Working withs stack data structures... Need help producing a value of arithmetic
ID: 3737516 • Letter: W
Question
Working withs stack data structures... Need help producing a value of arithmetic expressions, using the following requirements:
All operands are positive integers.
Only the following operators are allowed: +, -, *, and /
Expression is fully parenthesized. In other words, each operator has a pair of "(" and")" surrounding its two operands.
Each line has one arithmetic expression. Operands, operators, and parentheses are separated by a space(s). Spaces or spaces can be used, as a delimiter when tokenizing an expression.
Explanation / Answer
I have done the code in JAVA langauage. However it shall be similar if you neede the code in C++.
import java.util.Stack;
public class EvaluateString
{
public static int evaluate(String expression)
{
char[] tokens = expression.toCharArray();
// Stack for numbers: 'values'
Stack<Integer> values = new Stack<Integer>();
// Stack for Operators: 'ops'
Stack<Character> ops = new Stack<Character>();
for (int i = 0; i < tokens.length; i++)
{
// Current token is a whitespace, skip it
if (tokens[i] == ' ')
continue;
// Current token is a number, push it to stack for numbers
if (tokens[i] >= '0' && tokens[i] <= '9')
{
StringBuffer sbuf = new StringBuffer();
// There may be more than one digits in number
while (i < tokens.length && tokens[i] >= '0' && tokens[i] <= '9')
sbuf.append(tokens[i++]);
values.push(Integer.parseInt(sbuf.toString()));
}
// Current token is an opening brace, push it to 'ops'
else if (tokens[i] == '(')
ops.push(tokens[i]);
// Closing brace encountered, solve entire brace
else if (tokens[i] == ')')
{
while (ops.peek() != '(')
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
ops.pop();
}
// Current token is an operator.
else if (tokens[i] == '+' || tokens[i] == '-' ||
tokens[i] == '*' || tokens[i] == '/')
{
// While top of 'ops' has same or greater precedence to current
// token, which is an operator. Apply operator on top of 'ops'
// to top two elements in values stack
while (!ops.empty() && hasPrecedence(tokens[i], ops.peek()))
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
// Push current token to 'ops'.
ops.push(tokens[i]);
}
}
// Entire expression has been parsed at this point, apply remaining
// ops to remaining values
while (!ops.empty())
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
// Top of 'values' contains result, return it
return values.pop();
}
// Returns true if 'op2' has higher or same precedence as 'op1',
// otherwise returns false.
public static boolean hasPrecedence(char op1, char op2)
{
if (op2 == '(' || op2 == ')')
return false;
if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-'))
return false;
else
return true;
}
// A utility method to apply an operator 'op' on operands 'a'
// and 'b'. Return the result.
public static int applyOp(char op, int b, int a)
{
switch (op)
{
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
if (b == 0)
throw new
UnsupportedOperationException("Cannot divide by zero");
return a / b;
}
return 0;
}
// Driver method to test above methods
public static void main(String[] args)
{
System.out.println(EvaluateString.evaluate("10 + 2 * 6"));
System.out.println(EvaluateString.evaluate("100 * 2 + 12"));
System.out.println(EvaluateString.evaluate("100 * ( 2 + 12 )"));
System.out.println(EvaluateString.evaluate("100 * ( 2 + 12 ) / 14"));
}
}
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