Question

Write a java code to implement the infix to postfix algorithm as described below:

Algorithm convertTo Post fix ( infix) // Converts an infix expression to an equivalent postfix expression operatorStack = a new empty stack postfix = anew empty string while (infix has characters left to parse) nextCharacter =next nonblank character of infix switch (nextCharacter) { case variable: Append nextCharacter to postfix break case 'A' operatorStack.push (nextCharacter) break case '+ case '-' : case '*' : case '/' while (!operatorStack . i s Empty () and precedence of next Character < precedence of operatorStack . peek()) { Append operatorStack . peek () to postfix operatorStack.pop () operatorStack.push (nextCharacter) break case operatorStack.push (nextCharacter) break case ')'//Stack is not empty if infix expression is valid topOperator = operatorStack. pop () while (top0pe rator != '(' Append topOperator to postfix top0perator } break operatorStack . pop ( default: break // Ignore unexpected characters } } while (!operatorStack . i s Empty () operatorStack.pop () top0perator Append topOperator to postfix } return postfix

Answer 1

`Hey,

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import java.util.Stack;

public class InfixToPostFix {

static int precedence(char c){
switch (c){
case '+':
case '-':
return 1;
case '*':
case '/':
return 2;
case '^':
return 3;
}
return -1;
}

static String infixToPostFix(String expression){

String result = "";
Stack<Character> stack = new Stack<>();
for (int i = 0; i <expression.length() ; i++) {
char c = expression.charAt(i);

//check if char is operator
if(precedence(c)>0){
while(stack.isEmpty()==false && precedence(stack.peek())>=precedence(c)){
result += stack.pop();
}
stack.push(c);
}else if(c==')'){
char x = stack.pop();
while(x!='('){
result += x;
x = stack.pop();
}
}else if(c=='('){
stack.push(c);
}else{
//character is neither operator nor (
result += c;
}
}
for (int i = 0; i <=stack.size() ; i++) {
result += stack.pop();
}
return result;
}

public static void main(String[] args) {
String exp = "A+B*(C^D-E)";
System.out.println("Infix Expression: " + exp);
System.out.println("Postfix Expression: " + infixToPostFix(exp));
}
}

S How do you verify the identity: x C Computer Science question X <> Online Java Compiler - Online. X + f → C tutorialspoint.com/compile_java_online.php Á : : Incognito A : o codingground Compile and Execute Java Online (JDK 1.8.0) Fork A. Project Edit » Setting - Login SIMPLY 0.Execute > Share Source File STDIN L.lt Result ES $javac InfixToPostFix.java Sjava -Xmx128M -Xms16M InfixToPostFix Infix Expression: A+B* (CD-E) Postfix Expression: ABCD E-*+ static int precedence(char c){ switch (c){ case '+': case '-': return 1; case '*': case '/': return 2; case 'A': return 3; return -1; static String infixToPostFix(String expression) { String result = ""; Stack<Character> stack = new StackO ; for (int i = 0; i <expression.length(); i++) { char c = expression.charAt(i); //check if char is operator if(precedence(C)>0){ while(stack.isEmpty()==false && precedence(stack.peek O)>=precedence()){ result += stack.pop(); stack.push(); }else if(c==''){ char x = stack.pop(); while(x!='(){ result += x; X = stack.popO; We use cookies to provide and improve our services. By using our site, you consent to our Cookies Policy. Accept Learn more

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