1. Stacks 24.06.20161

2. The Stack ADT stores arbitrary objects. Insertions and deletions follow the last-in first-out (LIFO) scheme. The last item placed on the stack will be the first item removed The Stack ADT 24.06.20162

3. ADT Stack Operations Create an empty stack Destroy a stack Determine whether a stack is empty Add a new item -- push Remove the item that was added most recently -- pop Retrieve the item that was added most recently – top/getTop pushpop Stack top 24.06.20163

4. ADT Stack Operations‏ createStack() ‏ creates a an empty stack destroyStack() ‏ destroys a stack isEmpty():boolean determines whether a stack is empty or not push(in newItem:StackItemType) throw StackException Adds newItem to the top of a stack pop() throw StackException pop(out stackTop:StackItemType) throw StackException Removes the top of a stack (i.e., removes the item that was added most recently) getTop(out stackTop:StackItemType) throw StackException Retrieves the top of stack into stackTop 24.06.20164

5. Implementations of the ADT Stack The ADT stack can be implemented using An array A linked list The ADT list All three implementations use a StackException class to handle possible exceptions 24.06.20165

6. 2 1 0 Implementations of the ADT Stack 24.06.20166

7. An Array-Based Implementation of the ADT Stack Private data fields An array of items of type StackItemType The index top Compiler-generated destructor and copy constructor 24.06.20167

8. An Array-Based Implementation: Header File #include "StackException.h" const int MAX_STACK = maximum-size-of-stack; typedef desired-type-of-stack-item StackItemType; class Stack { public: Stack(); // default constructor; copy constructor and destructor are supplied by the compiler // stack operations: bool isEmpty() const; // Determines whether a stack is empty. void push(StackItemType newItem) throw(StackException); // Adds an item to the top of a stack. void pop() throw(StackException);// Removes the top of a stack. void topAndPop(StackItemType& stackTop) throw(StackException); void getTop(StackItemType& stackTop) const throw(StackException); // Retrieves top of stack. private: StackItemType items[MAX_STACK]; // array of stack items int top; // index to top of stack }; 24.06.20168

9. An Array-Based Implementation: isEmpty, push #include "StackA.h" // Stack class specification file Stack::Stack(): top(-1) {} // default constructor bool Stack::isEmpty() const { return top < 0; } void Stack::push(StackItemType newItem) throw(StackException) { // if stack has no more room for another item if (top >= MAX_STACK-1) ‏ throw StackException("StackException: stack full on push"); else { ++top; items[top] = newItem; }} 24.06.20169

10. An Array-Based Implementation: pop void Stack::pop() throw(StackException) { if (isEmpty()) ‏ throw StackException("StackException: stack empty on pop"); else --top; // stack is not empty; pop top } void Stack::topAndPop(StackItemType& stackTop) throw(StackException) { if (isEmpty()) ‏ throw StackException("StackException: stack empty on pop"); else { // stack is not empty; retrieve top stackTop = items[top]; --top; // pop top } 24.06.201610

11. An Array-Based Implementation: getTop void Stack::getTop(StackItemType& stackTop) const throw(StackException) { if (isEmpty()) ‏ throw StackException("StackException: stack empty on getTop"); else // stack is not empty; retrieve top stackTop = items[top]; } 24.06.201611

12. An Array-Based Implementation: main #include #include "StackA.h" int main() { StackItemType anItem; Stack aStack; cin >> anItem; // read an item aStack.push(anItem); // push it onto stack return 0; } 24.06.201612

13. An Array-Based Implementation: Stack Exception Class #include using namespace std; class StackException: public exception{ public: StackException(const string &message=""): exception(message.c_str()) { } }; 24.06.201613

14. A Pointer-Based Implementation of the ADT Stack A pointer-based implementation Required when the stack needs to grow and shrink dynamically top is a reference to the head of a linked list of items A copy constructor and destructor must be supplied 24.06.201614

15. A Pointer-Based Implementation: Header File #include "StackException.h" typedef desired-type-of-stack-item StackItemType; class Stack{ public: Stack(); // default constructor Stack(const Stack& aStack); // copy constructor ~Stack(); // destructor bool isEmpty() const; void push(StackItemType newItem); void pop() throw(StackException); void topAndPop(StackItemType& stackTop) throw(StackException); void getTop(StackItemType& stackTop) const throw(StackException); 24.06.201615

16. A Pointer-Based Implementation: Header File private: struct StackNode { // a node on the stack StackItemType item; // a data item on the stack StackNode *next; // pointer to next node }; StackNode *topPtr; // pointer to first node in the stack }; 24.06.201616

17. A Pointer-Based Implementation: constructor #include "StackP.h" // header file #include // for NULL Stack::Stack() : topPtr(NULL) {} // default constructor Stack::Stack(const Stack& aStack) {// copy constructor if (aStack.topPtr == NULL) ‏ topPtr = NULL; // original list is empty else { // copy first node topPtr = new StackNode; topPtr->item = aStack.topPtr->item; // copy rest of list StackNode *newPtr = topPtr; // new list pointer for (StackNode *origPtr = aStack.topPtr->next; origPtr != NULL; origPtr = origPtr->next) { newPtr->next = new StackNode; newPtr = newPtr->next; newPtr->item = origPtr->item; } newPtr->next = NULL; } 24.06.201617

18. A Pointer-Based Implementation: destructor, isEmpty Stack::~Stack() { // pop until stack is empty while (!isEmpty()) ‏ pop(); } bool Stack::isEmpty() const { return topPtr == NULL; } 24.06.201618

19. A Pointer-Based Implementation: push void Stack::push(StackItemType newItem) { // create a new node StackNode *newPtr = new StackNode; // set data portion of new node newPtr->item = newItem; // insert the new node newPtr->next = topPtr; topPtr = newPtr; } 24.06.201619

20. A Pointer-Based Implementation: pop void Stack::pop() throw(StackException) { if (isEmpty()) ‏ throw StackException("StackException: stack empty on pop"); else { // stack is not empty; delete top StackNode *temp = topPtr; topPtr = topPtr->next; // return deleted node to system temp->next = NULL; // safeguard delete temp; } 11/20/08 24.06.201620

21. A Pointer-Based Implementation: topandPop void Stack::topAndPop(StackItemType& stackTop) throw (StackException) { if (isEmpty()) ‏ throw StackException("StackException: stack empty on pop"); else { // stack is not empty; retrieve and delete top stackTop = topPtr->item; StackNode *temp = topPtr; topPtr = topPtr->next; // return deleted node to system temp->next = NULL; // safeguard delete temp; } 24.06.201621

22. A Pointer-Based Implementation: getTop void Stack::getTop(StackItemType& stackTop) const throw(StackException) { if (isEmpty()) ‏ throw StackException("StackException: stack empty on getTop"); else // stack is not empty; retrieve top stackTop = topPtr->item; } 24.06.201622

23. An Implementation That Uses the ADT List The ADT list can be used to represent the items in a stack If the item in position 1 is the top push(newItem)‏ insertFirst(newItem)‏ pop() remove(first().retrieve())‏ getTop(stackTop) first().retrieve()‏ 24.06.201623

24. Comparing Implementations Fixed size versus dynamic size An array-based implementation Prevents the push operation from adding an item to the stack if the stack’s size limit has been reached A pointer-based implementation Does not put a limit on the size of the stack An implementation that uses a linked list versus one that uses a pointer-based implementation of the ADT list Linked list approach is more efficient ADT list approach reuses an already implemented class Much simpler to write Saves time When the ADT stack is used to solve a problem, the use of the ADT’s operations should not depend on its implementation 24.06.201624

25. A Simple Stack Application: Undo sequence in a text editor Problem: abcd   fg   abf // Reads an input line. Enter a character or a backspace character (  ) ‏ readAndCorrect(out aStack:Stack) ‏ aStack.createStack() ‏ read newChar while (newChar is not end-of-line symbol) { if (newChar is not backspace character) ‏ aStack.push(newChar) ‏ else if (!aStack.isEmpty()) ‏ aStack.pop() ‏ read newChar } 24.06.201625

26. A Simple Stack Application: Display Backward Display the input line in reversed order by writing the contents of stack, named aStack. displayBackward(in aStack:Stack) ‏ while (!aStack.isEmpty())) { aStack.pop(newChar) ‏ write newChar } 24.06.201626

27. Checking for Balanced Braces A stack can be used to verify whether a program contains balanced braces An example of balanced braces abc{defg{ijk}{l{mn}}op}qr An example of unbalanced braces abc{def}}{ghij{kl}m Requirements for balanced braces Each time we encounter a “}”, it matches an already encountered “{” When we reach the end of the string, we have matched each “{” 24.06.201627

28. 11/20/08 Checking for Balanced Braces 24.06.201628

29. Checking for Balanced Braces: Algorithm aStack.createStack(); balancedSoFar = true; i=0; while (balancedSoFar and i < length of aString) { ch = character at position i in aString;i++; if (ch is ‘{‘)// push an open brace aStack.push(‘{‘); else if (ch is ‘}’)// close brace if (!aStack.isEmpty()) ‏ aStack.pop();// pop a matching open brace else// no matching open brace balancedSoFar = false; // ignore all characters other than braces } if (balancedSoFar and aStack.isEmpty()) ‏ aString has balanced braces else aString does not have balanced braces 24.06.201629

30. Recognizing Strings in a Language L = {w$w’ : w is a possible empty string of characters other than $, w’ = reverse(w) } abc$cba, a$a, $ are in the language L abc$abc, a$b, a$ are not in the language L Problem: Deciding whether a given string in the language L or not. A solution using a stack Traverse the first half of the string, pushing each character onto a stack Once you reach the $, for each character in the second half of the string, match a popped character off the stack 24.06.201630

31. Recognizing Strings in a Language -- Algorithm aStack.createStack(); i=0; ch = character at position i in aString; while (ch is not ‘$’) { // push the characters before $ (w) onto the stack aStack.push(ch); i++; ch = character at position i in aString; } i++; inLanguage = true; // skip $; assume string in language while (inLanguage and i <length of aString) // match the reverse of if (aStack.isEmpty()) inLanguage = false; // first part shorter than second part else { aStack.pop(stackTop); ch = character at position i in aString; if (stackTop equals to ch) i++;// characters match else inLanguage = false;// characters do not match } if (inLanguage and aStack.isEmpty()) ‏ aString is in language else aString is not in language 24.06.201631

32. Infix/Postfix/Prefix Notation: (consider conventional priority ordering among arithmetical operations) ‏ 2 + 4 * 5 -8 / 4 (infix) ‏ 2 4 5 * + 8 4 / - (postfix – reverse polish notation) ‏ - + * 4 5 2 / 8 4 (prefix – polish notation) ‏ postfix/prefix no need for parenthesis and priority rules Application: Algebraic Expressions 24.06.201632

33. Application: Algebraic Expressions To evaluate an infix expression Convert the infix expression to postfix form Evaluate the postfix expression Infix ExpressionPostfix Expression 5 + 2 * 35 2 3 * + 5 * 2 + 35 2 * 3 + 5 * ( 2 + 3 ) - 45 2 3 + * 4 - 24.06.201633

34. Evaluating Postfix Expressions When an operand is entered, the calculator Pushes it onto a stack When an operator is entered, the calculator Applies it to the top two operands of the stack Pops the operands from the stack Pushes the result of the operation on the stack 24.06.201634

35. Evaluating Postfix Expressions 24.06.201635

36. Converting Infix Expressions to Postfix Expressions Facts about converting from infix to postfix Operands always stay in the same order with respect to one another An operator will move only “to the right” with respect to the operands All parentheses are removed 24.06.201636

37. Infix to Postfix Conversion: start reading chars until the end of expression if a right parenthesis, pop and write until a left parenthesis if an operator, pop and write until a lower precedence operator is seen, then push the operator if an operand, write if left parenthesis, push pop the stack and write until empty Converting Infix Expressions to Postfix Expressions 24.06.201637

38. Infix to Postfix Conversion: e.g. 2 + 4 * 5 -8 / 4 a+ b * c + (d * e + f) * g Converting Infix Expressions to Postfix Expressions 24.06.201638

39. Converting Infix Expr. to Postfix Expr.: Algorithm for (each character ch in the infix expression) { switch (ch) { case operand: // append operand to end of postfixExpr postfixExpr=postfixExpr+ch; break; case ‘(‘:// save ‘(‘ on stack aStack.push(ch); break; case ‘)’: // pop stack until matching ‘(‘, and remove ‘(‘ while (top of stack is not ‘(‘) { postfixExpr=postfixExpr+(top of stack); aStack.pop(); } aStack.pop(); break; 24.06.201639

40. Converting Infix Expr. to Postfix Expr.: Algorithm case operator: // process stack operators of greater precedence while (!aStack.isEmpty() and top of stack is not ‘(‘ and precedence(ch) <= precedence(top of stack) ) { postfixExpr=postfixExpr+(top of stack); aStack(pop); } aStack.push(); break;// save new operator } } // end of switch and for // append the operators in the stack to postfixExpr while (!isStack.isEmpty()) { postfixExpr=postfixExpr+(top of stack); aStack(pop); } 24.06.201640

41. a - (b + c * d)/ e  a b c d * + e / - Converting Infix Expressions to Postfix Expressions 24.06.201641

42. Application: A Search Problem Problem: For each customer request, indicate whether a sequence of flights exists from the origin city to the destination city The flight map is a directed graph Adjacent vertices are two vertices that are joined by an edge A directed path is a sequence of directed edges A Flight Map 24.06.201642

43. A Nonrecursive Solution That Uses a Stack The solution performs an exhaustive search Beginning at the origin city, the solution will try every possible sequence of flights until either It finds a sequence that gets to the destination city It determines that no such sequence exists Backtracking can be used to recover from a wrong choice of a city 24.06.201643

44. 11/20/08 P is origin city; Z is destination city A Nonrecursive Solution - Trace 24.06.201644

45. A Nonrecursive Solution -- Algorithm searchS(in originCity:City, in destinationCity:City) : boolean // searches for a sequence of flights from originCity to destinationCity aStack.createStack(); clear marks on all cities; aStack.push(originCity); mark origin as visited; while (!aStack.isEmpty() and destinationCity is not at top of stack) { if (no flights exist from city on top of stack to unvisited cities) ‏ aStack.pop();// backtrack else { select an unvisited destination city C for a flight from city on top of stack; aStack.push(C); mark C as visited; }} if (aStack.isEmpty()) return false;// no path exists else return true;// path exists 24.06.201645

46. A Recursive Solution for Search Problem searchR(in originCity:City, in destinationCity:City):boolean mark originCity as visited; if (originCity is destinationCity) Terminate -- the destination is reached else for (each unvisited city C adjacent to originCity) searchR(C, destinationCity); 24.06.201646

47. The Relationship Between Stacks and Recursion A strong relationship exists between recursion and stacks Typically, stacks are used by compilers to implement recursive methods During execution, each recursive call generates an activation record that is pushed onto a stack Stacks can be used to implement a nonrecursive version of a recursive algorithm 24.06.201647

48. The C++ run-time system keeps track of the chain of active functions with a stack When a function is called, the run-time system pushes on the stack a frame containing –Local variables and return value When a function returns, its frame is popped from the stack and control is passed to the method on top of the stack main() { int i = 5; foo(i); } foo(int j) { int k; k = j+1; bar(k); } bar(int m) { … } main i = 5 foo j = 5 k = 6 bar m = 6 Run-time Stack C++ Runtime Stack 24.06.201648

49. Stacks -- Summary ADT stack operations have a last-in, first-out (LIFO) behavior There are different stack implementations Two Major Implementations: Array-Based, Pointer-Based Many stack applications expression processing language recognition search problem A strong relationship exists between recursion and stacks Any recursive program can be rewritten as a nonrecursive program using stacks. 24.06.201649