

Evaluation of an infix expression that is fully parenthesized using stack in java.
Infix notation is the common arithmetic and logical formula notation, in which operators are written
infixstyle between the operands they act on (e.g. 2 + 2). It is not as simple to parse by computers as prefix notation
( e.g. + 2 2 ) or postfix notation ( e.g. 2 2 + ), but many programming languages use it due to its familiarity.
Here are the steps to evaluate infix expression which is fully parenthesized using stack.
1. Read one input character
2. Actions at end of each input
Opening brackets (2.1) Push into stack and then Go to step (1)
Number (2.2) Push into stack and then Go to step (1)
Operator (2.3) Push into stack and then Go to step (1)
Closing brackets (2.4) Pop from character stack
(2.4.1) Pop is used four times
The first popped element is assigned to op2
The second popped element is assigned to op
The third popped element is assigned to op1
The fourth popped element is the remaining opening bracket,
which can be discarded.
Evaluate op1 op op2
Convert the result into character and
push into the stack
New line character (2.5) Pop from stack and print the answer
STOP
package com.java2novice.ds.stack;
import java.util.StringTokenizer;
public class InfixFullParantEx {
public static String evaluateInfix(String exps){
/**remove if any spaces from the expression**/
exps = exps.replaceAll("\\s+", "");
/**we assume that the expression is in valid format**/
MyGenericsStack<String> stack = new MyGenericsStack<String>(exps.length());
/**break the expression into tokens**/
StringTokenizer tokens = new StringTokenizer(exps, "{}()*/+", true);
while(tokens.hasMoreTokens()){
String tkn = tokens.nextToken();
/**read each token and take action**/
if(tkn.equals("(")
 tkn.equals("{")
 tkn.matches("[09]+")
 tkn.equals("*")
 tkn.equals("/")
 tkn.equals("+")
 tkn.equals("")){
/**push token to the stack**/
stack.push(tkn);
} else if(tkn.equals("}")  tkn.equals(")")){
try {
int op2 = Integer.parseInt(stack.pop());
String oprnd = stack.pop();
int op1 = Integer.parseInt(stack.pop());
/**Below pop removes either } or ) from stack**/
if(!stack.isStackEmpty()){
stack.pop();
}
int result = 0;
if(oprnd.equals("*")){
result = op1*op2;
} else if(oprnd.equals("/")){
result = op1/op2;
} else if(oprnd.equals("+")){
result = op1+op2;
} else if(oprnd.equals("")){
result = op1op2;
}
/**push the result to the stack**/
stack.push(result+"");
} catch (Exception e) {
e.printStackTrace();
break;
}
}
}
String finalResult = "";
try {
finalResult = stack.pop();
} catch (Exception e) {
e.printStackTrace();
}
return finalResult;
}
public static void main(String a[]){
String expr = "((2*5)+(6/2))";
System.out.println("Expression: "+expr);
System.out.println("Final Result: "+evaluateInfix(expr));
expr = "(((2 * 5)  (1 * 2)) / (11  9))";
System.out.println("Expression: "+expr);
System.out.println("Final Result: "+evaluateInfix(expr));
}
}
/**
* Stack implementation
*/
public class MyGenericsStack<T extends Object> {
private int stackSize;
private T[] stackArr;
private int top;
/**
* constructor to create stack with size
* @param size
*/
@SuppressWarnings("unchecked")
public MyGenericsStack(int size) {
this.stackSize = size;
this.stackArr = (T[]) new Object[stackSize];
this.top = 1;
}
/**
* This method adds new entry to the top
* of the stack
* @param entry
* @throws Exception
*/
public void push(T entry){
if(this.isStackFull()){
System.out.println(("Stack is full. Increasing the capacity."));
this.increaseStackCapacity();
}
System.out.println("Adding: "+entry);
this.stackArr[++top] = entry;
}
/**
* This method removes an entry from the
* top of the stack.
* @return
* @throws Exception
*/
public T pop() throws Exception {
if(this.isStackEmpty()){
throw new Exception("Stack is empty. Can not remove element.");
}
T entry = this.stackArr[top];
System.out.println("Removed entry: "+entry);
return entry;
}
/**
* This method returns top of the stack
* without removing it.
* @return
*/
public T peek() {
return stackArr[top];
}
private void increaseStackCapacity(){
@SuppressWarnings("unchecked")
T[] newStack = (T[]) new Object[this.stackSize*2];
for(int i=0;i<stackSize;i++){
newStack[i] = this.stackArr[i];
}
this.stackArr = newStack;
this.stackSize = this.stackSize*2;
}
/**
* This method returns true if the stack is
* empty
* @return
*/
public boolean isStackEmpty() {
return (top == 1);
}
/**
* This method returns true if the stack is full
* @return
*/
public boolean isStackFull() {
return (top == stackSize  1);
}
}


Output: 
Expression: ((2*5)+(6/2))
Adding: (
Adding: (
Adding: 2
Adding: *
Adding: 5
Removed entry: 5
Removed entry: *
Removed entry: 2
Removed entry: (
Adding: 10
Adding: +
Adding: (
Adding: 6
Adding: /
Adding: 2
Removed entry: 2
Removed entry: /
Removed entry: 6
Removed entry: (
Adding: 3
Removed entry: 3
Removed entry: +
Removed entry: 10
Removed entry: (
Adding: 13
Removed entry: 13
Final Result: 13
Expression: (((2 * 5)  (1 * 2)) / (11  9))
Adding: (
Adding: (
Adding: (
Adding: 2
Adding: *
Adding: 5
Removed entry: 5
Removed entry: *
Removed entry: 2
Removed entry: (
Adding: 10
Adding: 
Adding: (
Adding: 1
Adding: *
Adding: 2
Removed entry: 2
Removed entry: *
Removed entry: 1
Removed entry: (
Adding: 2
Removed entry: 2
Removed entry: 
Removed entry: 10
Removed entry: (
Adding: 8
Adding: /
Adding: (
Adding: 11
Adding: 
Adding: 9
Removed entry: 9
Removed entry: 
Removed entry: 11
Removed entry: (
Adding: 2
Removed entry: 2
Removed entry: /
Removed entry: 8
Removed entry: (
Adding: 4
Removed entry: 4
Final Result: 4





List of Stack Data Structure Examples
 Stack introduction & implementation
 Java Dynamic Stack Implementation
 Stack implementation using generics bounded type.
 Reverse a word or string using Stack data structure.
 Write a program to find out delimiter matching using stack.
 Convert a decimal into a binary number using stack.
 Towers of Hanoi implementation using stack.
 Evaluation of an infix expression that is fully parenthesized using stack in java.



Procedural Vs Objectoriented Programs
In procedural program, programming logic follows certain procedures and the instructions are executed one after another.
In OOP program, unit of program is object, which is nothing but combination of data and code.
In procedural program, data
is exposed to the whole program whereas in OOPs program, it is accessible with in the object and which in turn assures the
security of the code.
It’s not that I’m so smart, it’s just that I stay with problems longer.
 Albert Einstein
