Answers for "infix to postfix conversion"

1

infix to postfix conversion

Best Solution
-------------------------------------------------------------------
#include<bits/stdc++.h>
using namespace std;

int prec(char c) {
    if(c == '^') {
        return 3;
    }
    else if(c == '*' || c == '/') {
        return 2;
    }
    else if(c == '+' || c =='-') {
        return 1;
    }
    else {
        return -1;
    }
}

string infixToPostfix(string s) {
    stack<char> st;
    string res;

    for (int i = 0; i < s.length(); i++)
    {
        if((s[i] >= 'a' && s[i] <= 'z') || (s[i] >= 'A' && s[i] <= 'Z')) {
            res += s[i];
        }
        else if(s[i] == '(') {
            st.push(s[i]);
        }
        else if(s[i] == ')') {
            while (!st.empty() && st.top() != '(')
            {
                res += st.top();
                st.pop();
            }
            if(!st.empty()) {
                st.pop(); // Popping '(' here
            }
        }
        else {
            while (!st.empty() && prec(st.top()) >= prec(s[i]))
            {
                res += st.top();
                st.pop();
            }
            st.push(s[i]);
        }
    }
    
    while (!st.empty())
    {
       res += st.top();
       st.pop();
    }
    
    return res;
}

int main() {
    string exp = "a+b*(c^d-e)^(f+g*h)-i";
    cout<<infixToPostfix(exp);
    return 0;
}
Posted by: Guest on July-02-2021
0

infix to postfix conversion

Begin
   initially push some special character say # into the stack
   for each character ch from infix expression, do
      if ch is alphanumeric character, then
         add ch to postfix expression
      else if ch = opening parenthesis (, then
         push ( into stack
      else if ch = ^, then            //exponential operator of higher precedence
         push ^ into the stack
      else if ch = closing parenthesis ), then
         while stack is not empty and stack top ≠ (,
            do pop and add item from stack to postfix expression
         done

         pop ( also from the stack
      else
         while stack is not empty AND precedence of ch <= precedence of stack top element, do
            pop and add into postfix expression
         done

         push the newly coming character.
   done

   while the stack contains some remaining characters, do
      pop and add to the postfix expression
   done
   return postfix
End
Posted by: Guest on April-17-2021

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