《基本数据结构》PPT课件.ppt
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1、Elementary Data Structures,Stacks,Queues,&ListsAmortized analysisTrees,Elementary Data Structures,2,The Stack ADT(2.1.1),The Stack ADT stores arbitrary objectsInsertions and deletions follow the last-in first-out schemeThink of a spring-loaded plate dispenserMain stack operations:push(object):insert
2、s an elementobject pop():removes and returns the last inserted element,Auxiliary stack operations:object top():returns the last inserted element without removing itinteger size():returns the number of elements storedboolean isEmpty():indicates whether no elements are stored,Elementary Data Structure
3、s,3,Applications of Stacks,Direct applicationsPage-visited history in a Web browserUndo sequence in a text editorChain of method calls in the Java Virtual Machine or C+runtime environmentIndirect applicationsAuxiliary data structure for algorithmsComponent of other data structures,Elementary Data St
4、ructures,4,Array-based Stack(2.1.1),A simple way of implementing the Stack ADT uses an arrayWe add elements from left to rightA variable t keeps track of the index of the top element(size is t+1),Algorithm pop():if isEmpty()thenthrow EmptyStackException else t t 1return St+1,Algorithm push(o)if t=S.
5、length 1 thenthrow FullStackException else t t+1St o,Elementary Data Structures,5,Growable Array-based Stack(1.5),In a push operation,when the array is full,instead of throwing an exception,we can replace the array with a larger oneHow large should the new array be?incremental strategy:increase the
6、size by a constant cdoubling strategy:double the size,Algorithm push(o)if t=S.length 1 thenA new array ofsize for i 0 to t do Ai Si S At t+1St o,Elementary Data Structures,6,Comparison of the Strategies,We compare the incremental strategy and the doubling strategy by analyzing the total time T(n)nee
7、ded to perform a series of n push operationsWe assume that we start with an empty stack represented by an array of size 1We call amortized time of a push operation the average time taken by a push over the series of operations,i.e.,T(n)/n,Elementary Data Structures,7,Analysis of the Incremental Stra
8、tegy,We replace the array k=n/c timesThe total time T(n)of a series of n push operations is proportional ton+c+2c+3c+4c+kc=n+c(1+2+3+k)=n+ck(k+1)/2Since c is a constant,T(n)is O(n+k2),i.e.,O(n2)The amortized time of a push operation is O(n),Elementary Data Structures,8,Direct Analysis of the Doublin
9、g Strategy,We replace the array k=log2 n timesThe total time T(n)of a series of n push operations is proportional ton+1+2+4+8+2k=n+2k+1-1=2n-1T(n)is O(n)The amortized time of a push operation is O(1),Elementary Data Structures,9,The accounting method determines the amortized running time with a syst
10、em of credits and debitsWe view a computer as a coin-operated device requiring 1 cyber-dollar for a constant amount of computing.,Accounting Method Analysis of the Doubling Strategy,We set up a scheme for charging operations.This is known as an amortization scheme.The scheme must give us always enou
11、gh money to pay for the actual cost of the operation.The total cost of the series of operations is no more than the total amount charged.(amortized time)(total$charged)/(#operations),Elementary Data Structures,10,Amortization Scheme for the Doubling Strategy,Consider again the k phases,where each ph
12、ase consisting of twice as many pushes as the one before.At the end of a phase we must have saved enough to pay for the array-growing push of the next phase.At the end of phase i we want to have saved i cyber-dollars,to pay for the array growth for the beginning of the next phase.,We charge$3 for a
13、push.The$2 saved for a regular push are“stored”in the second half of the array.Thus,we will have 2(i/2)=i cyber-dollars saved at then end of phase i.Therefore,each push runs in O(1)amortized time;n pushes run in O(n)time.,Elementary Data Structures,11,The Queue ADT(),The Queue ADT stores arbitrary o
14、bjectsInsertions and deletions follow the first-in first-out schemeInsertions are at the rear of the queue and removals are at the front of the queueMain queue operations:enqueue(object):inserts an element at the end of the queueobject dequeue():removes and returns the element at the front of the qu
15、eue,Auxiliary queue operations:object front():returns the element at the front without removing itinteger size():returns the number of elements storedboolean isEmpty():indicates whether no elements are storedExceptionsAttempting the execution of dequeue or front on an empty queue throws an EmptyQueu
16、eException,Elementary Data Structures,12,Applications of Queues,Direct applicationsWaiting linesAccess to shared resources(e.g.,printer)MultiprogrammingIndirect applicationsAuxiliary data structure for algorithmsComponent of other data structures,Elementary Data Structures,13,Singly Linked List,A si
17、ngly linked list is a concrete data structure consisting of a sequence of nodesEach node storeselementlink to the next node,next,elem,node,A,B,C,D,Elementary Data Structures,14,Queue with a Singly Linked List,We can implement a queue with a singly linked listThe front element is stored at the first
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