaining
operator and output
resulting expression is: W X Y /- Z +
Converting Infix into Postfix
stack.push(EOL marker)
for (each character ch in the infix expression) {
switch (ch) {
case operand:
//append ch to output postfix expression
postfixExpr = postfixExpr + ch;
case '(':
stack.push(ch); // push ch onto operator stack
case ')':
while (top of stack is not '(' ) {
postfixExpr = postfixExpr + stack.pop(); // pop and append expr.
}
stack.pop();
case operator:
while (!stack.isEmpty() && inputPrec(ch) <= stackPrec(stack.top())) {
postfixExpr = postfixExpr + stack.pop();
}
stack.push(ch);
} // end for
while (!stack.isEmpty() { // pop all remaining operators off stack and append
17 postfixExpr = postfixExpr + stack.pop();
Conversion Example
2 + 3 * [(7-5)/2]
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2 3 7 5
2 copy into output
‘+' is pushed onto operator stack
3 copy into output
‘*' has higher precedence than ‘+' on stack - it gets
pushed
‘[‘ is left bracket, push it onto stack
‘(‘is left bracket, push it onto stack
7 is copied into output
‘-' has higher precedence than ‘(‘, push onto stack
Conversion Example
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2 + 3 * [(7-5)/2]
Prefix Expressions
Infix
6-1
(4+3)*2
(2+3)/(9-4)
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Prefix
-6 1
*+432
/+23–94
Review
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Which of the following is NOT true about
converting infix expressions to postfix
expressions?
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the operands always stay in the same order
with respect to one another
the operators always stay in the same order
with respect to one another
an operator will move only “to the right” with
respect to the operands
all parentheses are removed
Review
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Which of the following is the postfix form
of the infix expression: (a + b) * c / d
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ab+c*d/
ab*c/d+
a+b*c/d
ab+cd*/
Review
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StackInterface provides the specifications
for ______.
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only the array-based implementation of a
stack
only the reference-based implementation of a
stack
only an implementation of a stack that uses
the ADT list
all the implementations of a stack
Review
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In the StackInterface class, the push
method accepts as its parameter an item
that is an instance of ______.
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Integer
Double
String
Object