public int check(string exp)
{
stack stk = new stack(exp.Length);
int i = 0;
while (i < exp.Length)
{
char ch = exp[i];
if (ch == '(' || ch == '{' || ch == '[')
{
stk.push(ch);
}
else if (ch == ')' || ch == '}' || ch == ']')
{
char sp = stk.pop();
if (sp == 'e')
{
return(0);
}
else if (ch == ')' && sp == '(')
{
}
else if (ch == '}' && sp == '{')
{
}
else if (ch == ']' && sp == '[')
{
}
else
{
return(0);
}
}
i++;
}
if (stk.pop() == 'e')
{
return(1);
}
else
{
return(0);
}
}
public string conversion(string infix)//geting the infix expression and returning the reverse polish notation/post fix expression
{
int c = 0;
string rpn = ""; //initaly string will empty
int size = infix.Length, i = 0; //geting the length of the experasion(infix string) and i variable will be use for counting in while loop
stack stk = new stack(size); //crating a stack according to the size of expression
while (i < size) //this loop is used for examining the expression
{
char ch = infix[i]; //geting a single character of the string for expresion
//Console.WriteLine(ch);
//return "a";
if (ch == '(' || ch == '{' || ch == '[' || ch == ')' || ch == '}' || ch == ']')
{
if (ch == '(' || ch == '{' || ch == '[')
{
c++;
stk.push(ch);
}
else if (ch == ')' || ch == '}' || ch == ']')
{
if (ch == ')')
{
while (true)
{
char sp = stk.pop();
if (sp == '(' || sp == 'e')
{
break;
}
rpn = rpn + "" + sp;
}
}
else if (ch == '}')
{
while (true)
{
char sp = stk.pop();
if (sp == 'e')
{
break;
}
if (sp == '{')
{
break;
}
rpn = rpn + "" + sp;
}
}
else
{
while (true)
{
char sp = stk.pop();
if (sp == '[' || sp == 'e')
{
break;
}
//Console.WriteLine(sp);
rpn = rpn + "" + sp;
sp = stk.pop();
}
}
}
}
else if (ch == '#')
{
rpn = rpn + "" + ch;
rpn = rpn + "" + infix[++i];
}
else if (ch >= '0' && ch <= '9') //for operand
{
rpn = rpn + "" + ch; //appending operand with the string
}
else if (ch == '+' || ch == '-' || ch == '*' || ch == '/') //for operator examining
{
while (true) //make it infinite because dont know the operator precedence
{
char sp = stk.pop(); //geting the latest store operand in the que for checking the operator precedence and type(+,-,*or/ )
if (ch == '+' || ch == '-') //actions related to plus and minus operator
{
if (sp == 'e' || sp == '(' || sp == '{' || sp == '[') //if the que will empty
{
if (sp == '(' || sp == '{' || sp == '[')
{
stk.push(sp);
}
stk.push(ch); //pushing the latest read operator in the expression
break; //breaking the infinite loop for geting the next operator in the expression
}
else if (sp == '/' || sp == '*') //for equal precedence
{
rpn = rpn + "" + sp; //When same precedence then solve from left to right
stk.push(ch); //now pushing the latest same presendece operator
break; //after it will break obviously before it lies a low precedence
}
else if (sp == '+' || sp == '-') //for low precedence
{
//stk.push(sp);//because it will not append with post fix string this time and its presedence is less so again push in stack
rpn = rpn + "" + sp;
stk.push(ch); //pushing high precedecne operator
break; //after it will break obviously before it lies a low precedence
}
else //appending with array because these operator have high precedence then the latest(ch) operator
{
rpn = rpn + "" + sp;//appending with string
}
//rpn = rpn + "" + sp;//appending the operator with the post fix string
}
else if (ch == '*' || ch == '/') //for division and multiplication
{
if (sp == 'e' || sp == '(' || sp == '{' || sp == '[') //when stack will empty means this operator will perform after the first save operator
{
if (sp == '(' || sp == '{' || sp == '[')
{
stk.push(sp);
}
stk.push(ch); //push that operator
break; //obviously for next operator and stack is now waiting for next opertaor
}
else if (sp == '/' || sp == '*') //for equal precedence
{
rpn = rpn + "" + sp; //When same precedence then solve from left to right
stk.push(ch); //now pushing the latest same presendece operator
break; //after it will break obviously before it lies a low precedence
}
else if (sp == '+' || sp == '-') //for low precedence
{
stk.push(sp); //because it will not append with post fix string this time and its presedence is less so again push in stack
stk.push(ch); //pushing high precedecne operator
break; //after it will break obviously before it lies a low precedence
}
else //appending with array because these operator have high precedence then the latest(ch) operator
{
rpn = rpn + "" + sp;//appending with string
}
}
}
}
i++; //incrementing for next character in the string and termenating the loop
}
while (true) //for appending the remaining operator in the que it is infinite because dont know the number of the remaining operator in the que
{
char sp = stk.pop(); //get the remaining operator from the que
if (sp == 'e') //checking the que for empty
{
break; //terminating the loop because now que is empty
}
if (sp == '[')
{
Console.WriteLine("Rwanda");
}
rpn = rpn + "" + sp; //appending the operators with the post fix expression
}
return(rpn); //returnin the reverse polish notation or post fix expression
}