PostgreSQL and coding
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Programming
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https://codility.com/c/intro/KPPACJ-GY7
Its a 150 minutes test about PostgreSQL and coding. You can Pause whenever you want
There are 4 question on SQL and coding
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Algorithm convertToPostfix(infix)
// Converts an infix expression to an equivalent postfix expression.
operatorStack = a new empty stack
postfix = a new empty string
while (infix has characters left to parse)
{
nextCharacter = next nonblank character of infix
switch (nextCharacter)
{
case variable:
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break
case '^' :
operatorStack.push(nextCharacter)
break
case '+' : case '-' : case '*' : case '/' :
while (!operatorStack.isEmpty() and precedence of nextCharacter <= precedence of operatorStack.peek())
{
Append operatorStack.peek() to postfix
operatorStack.pop()
}
operatorStack.push(nextCharacter)
break
case '( ' :
operatorStack.push(nextCharacter)
break
case ')' : // Stack is not empty if infix expression is valid topOperator = operatorStack.pop()
while (topOperator != '(')
{
Append topOperator to postfix topOperator = operatorStack.pop()
}
break
default: break // Ignore unexpected characters
}
}
while (!operatorStack.isEmpty())
{
topOperator = operatorStack.pop() Append topOperator to postfix
}
return postfix
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Northern Virginia Community College Area and Perimeter Program
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CSC400 Program 2 (Java Postfix Converter)
Create a Java Postfix converter using the algorithm provided in Segment& ...
CSC400 Program 2 (Java Postfix Converter)
Create a Java Postfix converter using the algorithm provided in Segment 5.16 on page 167 (Chapter 5 of Carrano & Henry, 2019, p. 167).
Ensure that your program has the required class and a test class. Submit Exercise 7 and screenshots of your program's execution and output compiled into a single document. Also attach all appropriate source code in a zip file.
Reference:
Carrano, F. M., & Henry, T. (2019). Data structures and abstractions with Java (5th ed.). Pearson.
5.16 The infix-to-postfix algorithm. The following algorithm encompasses the previous observations about the conversion process. For simplicity, all operands in our expression are single-letter variables.
Algorithm convertToPostfix(infix)
// Converts an infix expression to an equivalent postfix expression.
operatorStack = a new empty stack
postfix = a new empty string
while (infix has characters left to parse)
{
nextCharacter = next nonblank character of infix
switch (nextCharacter)
{
case variable:
Append nextCharacter to postfix
break
case '^' :
operatorStack.push(nextCharacter)
break
case '+' : case '-' : case '*' : case '/' :
while (!operatorStack.isEmpty() and precedence of nextCharacter <= precedence of operatorStack.peek())
{
Append operatorStack.peek() to postfix
operatorStack.pop()
}
operatorStack.push(nextCharacter)
break
case '( ' :
operatorStack.push(nextCharacter)
break
case ')' : // Stack is not empty if infix expression is valid topOperator = operatorStack.pop()
while (topOperator != '(')
{
Append topOperator to postfix topOperator = operatorStack.pop()
}
break
default: break // Ignore unexpected characters
}
}
while (!operatorStack.isEmpty())
{
topOperator = operatorStack.pop() Append topOperator to postfix
}
return postfix
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