Use a stack in a program that converts infix notation to postfix notation. Infix is the...

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Use a stack in a program that converts infix notation to postfix notation. Infix is the language we use to learn math, computers may convert of a + b as a b +, Postfix. This requires stack use to perform calculations in a + manner. With this assignment, you practice this conversion and apply stacks. For this assignment you must the built-in stack library for your language: Java: import java.util.Stack; ... Stack postfixStack = new Stack(); www postfixStack.push ("A"); String item = postfixStack.pop(); Python: postfixStack = [] postfixStack.append("A") item= postfixStack.pop () C++: #include <stack> stack postfixStack; postfixStack.push("A"); string item = postfixStack.pop (); C#: using System.Collections.Generic; Stack postfixStack = new Stack (); postfixStack. Push ("A"); string item = postfixStack. Pop (); Assume numbers are integers and standard operations are +, -, *, I, () and exponents. 2+3*4 a*b+5 (1+2)*7 a*b/c (a/(b-c+d))*(e-a)*cabc-d+/ea-*c* a/b-c+d*e-a c ab/c-de +ac"- First print all solutions from table above and require one user input. Expected output: Infix: 2+3*4 Postfix: 234*+ Infix: a"b+5 Postfix: ab 5+ Infix: (1+2)*7 Postfix: 12+7* 234*+ ab*5+ 12+7* ab*c/ Infix: a*b/c Postfix: ab*c/ *********** Infix: (a/(b-c+d))*(e-a)*c Postfix: abc-d+/ea-*c* Infix: a/b-c+d*e-a*c Postfix: ab/c-de*+ac*- Enter infix notation with no spaces: (a/(b-c+d))*(e-a)*c Infix: (a/(b-c+d))*(e-a)*c Postfix: abc-d+/ea-*c* Use a stack in a program that converts infix notation to postfix notation. Infix is the language we use to learn math, computers may convert of a + b as a b +, Postfix. This requires stack use to perform calculations in a + manner. With this assignment, you practice this conversion and apply stacks. For this assignment you must the built-in stack library for your language: Java: import java.util.Stack; ... Stack postfixStack = new Stack(); www postfixStack.push ("A"); String item = postfixStack.pop(); Python: postfixStack = [] postfixStack.append("A") item= postfixStack.pop () C++: #include <stack> stack postfixStack; postfixStack.push("A"); string item = postfixStack.pop (); C#: using System.Collections.Generic; Stack postfixStack = new Stack (); postfixStack. Push ("A"); string item = postfixStack. Pop (); Assume numbers are integers and standard operations are +, -, *, I, () and exponents. 2+3*4 a*b+5 (1+2)*7 a*b/c (a/(b-c+d))*(e-a)*cabc-d+/ea-*c* a/b-c+d*e-a c ab/c-de +ac"- First print all solutions from table above and require one user input. Expected output: Infix: 2+3*4 Postfix: 234*+ Infix: a"b+5 Postfix: ab 5+ Infix: (1+2)*7 Postfix: 12+7* 234*+ ab*5+ 12+7* ab*c/ Infix: a*b/c Postfix: ab*c/ *********** Infix: (a/(b-c+d))*(e-a)*c Postfix: abc-d+/ea-*c* Infix: a/b-c+d*e-a*c Postfix: ab/c-de*+ac*- Enter infix notation with no spaces: (a/(b-c+d))*(e-a)*c Infix: (a/(b-c+d))*(e-a)*c Postfix: abc-d+/ea-*c*

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def infixtopostfixexpression precedence 1 1 2 2 3 postfix stac... View the full answer