Requisito previo – Apilar | Conjunto 1 (Introducción)
Expresión infija: La expresión de la forma a op b. Cuando un operador está entre cada par de operandos.
Postfijo expresión: La expresión de la forma ab op. Cuando se sigue un operador para cada par de operandos.
¿Por qué posfijar la representación de la expresión?
El compilador escanea la expresión de izquierda a derecha o de derecha a izquierda.
Considere la siguiente expresión: a op1 b op2 c op3 d
ie : a + b * c + d
El compilador primero escanea la expresión para evaluar la expresión b * c, luego vuelve a escanear la expresión para agregarle a. Luego, el resultado se agrega a d después de otro escaneo.
El escaneo repetido lo hace muy ineficiente. Es mejor convertir la expresión a sufijo (o prefijo) antes de la evaluación.
La expresión correspondiente en forma de sufijo es abc*+d+. Las expresiones de sufijo se pueden evaluar fácilmente usando una pila. Cubriremos la evaluación de la expresión postfix en una publicación separada.
Algoritmo
1. Escanea la expresión infija de izquierda a derecha.
2. Si el carácter escaneado es un operando, envíelo.
3. Si no,
1Si la precedencia y asociatividad del operador escaneado es mayor que la precedencia y asociatividad del operador en la pila (o la pila está vacía o la pila contiene un ‘(‘ ), empújelo.
C++14
/* C++ implementation to convert
infix expression to postfix*/
#include <bits/stdc++.h>
using namespace std;
// Function to return precedence of operators
int 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 expression
void 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 functions
int main()
{
string exp = "a+b*(c^d-e)^(f+g*h)-i";
infixToPostfix(exp);
return 0;
}
C
// C program to convert infix expression to postfix
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
// Stack type
struct Stack
{
int top;
unsigned capacity;
int* array;
};
// Stack Operations
struct 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 operand
int 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 precedence
int 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 functions
int main()
{
char exp[] = "a+b*(c^d-e)^(f+g*h)-i";
infixToPostfix(exp);
return 0;
}
Java
/* 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
# 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 function
exp = "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)
C#
using System;
using System.Collections.Generic;
/* c# implementation to convert
infix 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
Javascript
<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>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA