/**
* Checks if a sequence of characters have the correct order to do an arithmetic operation.
*
* @param pila The stack that is going to be analyzed
* @return true if only if the expression have the correct order
*/
bool comprovaExpresio (ArrayStack<char> &pila) {
//If the stack is empty, it returns true because there isn't elements to analyze.
if(!pila.empty()) {
/**
* If the stack is not empty, we proceed to do the next algorithm:
* Two elements: pila (the original stack) and pilaAux2 (an auxiliary stack
* to do the proper operations)
* 1) If "pilaAux2" is empty, we transfer the last char of "pila" to "pilaAux2"
* and we delete this char from "pila". We go directly to step 4.
* If "pilaAux2" is not empty, we go to the step 2:
* 2) We check if the top of pilaAux2 is equal to the opposite of the top
* element of "pila" (Note that we only check the closing expression).
* If it is true, we delete the char from "pilaAux2" and "pila" and
* we go directly to step 4. If it is not true, we go to the step 3.
* 3) We push the top element of "pila" to "pilaAux". We go to the step 4.
* 4) If pila is not empty, we do again the process deleting always the top
* element of "pila".
*
* We can prove that the loop finishes. If the size of the stack is n and
* also is a positive integer, this implies that the stack has a size of n
* that follows the next rule, 0 <= n.
* If n = 0 we stop. If not, we decrement the size of stack in 1 in each
* loop, so we have a size of n-1. If n-1 = 0 pila is empty, and the algorithm
* stops, if not, we do again the same process until we have size=0. QED
*
* If "pilaAux2" is empty, implies that the expression is correct.
* The complexity of this algorithm is O(n). Because we have O(n) (we create
* the first stack) + O(n) (we check all elements of first stack) + O(1)
* (const senteces) = 2*O(n) + O(1) = O(n)
*/
ArrayStack<char> pilaAux2;
while(!pila.empty()) { //Checks if "pila" is empty
if(pilaAux2.empty()) { //Checks if "pilaAux2" is empty
pilaAux2.push(pila.top()); //Adds the top element of "pila" to "pilaAux2"
}
else {
if(topAparellat(pila.top() , pilaAux2.top())) { //Checks if the tops elemets of the stacks are opposite
pilaAux2.pop(); //Deletes the last element of "pilaAux2"
}
else {
pilaAux2.push(pila.top()); //Adds the top element of "pila" to "pilaAux2"
}
}
pila.pop(); //Deletes the top element of "pila"
}
if (pilaAux2.empty()) { //Checks if "pilaAux2" is empty to know if the expression is well done or not
return true;
}
else {
return false;
}
}
else {
return true;
}
}
void Sort(ArrayStack<int> &stack) {
ArrayStack<int> helper;
while (!stack.empty()) {
int cur = stack.pop();
int count = 0;
while (!helper.empty() && helper.peek() < cur) {
count++;
stack.push(helper.pop());
}
helper.push(cur);
while (count--)
helper.push(stack.pop());
}
while (!helper.empty())
stack.push(helper.pop());
}
void array_stack(){
ArrayStack<char,100> charStack;
for(int i = 0;i<100;i++)
charStack.push('a');
while(!charStack.empty()){
char tmp = charStack.top();
charStack.pop();
}
}