© 2004 Goodrich, Tamassia Stacks. © 2004 Goodrich, Tamassia Stacks2 Abstract Data Types (ADTs) An abstract data type (ADT) is an abstraction of a data.

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Presentation transcript:

© 2004 Goodrich, Tamassia Stacks

© 2004 Goodrich, Tamassia Stacks2 Abstract Data Types (ADTs) An abstract data type (ADT) is an abstraction of a data structure An ADT specifies: Data stored Operations on the data Error conditions associated with operations Example: ADT modeling a simple stock trading system The data stored are buy/sell orders The operations supported are  order buy(stock, shares, price)  order sell(stock, shares, price)  void cancel(order) Error conditions:  Buy/sell a nonexistent stock  Cancel a nonexistent order

© 2004 Goodrich, Tamassia Stacks3 The Stack ADT (§4.2) The Stack ADT stores arbitrary objects Insertions and deletions follow the last-in first-out scheme Think of a spring-loaded plate dispenser Main stack operations: push(object): inserts an element object pop(): removes and returns the last inserted element Auxiliary stack operations: object top(): returns the last inserted element without removing it integer size(): returns the number of elements stored boolean isEmpty(): indicates whether no elements are stored

© 2004 Goodrich, Tamassia Stacks4 Stack Interface in Java Java interface corresponding to our Stack ADT Requires the definition of class EmptyStackException Different from the built-in Java class java.util.Stack public interface Stack { public int size(); public boolean isEmpty(); public Object top() throws EmptyStackException; public void push(Object o); public Object pop() throws EmptyStackException; }

© 2004 Goodrich, Tamassia Stacks5 Exceptions Attempting the execution of an operation of ADT may sometimes cause an error condition, called an exception Exceptions are said to be “thrown” by an operation that cannot be executed In the Stack ADT, operations pop and top cannot be performed if the stack is empty Attempting the execution of pop or top on an empty stack throws an EmptyStackException

© 2004 Goodrich, Tamassia Stacks6 Applications of Stacks Direct applications Page-visited history in a Web browser Undo sequence in a text editor Chain of method calls in the Java Virtual Machine Indirect applications Auxiliary data structure for algorithms Component of other data structures

© 2004 Goodrich, Tamassia Stacks7 Method Stack in the JVM The Java Virtual Machine (JVM) keeps track of the chain of active methods with a stack When a method is called, the JVM pushes on the stack a frame containing Local variables and return value Program counter, keeping track of the statement being executed When a method ends, its frame is popped from the stack and control is passed to the method on top of the stack Allows for recursion main() { int i = 5; foo(i); } foo(int j) { int k; k = j+1; bar(k); } bar(int m) { … } bar PC = 1 m = 6 foo PC = 3 j = 5 k = 6 main PC = 2 i = 5

© 2004 Goodrich, Tamassia Stacks8 Array-based Stack A simple way of implementing the Stack ADT uses an array We add elements from left to right A variable keeps track of the index of the top element S 012 t … Algorithm size() return t + 1 Algorithm pop() if isEmpty() then throw EmptyStackException else t  t  1 return S[t + 1]

© 2004 Goodrich, Tamassia Stacks9 Array-based Stack (cont.) The array storing the stack elements may become full A push operation will then throw a FullStackException Limitation of the array- based implementation Not intrinsic to the Stack ADT S 012 t … Algorithm push(o) if t = S.length  1 then throw FullStackException else t  t + 1 S[t]  o

© 2004 Goodrich, Tamassia Stacks10 Performance and Limitations Performance Let n be the number of elements in the stack The space used is O(n) Each operation runs in time O(1) Limitations The maximum size of the stack must be defined a priori and cannot be changed Trying to push a new element into a full stack causes an implementation-specific exception

© 2004 Goodrich, Tamassia Stacks11 Array-based Stack in Java public class ArrayStack implements Stack { // holds the stack elements private Object S[ ]; // index to top element private int top = -1; // constructor public ArrayStack(int capacity) { S = new Object[capacity]); } public Object pop() throws EmptyStackException { if isEmpty() throw new EmptyStackException (“Empty stack: cannot pop”); Object temp = S[top]; // facilitates garbage collection S[top] = null; top = top – 1; return temp; }

© 2004 Goodrich, Tamassia Stacks12 Parentheses Matching Each “(”, “{”, or “[” must be paired with a matching “)”, “}”, or “[” correct: ( )(( )){([( )])} correct: ((( )(( )){([( )])} incorrect: )(( )){([( )])} incorrect: ({[ ])} incorrect: (

© 2004 Goodrich, Tamassia Stacks13 Parentheses Matching Algorithm Algorithm ParenMatch( X, n ) : Input: An array X of n tokens, each of which is either a grouping symbol, a variable, an arithmetic operator, or a number Output: true if and only if all the grouping symbols in X match Let S be an empty stack for i = 0 to n - 1 do if X [ i ] is an opening grouping symbol then S. push (X [ i ] ) else if X [ i ] is a closing grouping symbol then if S. isEmpty () then return false { nothing to match with } if S. pop () does not match the type of X [ i ] then return false { wrong type } if S. isEmpty () then return true { every symbol matched } else return false { some symbols were never matched }

© 2004 Goodrich, Tamassia Stacks14 HTML Tag Matching The Little Boat The storm tossed the little boat like a cheap sneaker in an old washing machine. The three drunken fishermen were used to such treatment, of course, but not the tree salesman, who even as a stowaway now felt that he had overpaid for the voyage. Will the salesman die? What color is the boat? And what about Naomi? The Little Boat The storm tossed the little boat like a cheap sneaker in an old washing machine. The three drunken fishermen were used to such treatment, of course, but not the tree salesman, who even as a stowaway now felt that he had overpaid for the voyage. 1. Will the salesman die? 2. What color is the boat? 3. And what about Naomi? For fully-correct HTML, each should pair with a matching

© 2004 Goodrich, Tamassia Stacks15 Growable Array-based Stack In a push operation, when the array is full, instead of throwing an exception, we can replace the array with a larger one How large should the new array be? incremental strategy: increase the size by a constant c doubling strategy: double the size Algorithm push(o) if t = S.length  1 then A  new array of size … for i  0 to t do A[i]  S[i] S  A t  t + 1 S[t]  o

© 2004 Goodrich, Tamassia Stacks16 Comparison of the Strategies We compare the incremental strategy and the doubling strategy by analyzing the total time T(n) needed to perform a series of n push operations We assume that we start with an empty stack represented by an array of size 1 We call amortized time of a push operation the average time taken by a push over the series of operations, i.e., T(n)/n

© 2004 Goodrich, Tamassia Stacks17 Incremental Strategy Analysis We replace the array k = n / c times The total time T(n) of a series of n push operations is proportional to n + c + 2c + 3c + 4c + … + kc = n + c( … + k) = n + ck(k + 1)/2 Since c is a constant, T(n) is O(n + k 2 ), i.e., O(n 2 ) The amortized time of a push operation is O(n)

© 2004 Goodrich, Tamassia Stacks18 Doubling Strategy Analysis We replace the array k = log 2 n times The total time T(n) of a series of n push operations is proportional to n …+ 2 k = n  2 k + 1  1 = 2n  1 T(n) is O(n) The amortized time of a push operation is O(1) geometric series

© 2004 Goodrich, Tamassia Stacks19 Computing Spans (not in book) We show how to use a stack as an auxiliary data structure in an algorithm Given an an array X, the span S[i] of X[i] is the maximum number of consecutive elements X[j] immediately preceding X[i] and such that X[j]  X[i] Spans have applications to financial analysis E.g., stock at 52-week high X S

© 2004 Goodrich, Tamassia Stacks20 Quadratic Algorithm Algorithm spans1(X, n) Input array X of n integers Output array S of spans of X # S  new array of n integers n for i  0 to n  1 don s  1n while s  i  X[i  s]  X[i] 1  2  …  (n  1) s  s  11  2  …  (n  1) S[i]  s n return S 1 Algorithm spans1 runs in O(n 2 ) time

© 2004 Goodrich, Tamassia Stacks21 Computing Spans with a Stack We keep in a stack the indices of the elements visible when “looking back” We scan the array from left to right Let i be the current index We pop indices from the stack until we find index j such that X[i]  X[j] We set S[i]  i  j We push x onto the stack

© 2004 Goodrich, Tamassia Stacks22 Linear Algorithm Algorithm spans2(X, n) # S  new array of n integers n A  new empty stack 1 for i  0 to n  1 don while (  A.isEmpty()  X[top()]  X[i] ) don A.pop() n if A.isEmpty() then n S[i]  i  1n else S[i]  i  top()n A.push(i)n return S 1 Each index of the array Is pushed into the stack exactly one Is popped from the stack at most once The statements in the while-loop are executed at most n times Algorithm spans2 runs in O(n) time