Answer:
Following are the solution to this question:
Explanation:
This algorithm uses the divide and rule approach works, which splits the list into two sublists depending on if they are smaller or larger than the pivotal factor. It has O(n*log n) complexity.
It splits down the list in more than one version frequently till every sublist becomes based on a single element and fuses it to offer an ordered array. The merge type works according. It has O(n*log n) complexity.
Please find the attachment file of the sorting.
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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