Stack | Set 2 (Infix to Postfix) - GeeksforGeeks

12 Apr, 2022
Prerequisite – Stack | Set 1 (Introduction) Infix expression: The expression of the form a op b. When an operator is in-between every pair of operands.Postfix expression: The expression of the form a b op. When an operator is followed for every pair of operands.Why postfix representation of the expression? The compiler scans the expression either from left to right or from right to left. Consider the below expression: a op1 b op2 c op3 d If op1 = +, op2 = *, op3 = +The compiler first scans the expression to evaluate the expression b * c, then again scans the expression to add a to it. The result is then added to d after another scan.The repeated scanning makes it very in-efficient. It is better to convert the expression to postfix(or prefix) form before evaluation.The corresponding expression in postfix form is abc*+d+. The postfix expressions can be evaluated easily using a stack. We will cover postfix expression evaluation in a separate post.Algorithm 1. Scan the infix expression from left to right. 2. If the scanned character is an operand, output it. 3. Else,       1 If the precedence and associativity of the scanned operator is greater than the precedence and associativity of the operator in the stack(or the stack is empty or the stack contains a ‘(‘ ), push it.
      2 ‘^’ operator is right associative and other operators like ‘+’,’-‘,’*’ and ‘/’ are left associative. Check especially for a condition when both top of the operator stack and scanned operator are ‘^’. In this condition the precedence of scanned operator is higher due to it’s right associativity. So it will be pushed in the operator stack. In all the other cases when the top of operator stack is same as scanned operator we will pop the operator from the stack because of left associativity due to which the scanned operator has less precedence.       3 Else, Pop all the operators from the stack which are greater than or equal to in precedence than that of the scanned operator. After doing that Push the scanned operator to the stack. (If you encounter parenthesis while popping then stop there and push the scanned operator in the stack.) 4. If the scanned character is an ‘(‘, push it to the stack. 5. If the scanned character is an ‘)’, pop the stack and output it until a ‘(‘ is encountered, and discard both the parenthesis. 6. Repeat steps 2-6 until infix expression is scanned. 7. Print the output 8. Pop and output from the stack until it is not empty.
Following is the implementation of the above algorithm  
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/* C++ implementation to convertinfix expression to postfix*/ #include <bits/stdc++.h>using namespace std; // Function to return precedence of operatorsint prec(char c){    if (c == '^')        return 3;    else if (c == '/' || c == '*')        return 2;    else if (c == '+' || c == '-')        return 1;    else        return -1;} // The main function to convert infix expression// to postfix expressionvoid infixToPostfix(string s){     stack<char> st; // For stack operations, we are using                    // C++ built in stack    string result;     for (int i = 0; i < s.length(); i++) {        char c = s[i];         // If the scanned character is        // an operand, add it to output string.        if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')            || (c >= '0' && c <= '9'))            result += c;         // If the scanned character is an        // ‘(‘, push it to the stack.        else if (c == '(')            st.push('(');         // If the scanned character is an ‘)’,        // pop and to output string from the stack        // until an ‘(‘ is encountered.        else if (c == ')') {            while (st.top() != '(') {                result += st.top();                st.pop();            }            st.pop();        }         // If an operator is scanned        else {            while (!st.empty()                   && prec(s[i]) <= prec(st.top())) {                if (c == '^' && st.top() == '^')                    break;                else {                    result += st.top();                    st.pop();                }            }            st.push(c);        }    }     // Pop all the remaining elements from the stack    while (!st.empty()) {        result += st.top();        st.pop();    }     cout << result << endl;} // Driver program to test above functionsint main(){    string exp = "a+b*(c^d-e)^(f+g*h)-i";    infixToPostfix(exp);    return 0;}
// C program to convert infix expression to postfix#include <stdio.h>#include <string.h>#include <stdlib.h> // Stack typestruct Stack{    int top;    unsigned capacity;    int* array;}; // Stack Operationsstruct Stack* createStack( unsigned capacity ){    struct Stack* stack = (struct Stack*)           malloc(sizeof(struct Stack));     if (!stack)        return NULL;     stack->top = -1;    stack->capacity = capacity;     stack->array = (int*) malloc(stack->capacity *                                   sizeof(int));     return stack;}int isEmpty(struct Stack* stack){    return stack->top == -1 ;}char peek(struct Stack* stack){    return stack->array[stack->top];}char pop(struct Stack* stack){    if (!isEmpty(stack))        return stack->array[stack->top--] ;    return '$';}void push(struct Stack* stack, char op){    stack->array[++stack->top] = op;}  // A utility function to check if// the given character is operandint isOperand(char ch){    return (ch >= 'a' && ch <= 'z') ||           (ch >= 'A' && ch <= 'Z');} // A utility function to return// precedence of a given operator// Higher returned value means// higher precedenceint Prec(char ch){    switch (ch)    {    case '+':    case '-':        return 1;     case '*':    case '/':        return 2;     case '^':        return 3;    }    return -1;}  // The main function that// converts given infix expression// to postfix expression.int infixToPostfix(char* exp){    int i, k;     // Create a stack of capacity    // equal to expression size    struct Stack* stack = createStack(strlen(exp));    if(!stack) // See if stack was created successfully        return -1 ;     for (i = 0, k = -1; exp[i]; ++i)    {                 // If the scanned character is        // an operand, add it to output.        if (isOperand(exp[i]))            exp[++k] = exp[i];                 // If the scanned character is an        // ‘(‘, push it to the stack.        else if (exp[i] == '(')            push(stack, exp[i]);                 // If the scanned character is an ‘)’,        // pop and output from the stack        // until an ‘(‘ is encountered.        else if (exp[i] == ')')        {            while (!isEmpty(stack) && peek(stack) != '(')                exp[++k] = pop(stack);            if (!isEmpty(stack) && peek(stack) != '(')                return -1; // invalid expression                        else                pop(stack);        }        else // an operator is encountered        {            while (!isEmpty(stack) &&                 Prec(exp[i]) <= Prec(peek(stack)))                exp[++k] = pop(stack);            push(stack, exp[i]);        }     }     // pop all the operators from the stack    while (!isEmpty(stack))        exp[++k] = pop(stack );     exp[++k] = '\0';    printf( "%s", exp );} // Driver program to test above functionsint main(){    char exp[] = "a+b*(c^d-e)^(f+g*h)-i";    infixToPostfix(exp);    return 0;}
/* Java implementation to convert infix expression to postfix*/// Note that here we use Stack class for Stack operations import java.util.Stack; class Test{         // A utility function to return    // precedence of a given operator    // Higher returned value means    // higher precedence    static int Prec(char ch)    {        switch (ch)        {        case '+':        case '-':            return 1;              case '*':        case '/':            return 2;              case '^':            return 3;        }        return -1;    }          // The main method that converts    // given infix expression    // to postfix expression.    static String infixToPostfix(String exp)    {        // initializing empty String for result        String result = new String("");                 // initializing empty stack        Stack<Character> stack = new Stack<>();                 for (int i = 0; i<exp.length(); ++i)        {            char c = exp.charAt(i);                         // If the scanned character is an            // operand, add it to output.            if (Character.isLetterOrDigit(c))                result += c;                          // If the scanned character is an '(',            // push it to the stack.            else if (c == '(')                stack.push(c);                         //  If the scanned character is an ')',            // pop and output from the stack            // until an '(' is encountered.            else if (c == ')')            {                while (!stack.isEmpty() &&                        stack.peek() != '(')                    result += stack.pop();                                     stack.pop();            }            else // an operator is encountered            {                while (!stack.isEmpty() && Prec(c)                         <= Prec(stack.peek())){                                       result += stack.pop();             }                stack.push(c);            }              }              // pop all the operators from the stack        while (!stack.isEmpty()){            if(stack.peek() == '(')                return "Invalid Expression";            result += stack.pop();         }        return result;    }       // Driver method    public static void main(String[] args)    {        String exp = "a+b*(c^d-e)^(f+g*h)-i";        System.out.println(infixToPostfix(exp));    }}
# Python program to convert infix expression to postfix # Class to convert the expression  class Conversion:     # Constructor to initialize the class variables    def __init__(self, capacity):        self.top = -1        self.capacity = capacity        # This array is used a stack        self.array = []        # Precedence setting        self.output = []        self.precedence = {'+': 1, '-': 1, '*': 2, '/': 2, '^': 3}     # check if the stack is empty    def isEmpty(self):        return True if self.top == -1 else False     # Return the value of the top of the stack    def peek(self):        return self.array[-1]     # Pop the element from the stack    def pop(self):        if not self.isEmpty():            self.top -= 1            return self.array.pop()        else:            return "$"     # Push the element to the stack    def push(self, op):        self.top += 1        self.array.append(op)     # A utility function to check is the given character    # is operand    def isOperand(self, ch):        return ch.isalpha()     # Check if the precedence of operator is strictly    # less than top of stack or not    def notGreater(self, i):        try:            a = self.precedence[i]            b = self.precedence[self.peek()]            return True if a <= b else False        except KeyError:            return False     # The main function that    # converts given infix expression    # to postfix expression    def infixToPostfix(self, exp):         # Iterate over the expression for conversion        for i in exp:            # If the character is an operand,            # add it to output            if self.isOperand(i):                self.output.append(i)             # If the character is an '(', push it to stack            elif i == '(':                self.push(i)             # If the scanned character is an ')', pop and            # output from the stack until and '(' is found            elif i == ')':                while((not self.isEmpty()) and                      self.peek() != '('):                    a = self.pop()                    self.output.append(a)                if (not self.isEmpty() and self.peek() != '('):                    return -1                else:                    self.pop()             # An operator is encountered            else:                while(not self.isEmpty() and self.notGreater(i)):                        # this is to pass cases like a^b^c                    # without if ab^c^                    # with if abc^^                    if i == "^" and self.array[-1] == i:                        break                    self.output.append(self.pop())                self.push(i)         # pop all the operator from the stack        while not self.isEmpty():            self.output.append(self.pop())         print "".join(self.output)  # Driver program to test above functionexp = "a+b*(c^d-e)^(f+g*h)-i"obj = Conversion(len(exp))obj.infixToPostfix(exp) # This code is contributed by Nikhil Kumar Singh(nickzuck_007)
using System;using System.Collections.Generic; /* c# implementation to convertinfix expression to postfix*/// Note that here we use Stack// class for Stack operations public  class Test{         // A utility function to return    // precedence of a given operator    // Higher returned value means higher precedence    internal static int Prec(char ch)    {        switch (ch)        {        case '+':        case '-':            return 1;         case '*':        case '/':            return 2;         case '^':            return 3;        }        return -1;    }     // The main method that converts given infix expression    // to postfix expression.     public static string infixToPostfix(string exp)    {        // initializing empty String for result        string result = "";         // initializing empty stack        Stack<char> stack = new Stack<char>();         for (int i = 0; i < exp.Length; ++i)        {            char c = exp[i];             // If the scanned character is an            // operand, add it to output.            if (char.IsLetterOrDigit(c))            {                result += c;            }             // If the scanned character is an '(',            // push it to the stack.            else if (c == '(')            {                stack.Push(c);            }             //  If the scanned character is an ')',            // pop and output from the stack             // until an '(' is encountered.            else if (c == ')')            {                while (stack.Count > 0 &&                        stack.Peek() != '(')                {                    result += stack.Pop();                }                 if (stack.Count > 0 && stack.Peek() != '(')                {                    return "Invalid Expression"; // invalid expression                }                else                {                    stack.Pop();                }            }            else // an operator is encountered            {                while (stack.Count > 0 && Prec(c) <=                                  Prec(stack.Peek()))                {                    result += stack.Pop();                }                stack.Push(c);            }         }         // pop all the operators from the stack        while (stack.Count > 0)        {            result += stack.Pop();        }         return result;    }     // Driver method     public static void Main(string[] args)    {        string exp = "a+b*(c^d-e)^(f+g*h)-i";        Console.WriteLine(infixToPostfix(exp));    }} // This code is contributed by Shrikant13
<script>    /* Javascript implementation to convert    infix expression to postfix*/         //Function to return precedence of operators    function prec(c) {        if(c == '^')            return 3;        else if(c == '/' || c=='*')            return 2;        else if(c == '+' || c == '-')            return 1;        else            return -1;    }     // The main function to convert infix expression    //to postfix expression    function infixToPostfix(s) {         let st = []; //For stack operations, we are using C++ built in stack        let result = "";         for(let i = 0; i < s.length; i++) {            let c = s[i];             // If the scanned character is            // an operand, add it to output string.            if((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') || (c >= '0' && c <= '9'))                result += c;             // If the scanned character is an            // ‘(‘, push it to the stack.            else if(c == '(')                st.push('(');             // If the scanned character is an ‘)’,            // pop and to output string from the stack            // until an ‘(‘ is encountered.            else if(c == ')') {                while(st[st.length - 1] != '(')                {                    result += st[st.length - 1];                    st.pop();                }                st.pop();            }             //If an operator is scanned            else {                while(st.length != 0 && prec(s[i]) <= prec(st[st.length - 1])) {                    result += st[st.length - 1];                    st.pop();                }                st.push(c);            }        }         // Pop all the remaining elements from the stack        while(st.length != 0) {            result += st[st.length - 1];            st.pop();        }         document.write(result + "</br>");    }         let exp = "a+b*(c^d-e)^(f+g*h)-i";    infixToPostfix(exp); // This code is contributed by decode2207.</script>
abcd^e-fgh*+^*+i-
https://youtu.be/ysDharaQXkw?list=PLqM7alHXFySF7Lap-wi5qlaD8OEBx9RMV Quiz: Stack Questions
 
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