Answer: Date
Explanation: We receive new information everyday, and things are always changing. If something is old, it may need to be updated with the correct information
Answer:
A and B
Explanation:
parsing with a rich grammar like TAG faces two main obstacles: low parsing speed and a lot of ambiguous syntactical parses.
An optimization problem is one in which you want to find, not just a solution, but the best solution •<span>A <span>“greedy algorithm” sometimes works </span></span><span>well for optimization problems </span>•<span>But only a few optimization problems can </span><span> be solved by the <span>greedy method</span></span>
Answer:
updating the computer and having an anti virus on your computer
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.
