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3 years ago

Convert infix to postfix

Computers and Technology

1 answer:

kvv77 [185]3 years ago

4 0

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.

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Answer:

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}

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