Answer:
Is there an early pay discount?
Explanation:
This determines and instructs what path the code should take,
if there is no early pay discount, it has different instructions to follow.
A) force justified.
If you open up Word (or probably any word processor) and hover over the alignment options in the ribbon (or format the paragraph), it tells you what each one is, and the option that ensures the characters are spread evenly between the margins is labeled "justified."
Answer:
static int checkSymbol(char ch)
{
switch (ch)
{
case '+':
case '-':
return 1;
case '*':
case '/':
return 2;
case '^':
return 3;
}
return -1;
}
static String convertInfixToPostfix(String expression)
{
String calculation = new String("");
Stack<Character> operands = new Stack<>();
Stack<Character> operators = new Stack<>();
for (int i = 0; i<expression.length(); ++i)
{
char c = expression.charAt(i);
if (Character.isLetterOrDigit(c))
operands.push(c);
else if (c == '(')
operators.push(c);
else if (c == ')')
{
while (!operators.isEmpty() && operators.peek() != '(')
operands.push(operators.pop());
if (!operators.isEmpty() && operators.peek() != '(')
return NULL;
else
operators.pop();
}
else
{
while (!operators.isEmpty() && checkSymbol(c) <= checkSymbol(operators.peek()))
operands.push(operators.pop());
operators.push(c);
}
}
while (!operators.isEmpty())
operands.push(operators.pop());
while (!operands.isEmpty())
calculation+=operands.pop();
calculation=calculation.reverse();
return calculation;
}
Explanation:
- Create the checkSymbol function to see what symbol is being passed to the stack.
- Create the convertInfixToPostfix function that keeps track of the operands and the operators stack.
- Use conditional statements to check whether the character being passed is a letter, digit, symbol or a bracket.
- While the operators is not empty, keep pushing the character to the operators stack.
- At last reverse and return the calculation which has all the results.
