Programming Assignment2 - Deque and Randomized Queues Review

Assignment Specification

课程笔记

Subtext: Modular Programming

• Stacks and Queues are fundamental data types
  • Value: collection of objects
  • Basic Operation: insert, remove, iterate.
  • Difference: which item do we move? -> Stack: LIFO(last in first out) Queue: FIFO(first in first out)
• Client, implementation, interface
  • Client: program using operations defined in interface
  • Implementation: actual code implementing operations
  • Interface: description of data type, basic operations

Stack Programming API:

public class StackOfStrings
StackOfStrings() //create an empty stack
void push(String item)  //insert a new string onto stack
String pop() //remove and return the string most recently added
boolean isEmpty()  //is the stack empty?

linked-list implementation

//Attention: Stack have only one exit -> only one pointer is enough
/*Corner Cases:client add a null item -> IllegalArgumentExceptionremove() from empty stack -> NoSuchElementException
*/
public class StackOfStrings {private Node first;private class Node {String item;Node next;}public boolean isEmpty() {return first == null;}public StackOfStrings {Node first = null;}public void push(String item) {//Improve: add exception to deal with invalid operationNode oldfirst = first;first = new Node(); //Attention: must craete a new instance herefirst.item = item;first.next = oldfirst;}public String pop() {String item = first.item;first = first.next;return item;}

Proposition: Every operation takes constant time in the worst case. A stack with N items uses 40N bytes
Object overhead (16) + inner class extra overhead(8) + item reference (8) + next reference(8) = 40

array implementation

/*
Underflow: throw exception if pop from an empty stack
Overflow: use resizing array for array implementation
*/
public class FixedCapacityStackOfStrings {private String[] s;private int N = 0;public FixedCapacityStackOfStrings (int capacity) {s = new String[capacity];}public String pop() {//Attention: declare null pointer to avoid loitering so garbage collector can reclaim memoryString item = s[--N];s[N] = null;return item;}public void push(String item) {s[N++] = item;}public boolean isEmpty() {return n == 0;}
}

• Resizing array
  • Problem: Require client to provide capacity does not implement API. Constructor should not have int input
  • Question: How to grow and shrink array?
  • Answer: grow: double shrink: quarter - > Why? ->
    • double array for grow-> cost of is Linear N + (2 + 4 + 8 + .... + N) ~ 3N Geometric sequence: Sn = (a1 - an * q) / 1 - q
    • quarter for shrink -> avoid thrashing push - pop - push - pop when sequence is full -> each operation takes time propotional to N
      
      

//Note: array is between 25% and 100% full
public class ResizingArrayStackOfStrings {private String[] s;public ResizaingArrayStackOfStrings() {s = new String[1];}public void push(String item) {if (N == s.length) resize (2 * s.length);s[N++] = item;}private void resize(int capacity) {//create a double size array, copy the element from the old String array, update the pointerString[] copy = new String[capacity];for (int i = 0; i < capacity; i++) {copy[i] = s[i];s = copy;}public String pop() {String item = s[--N];S[N] = null;if (N > 0 && N = s.length/4) resize(s.length / 2);return item;}
}

• Queue Programming API
  • QueueOfStrings()
  • void enqueue(String item)
  • String dequeue()
  • boolean isEmpty()
    Same API with stack, only name changed

*linked list implementation
/*Attention:
Queue has two exit, so it needs two pointers
*/
public class LinkedQueueOfStrings {public LinkedQueueOfStrings() {Node first, last;int N = 0;}private class Node {String item;Node next;}public boolean isEmpty() {return first == null;}//enqueue elements added to the last of the queuepublic void enqueue(String item) {Node oldlast = last; // here last already points to an exist instance//Create a totally new Nodelast = new Node();last.item = item;last.next = null;//linked back with the queueif (isEmpty()) {//there is only one element exist ->first = last;}else {oldlast.next = last;}}public String dequeue() {String item = first.item;first = first.next;if (isEmpty()) {last = null;}return item;}
}

• Generic data types: autoboxing and unboxing
  • Autoboxing: Automatic cast between a primitive type and its wrapper

  Stack<Integer> s = new Stack<Integer>();
  s.push(17);  //s.push(Integer.valueOf(17));
  int a = s.pop();  //int a = s.pop().intValue();

在写代码的过程当中，心里需要有转换角色的意识，当你在实现一个API的时候，需要考虑的是
* 实现某个方法所要使用的数据结构，
* 调用方法 or 自己写方法，
* API的性能要求 -> 使用哪种算法可以满足要求 查找，插入，删除 时间 + 空间

• Iterators
  • What is an Iterable?
  • What is an Iterator?

  public interface Iterator<Item> {boolean hasNext();Item next();
  }

  • Why make data structures Iterable ?
• Java collections library
  List Interface. java.util.List is API for an sequence of items
  • java.util.ArrayList uses resizing array -> only some operations are effieient
  • java.util.LinkedList uses linked list -> only some operations are effieient
    tip: 不清楚library的具体实现的时候，尽量避免调用相关的方法。可能效率会很低。
• Arithmetic expression evaluation
  ( 1 + ( ( 2 + 3 ) * ( 4 * 5 ) ) )
  Two-stack algorithm. 【E. W. Dijkstra】
  • value: push onto the value stack
  • Operator: push onto the operator stack
  • Left parenthesis: ignore
  • Right parenthesis: pop operator and two values; push the result of applying that operator to those values onto the operand stack

总结：
Stack的链表实现
Stack的数组实现（resize array)
Queue的链表实现
Queue的数组实现（resize array)

对于双向队列的理解有误，导致错误实现。
双向对别不应当是两个队列的水平叠加，见figure 1

作业总结：

1. 对于文件的读写基本操作命令不够熟悉
2. 对于问题的定义 出现了没有搞清楚题目要求的现象，包括Deque的基本操作 以及Permutation 类当中，应当是读取全部数据，输出k个数据，而不是读取k个数据，输出全部数据的问题