1. Chapter 24 Implementing Lists, Stacks, Queues, and Priority Queues CIS265/506 Cleveland State University – Prof. Victor Matos Adapted from: Introduction to Java Programming: Comprehensive Version, Eighth Edition by Y. Daniel Liang

2. 2 Objectives 1. To design a list with interface and abstract class (§24.2). 2. To design and implement a dynamic list using an array (§24.3). 3. To design and implement a dynamic list using a linked structure (§24.4). 4. To design and implement a stack using an array list (§24.5). 5. To design and implement a queue using a linked list (§24.6). 6. To evaluate expressions using stacks (§24.7).

3. 3 What is a Data Structure? A data structure is a collection of data organized in some fashion. A data structure not only stores data, but also supports the operations for manipulating data in the structure. For example: An array is a data structure that holds a collection of data in sequential order. You can find the size of the array, store, retrieve, and modify data in the array. Array is simple and easy to use, but it has some limitations: …

4. 4 Limitations of arrays Once an array is created, its size cannot be altered. Array provides inadequate support for: inserting, deleting, sorting, and searching operations.

5. 5 Object-Oriented Data Structure A data structure can be consider as a container object or a collection object. To define a data structure is similar to declare a class. The class for a data structure should 1. use data fields to store data and 2. provide methods to support operations such as insertion and deletion.

6. 6 Four Classic Data Structures Four classic dynamic data structures are lists, stacks, queues, and binary trees. 1. A list is a collection of data stored sequentially. It supports insertion and deletion anywhere in the list. 2. A stack can be perceived as a special type of the list where insertions and deletions take place only at the one end, referred to as the top of a stack. 3. A queue represents a waiting list, where insertions take place at the back (also referred to as the tail of) of a queue and deletions take place from the front (also referred to as the head of) of a queue. 4. A binary tree is a data structure in which each data item has one direct ancestor and up to two descendants.

7. 7 Lists A list is a common data structure to store data in sequential order. Operations on a list are usually the following: · Retrieve an element from this list. · Insert a new element to this list. · Delete an element from this list. · Find how many elements are in this list. · Find if an element is in this list. · Find if this list is empty.

8. 8 Two Ways to Implement Lists 1. Use an array to store the elements. The array is dynamically created. If the capacity of the array is exceeded, create a new larger array and copy all the elements from the current array to the new array. 2. Use a linked structure. A linked structure consists of nodes. Each node is dynamically created to hold an element. All the nodes are linked together to form a list.

9. 9 Design of ArrayList and LinkedList For convenience, let’s name these two classes: MyArrayList and MyLinkedList. These two classes have common operations, but different data fields. Their common operations can be generalized in an interface or an abstract class. Such an abstract class is known as a convenience class.

10. 10 MyList Interface and MyAbstractList Class

11. 11 MyList Interface public interface MyList { /** Add a new element at the end of this list */ public void add(E e); /** Add a new element at the specified index in this list */ public void add(int index, E e); /** Clear the list */ public void clear(); /** Return true if this list contains the element */ public boolean contains(E e); /** Return the element from this list at the specified index */ public E get(int index); /** Return the index of the first matching element in this list. Return -1 if no match. */ public int indexOf(E e); /** Return true if this list contains no elements */ public boolean isEmpty(); /** Return the index of the last matching element in this list. Return -1 if no match. */ public int lastIndexOf(E e); /** Remove the first occurrence of the element o from this list. * Shift any subsequent elements to the left. * Return true if the element is removed. */ public boolean remove(E e); /** Remove the element at the specified position in this list * Shift any subsequent elements to the left. * Return the element that was removed from the list. */ public E remove(int index); /** Replace the element at the specified position in this list * with the specified element and returns the new set. */ public Object set(int index, E e); /** Return the number of elements in this list */ public int size(); }

12. 12 MyAbstractList Class public abstract class MyAbstractList implements MyList { protected int size = 0; // The size of the list /** Create a default list */ protected MyAbstractList() { } /** Create a list from an array of objects */ protected MyAbstractList(E[] objects) { for (int i = 0; i < objects.length; i++) add(objects[i]); } /** Add a new element at the end of this list */ public void add(E e) { add(size, e); } /** Return true if this list contains no elements */ public boolean isEmpty() { return size == 0; } /** Return the number of elements in this list */ public int size() { return size; } /** Remove the first occurrence of the element o from this list. * Shift any subsequent elements to the left. * Return true if the element is removed. */ public boolean remove(E e) { if (indexOf(e) >= 0) { remove(indexOf(e)); return true; } else return false; }

13. 13 Array Lists Array is a fixed-size data structure. Once an array is created, its size cannot be changed. You can still use array to implement dynamic data structures. The trick is to create a new larger array to replace the current array if the current array cannot hold new elements in the list. 1. Initially, an array, say data of ObjectType[], is created with a default size. 2. When inserting a new element into the array, first ensure there is enough room in the array. 3. If not, create a new array with the size as twice as the current one. Copy the elements from the current array to the new array. The new array now becomes the current array.

14. 14 Array List Animation www.cs.armstrong.edu/liang/animation/ArrayListAnimation.html

15. 15 Insertion Before inserting a new element at a specified index, shift all the elements after the index to the right and increase the list size by 1.

16. 16 Deletion To remove an element at a specified index, shift all the elements after the index to the left by one position and decrease the list size by 1.

17. 17 Implementing MyArrayList

18. 18 Implementing MyArrayList public class MyArrayList extends MyAbstractList { public static final int INITIAL_CAPACITY = 16; private E[] data = (E[])new Object[INITIAL_CAPACITY]; /** Create a default list */ public MyArrayList() { } /** Create a list from an array of objects */ public MyArrayList(E[] objects) { for (int i = 0; i < objects.length; i++) add(objects[i]); // Warning: don’t use super(objects)! } /** Add a new element at the specified index in this list */ public void add(int index, E e) { ensureCapacity(); // Move the elements to the right after the specified index for (int i = size - 1; i >= index; i--) data[i + 1] = data[i]; // Insert new element to data[index] data[index] = e; // Increase size by 1 size++; }

19. 19 Implementing MyArrayList /** Create a new larger array, double the current size */ private void ensureCapacity() { if (size >= data.length) { E[] newData = (E[])(new Object[size * 2 + 1]); System.arraycopy(data, 0, newData, 0, size); data = newData; } /** Clear the list */ public void clear() { data = (E[])new Object[INITIAL_CAPACITY]; size = 0; }

20. 20 Implementing MyArrayList cont. 2/4 /** Remove the element at the specified position in this list * Shift any subsequent elements to the left. * Return the element that was removed from the list. */ public E remove(int index) { E e = data[index]; // Shift data to the left for (int j = index; j < size - 1; j++) data[j] = data[j + 1]; data[size - 1] = null; // This element is now null // Decrement size size--; return e; }

21. 21 Implementing MyArrayList cont. 3/4 // Replace element at specified position in this list with the specified element public E set(int index, E e) { E old = data[index]; data[index] = e; return old; } /** Override toString() to return elements in the list */ public String toString() { StringBuilder result = new StringBuilder("["); for (int i = 0; i < size; i++) { result.append(data[i]); if (i < size - 1) result.append(", "); } return result.toString() + "]"; } /** Trims the capacity to current size */ public void trimToSize() { if (size != data.length) { // If size == capacity, no need to trim E[] newData = (E[])(new Object[size]); System.arraycopy(data, 0, newData, 0, size); data = newData; }

22. 22 Test MyArrayList cont. 4/4 public class TestList { public static void main(String[] args) { // Create a list MyList list = new MyArrayList (); // Add elements to the list list.add("America"); // Add it to the list System.out.println("(1) " + list); list.add(0, "Canada"); // Add it to the beginning of the list System.out.println("(2) " + list); list.add("Russia"); // Add it to the end of the list System.out.println("(3) " + list); list.add("France"); // Add it to the end of the list System.out.println("(4) " + list); list.add(2, "Germany"); // Add it to the list at index 2 System.out.println("(5) " + list); list.add(5, "Norway"); // Add it to the list at index 5 System.out.println("(6) " + list); // Remove elements from the list list.remove("Canada"); // Same as list.remove(0) in this case System.out.println("(7) " + list); list.remove(2); // Remove the element at index 2 System.out.println("(8) " + list); list.remove(list.size() - 1); // Remove the last element System.out.println("(9) " + list); }

23. 23 Linked Lists Observations: Since MyArrayList is implemented using an array, the methods get(int index), set(int index, Object o) and the add(Object o) for adding an element at the end of the list are efficient. However, the methods add(int index, Object o) and remove(int index) are inefficient because they require shifting potentially a large number of elements. You can use a linked structure to improve efficiency for adding and removing an element anywhere in a list.

24. 24 Linked List Animation www.cs.armstrong.edu/liang/animation/LinkedListAnimation.html

25. 25 Nodes in Linked Lists A linked list consists of nodes. Each node contains a data element, and each node is linked to its next neighbor. Thus a node can be defined as a class, as follows: class Node { E element; Node next; public Node(E obj) { element = obj; }

26. 26 Adding Three Nodes The variable head refers to the first node in the list, and the variable tail refers to the last node in the list. If the list is empty, both are null. Example. You can create three nodes to store three strings in a list, as follows: Step 1: Declare head and tail : Node head = null; Node tail = null; The list is empty now

27. 27 Adding Three Nodes, cont. Step 2: Create the first node and insert it to the list:

28. 28 Adding Three Nodes, cont. Step 3: Create the second node and insert it to the list:

29. 29 Adding Three Nodes, cont. Step 4: Create the third node and insert it to the list:

30. 30 Traversing All Elements in the List The last node has its next pointer data field set to null. You may use the following loop to traverse all the nodes in the list. Node current = head; while (current != null) { System.out.println(current.element); current = current.next; }

31. 31 MyLinkedList Designing/Implementing a generic, dynamic, singly-linked list.

32. 32 Implementing addFirst(E o) public void addFirst(E o) { Node newNode = new Node (o); newNode.next = head; head = newNode; size++; if (tail == null) tail = head; }

33. 33 Implementing addLast(E o) public void addLast(E o) { if (tail == null) { head = tail = new Node (element); } else { tail.next = new Node (element); tail = tail.next; } size++; }

34. 34 Implementing add(int index, E obj) public void add(int index, E obj) { if (index == 0) addFirst(obj); else if (index >= size) addLast(obj); else { Node current = head; for (int i = 1; i < index; i++) current = current.next; Node temp = current.next; current.next = new Node (obj); (current.next).next = temp; size++; }

35. 35 Implementing removeFirst() public E removeFirst() { if (size == 0) return null; else { Node temp = head; head = head.next; size--; if (head == null) tail = null; return temp.element; }

36. 36 Implementing removeLast() public E removeLast() { if (size == 0) return null; else if (size == 1) { Node temp = head; head = tail = null; size = 0; return temp.element; } else { Node current = head; for (int i = 0; i < size - 2; i++) current = current.next; Node temp = tail; tail = current; tail.next = null; size--; return temp.element; }

37. 37 Implementing remove(int index) public E remove(int index) { if (index = size) return null; else if (index == 0) return removeFirst(); else if (index == size - 1) return removeLast(); else { Node previous = head; for (int i = 1; i < index; i++) { previous = previous.next; } Node current = previous.next; previous.next = current.next; size--; return current.element; }

38. 38 Circular Linked Lists A circular, singly linked list is like a singly linked list, except that the pointer of the last node points back to the first node.

39. 39 Doubly Linked Lists A doubly linked list contains the nodes with two pointers. One points to the next node and the other points to the previous node. These two pointers are conveniently called a forward pointer and a backward pointer. So, a doubly linked list can be traversed forward and backward.

40. 40 Circular Doubly Linked Lists A circular, doubly linked list is doubly linked list, except that the forward pointer of the last node points to the first node and the backward pointer of the first pointer points to the last node.

41. 41 Stacks A stack can be viewed as a special type of list, where the elements are accessed, inserted, and deleted only from the end, called the top, of the stack.

42. 42 Queues A queue represents a waiting list. A queue can be viewed as a special type of list, where the elements are inserted into the end (tail) of the queue, and are accessed and deleted from the beginning (head) of the queue.

43. 43 Stack Animation www.cs.armstrong.edu/liang/animation/StackAnimation.html

44. 44 Queue Animation www.cs.armstrong.edu/liang/animation/QueueAnimation.html

45. 45 Implementing Stacks and Queues Suggestions 1. Use an array list to implement Stack 2. Use a linked list to implement Queue Since the insertion and deletion operations on a stack are made only at the end of the stack, using an array list to implement a stack is more efficient than a linked list. Since deletions are made at the beginning of the list, it is more efficient to implement a queue using a linked list than an array list.

46. 46 MyStack This section implements a stack class using an array list and a queue using a linked list.

47. 47 MyStack public class MyStack { private java.util.ArrayList list = new java.util.ArrayList(); public boolean isEmpty() { return list.isEmpty(); } public int getSize() { return list.size(); } public Object peek() { return list.get(getSize() - 1); } public Object pop() { Object o = list.get(getSize() - 1); list.remove(getSize() - 1); return o; } public void push(Object o) { list.add(o); } public int search(Object o) { return list.lastIndexOf(o); } /** Override the toString in the Object class */ public String toString() { return "stack: " + list.toString(); }

48. 48 MyQueue Implementing a queue using a custom-made linked list.

49. 49 MyQueue public class MyQueue { private MyLinkedList list = new MyLinkedList (); public void enqueue(E e) { list.addLast(e); } public E dequeue() { return list.removeFirst(); } public int getSize() { return list.size(); } public String toString() { return "Queue: " + list.toString(); }

50. 50 Example: Using Stacks and Queues public class TestStackQueue { public static void main(String[] args) { // Create a stack GenericStack stack = new GenericStack (); // Add elements to the stack stack.push(“aaa"); // Push it to the stack System.out.println("(1) " + stack); stack.push(“bbb"); // Push it to the the stack System.out.println("(2) " + stack); stack.push(“ccc"); // Push it to the stack stack.push(“ddd"); // Push it to the stack System.out.println("(3) " + stack); // Remove elements from the stack System.out.println("(4) " + stack.pop()); System.out.println("(5) " + stack.pop()); System.out.println("(6) " + stack); // Create a queue MyQueue queue = new MyQueue (); // Add elements to the queue queue.enqueue(“111"); // Add it to the queue System.out.println("(7) " + queue); queue.enqueue(“222"); // Add it to the queue System.out.println("(8) " + queue); queue.enqueue(“333"); // Add it to the queue queue.enqueue(“444"); // Add it to the queue System.out.println("(9) " + queue); // Remove elements from the queue System.out.println("(10) " + queue.dequeue()); System.out.println("(11) " + queue.dequeue()); System.out.println("(12) " + queue); }

51. 51 Priority Queue 1. A regular queue is a first-in and first-out data structure. 2. Elements are appended to the end of the queue and are removed from the beginning of the queue. 3. In a priority queue, elements are assigned with priorities. 4. When accessing elements, the element with the highest priority is removed first. 5. A priority queue has a largest-in, first-out behavior. 6. For example, the emergency room in a hospital assigns patients with priority numbers; the patient with the highest priority is treated first.

52. Case Study: Using Stacks to process arithmetical expressions

53. 53 Basic Definitions There are many ways to write (and evaluate) mathematical equations. The first, called infix notation, is what we are familiar with from elementary school: (5*2)-(((3+4*7)+8/6)*9) You would evaluate this equation from right to left, taking in to account precedence. So: 10 - (((3+28)+1.33)*9) 10 - ((31 + 1.33)*9) 10 - (32.33 * 9) 10 - 291 -281

54. 54 Basic Definitions An alternate method is postfix or Reverse Polish Notation (RPN). The corresponding RPN equation would be: 5 2 * 3 4 7 * + 8 6 / + 9 * - We’ll see how to evaluate this in a minute.

55. 55 Example: HP-65 TypeProgrammable Introduced1974 - MSRP $795 Calculator Entry modeRPN Display Type7-segment red LED Display Size10 digits CPU Processorproprietary Programming language(s) key codes Memory Register 8 (9) plus 4-level working stack Program Steps100 REFERENCE: http://en.wikipedia.org/wiki/HP-65 (Accessed on April 16, 2015) “… Like all Hewlett-Packard calculators of the era and most since, the HP-65 used reverse Polish notation (RPN) and a four-level automatic operand stack”

56. 56 Basic Definitions Note that in an infix expression, the operators appear in between the operands (1 + 2). Postfix equations have the operators after the equations (1 2 +). In Forward Polish Notation or prefix equations, the operators appear before the operands. The prefix form is rarely used (+ 1 2).

57. 57 Basic Definitions Reversed Polish Notation got its name from Jan Lukasiewicz, a Polish mathematician, who first published in 1951. Lukasiewicz was a pioneer in three-valued propositional calculus, he also was interested in developing a parenthesis-free method of representing logic expressions. Today, RPN is used in many compilers and interpreters as an intermediate form for representing logic. http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Lukasiewicz.html University of St. Andrews.

58. 58 Examples InfixPrefixPostfix A+B+ABAB+ A+B*C+A*BCABC*+ A*(B+C)*A+BCABC+* A*B+C+*ABCAB*C+ A+B*C+D-E*F-++A*BCD*EFABC*+D+EF*- (A+B)*(C+D-E)*F**+AB-+CDEFAB+CD+E-*F* RPN expressions

59. 59 Evaluating RPN Expressions We evaluate RPN using a left-to-right scan. An operator is preceded by two operands, so we store the first operand, then the second, and once the operator arrives, we use it to compute or evaluate the two operands we just stored. 3 5 + Store the value 3, then the value 5, then using the + operator, evaluate the pending calculation as 8.

60. 60 Evaluating RPN Expressions What happens if our equation has more than one operator? Now we’ll need a way to store the intermediate result as well: 3 5 + 10 * Store the 3, then 5. Evaluate with the +, getting 8. Store the 8, then store10, when * arrives evaluate the expression using the previous two arguments. The final result is 80.

61. 61 Evaluating RPN Expressions It starts to become apparent that we apply the operator to the last two operands we stored. Example: 3 5 2 * - Store the 3, then the 5, then the 2. Apply the * to the 5 and 2, getting 10. Store the value 10. Apply the - operator to the stored values 3 and 10 (3 - 10) getting -7.

62. 62 Evaluating RPN Expressions Algorithm to evaluate an RPN expression 1.We scan our input stream from left to right, removing the first character as we go. 2.We check the character to see if it is an operator or an operand. 3.If it is an operand, we push it on the stack. 4.If it is an operator, we remove the top two items from the stack, and perform the requested operation. 5.We then push the result back on the stack. 6.If all went well, at the end of the stream, there will be only one item on the stack - our final result.

63. 63 Evaluating RPN Expressions 1 2 3 4 5 6 7 8 9 10 3, 5,+ 2, 4 - * 6 * 5 + 2, 4 - * 6 * * 6 * + 2, 4 - * 6 * 2, 4 - * 6 * 4 - * 6 * - * 6 * 6 * * Step Stack RPN Expression 3 5353 8 2828 428428 -2 8 -16 6 -16 -96

64. 64 Evaluating RPN Expressions 1/3 package csu.matos; import java.util.Stack; import java.util.StringTokenizer; public class Driver { public static void main(String[] args) { // Taken from Daniel Liang – Intro to Java Prog. // the input is a correct postfix expression String expression = "1 2 + 3 *"; try { System.out.println( evaluateExpression(expression) ); } catch (Exception ex) { System.out.println("Wrong expression"); } /** Evaluate an expression **********************************************/ public static int evaluateExpression(String expression) { // Create operandStack to store operands Stack operandStack = new Stack (); // Extract operands and operators StringTokenizer tokens = new StringTokenizer(expression, " +-/*%", true);

65. 65 Evaluating RPN Expressions 2/3 // Phase 1: Scan tokens while (tokens.hasMoreTokens()) { String token = tokens.nextToken().trim(); // Extract a token if (token.length() == 0) { // Blank space continue; // Back to the while loop to extract the next token } else if (token.equals("+") || token.equals("-") || token.equals("*") || token.equals("/")) { processAnOperator(token, operandStack); } else { // An operand scanned // Push an operand to the stack operandStack.push(new Integer(token)); } // Return the result return ((Integer)(operandStack.pop())).intValue(); }

66. 66 Evaluating RPN Expressions 3/3 // Process one operator: Take an operator from operatorStack and // apply it on the operands in the operandStack public static void processAnOperator(String token, Stack operandStack) { char op = token.chatAt(0); if (op == '+') { int op1 = ((Integer)(operandStack.pop())).intValue(); int op2 = ((Integer)(operandStack.pop())).intValue(); operandStack.push(new Integer(op2 + op1)); } else if (op == '-') { int op1 = ((Integer)(operandStack.pop())).intValue(); int op2 = ((Integer)(operandStack.pop())).intValue(); operandStack.push(new Integer(op2 - op1)); } else if ((op == '*')) { int op1 = ((Integer)(operandStack.pop())).intValue(); int op2 = ((Integer)(operandStack.pop())).intValue(); operandStack.push(new Integer(op2 * op1)); } else if (op == '/') { int op1 = ((Integer)(operandStack.pop())).intValue(); int op2 = ((Integer)(operandStack.pop())).intValue(); operandStack.push(new Integer(op2 / op1)); }

67. 67 Converting Infix to Postfix Manual Transformation (Continued) Example: A + B * C Step 1: (A + ( B * C ) ) Change all infix notations in each parenthesis to postfix notation starting from the innermost expressions. This is done by moving the operator to the location of the expression’s closing parenthesis Step 2:( A + ( B C * ) ) Step 3:( A ( B C * ) + )

68. 68 Converting Infix to Postfix Manual Transformation (Continued) Example: A + B * C Step 2: (A + ( B * C ) ) Step 3: (A ( B C * ) + ) Remove all parentheses Step 4: A B C * +

69. 69 Converting Infix to Postfix Another Example (A + B ) * C + D + E * F - G Add Parentheses ( ( ( ( ( A + B ) * C ) + D ) + ( E * F ) ) - G ) Move Operators ( ( ( ( ( A B + ) C * ) D + ) ( E F * ) + ) G - ) Remove Parentheses A B + C * D + E F * + G -

70. 70 Converting Infix to Postfix Solution Version 2 - Parsing Strategy Example: A * B Write to output A, Store the * on a stack, Write the B, then Get the * from the stack and write it to output. So, solution is: A B *

71. 71 Converting Infix to Postfix Precedence of operators Highest 2: * / 1: + - Lowest:0: (

72. 72 Converting Infix to Postfix Conversion Algorithm while there is more data get the first symbol if symbol = ( put it on the stack if symbol = ) take item from top of stack while this item != ( add it to the end of the output string Cont....

73. 73 Converting Infix to Postfix if symbol is +, -, *, \ look at top of the stack while (stack is not empty AND the priority of the current symbol is less than OR equal to the priority of the symbol on top of the stack ) Get the stack item and add it to the end of the output string; put the current symbol on top of the stack if symbol is a character add it to the end of the output string End loop Cont....

74. 74 Converting Infix to Postfix Finally While ( stack is not empty ) Get the next item from the stack and place it at the end of the output string End

75. 75 Converting Infix to Postfix Function precedence_test (operator) case operator “*” OR “/” return 2; case operator “+” OR “-” return 1; case operator “(“ return 0; default return 99; //signals error condition!

76. 76 Converting Infix to Postfix Input Buffer *B-(C+D)+E B-(C+D)+E -(C+D)+E (C+D)+E C+D)+E +D)+E D)+E )+E +E E Operator Stack EMPTY * - -( -(+ - + EMPTY Output String A A B A B * A B * C A B * C D A B * C D + A B * C D + - A B * C D + - E A B * C D + - E + The line we are analyzing is: A*B-(C+D)+E

77. 77 Converting Infix to Postfix 1/4 public static void main(String[] args) { // Provide a correct infix expression to be converted String expression = "( 1 + 2 ) * 3"; try { System.out.println(infixToPostfix(expression)); } catch (Exception ex) { System.out.println("Wrong expression"); } public static String infixToPostfix(String expression) { // Result string String s = ""; // Create operatorStack to store operators Stack operatorStack = new Stack(); // Extract operands and operators StringTokenizer tokens = new StringTokenizer(expression, "()+-/*%", true);

78. 78 Converting Infix to Postfix 2/4 // Phase 1: Scan tokens while (tokens.hasMoreTokens()) { String token = tokens.nextToken().trim(); // Extract a token if (token.length() == 0) { // Blank space continue; // Back to the while loop to extract the next token } else if (token.charAt(0) == '+' || token.charAt(0) == '-') { // remove all +, -, *, / on top of the operator stack while (!operatorStack.isEmpty() && (operatorStack.peek().equals('+') || operatorStack.peek().equals('-') || operatorStack.peek().equals('*') || operatorStack.peek().equals('/') )) { s += operatorStack.pop() + " "; } // push the incoming + or - operator into the operator stack operatorStack.push(new Character(token.charAt(0))); }

79. 79 Converting Infix to Postfix 3/4 else if (token.charAt(0) == '*' || token.charAt(0) == '/') { // remove all *, / on top of the operator stack while (!operatorStack.isEmpty() && (operatorStack.peek().equals('*') || operatorStack.peek().equals('/') )) { s += operatorStack.pop() + " "; } // Push the incoming * or / operator into the operator stack operatorStack.push(new Character(token.charAt(0))); } else if (token.trim().charAt(0) == '(') { operatorStack.push(new Character('(')); // Push '(' to stack } else if (token.trim().charAt(0) == ')') { // remove all the operators from the stack until seeing '(' while (!operatorStack.peek().equals('(')) { s += operatorStack.pop() + " "; } operatorStack.pop(); // Pop the '(' symbol from the stack } else { // An operand scanned // Push an operand to the stack s += token + " "; }

80. 80 Converting Infix to Postfix 4/4 // Phase 2: remove the remaining operators from the stack while (!operatorStack.isEmpty()) { s += operatorStack.pop() + " "; } // Return the result return s; }