第二周作业
问题描述
详细描述请参见Queues
简单来讲,就是实现一个双头队列(Deque
. A double-ended queue or deque, pronounced “deck”),和一个随机队列RandomizedQueue
。双头队列要求实现可以任意选择在队列的头部或尾部添加元素,或者从头部或尾部删除元素。随机队列的入列和普通队列相同,出列要求是从队列中随机选取一个元素出列,同时提供一个sample()
方法,随机返回队列中的一个元素。
下面是教授要求的API:
public class Deque<Item> implements Iterable<Item> {
// construct an empty deque
public Deque()
// is the deque empty?
public boolean isEmpty()
// return the number of items on the deque
public int size()
// add the item to the front
public void addFirst(Item item)
// add the item to the back
public void addLast(Item item)
// remove and return the item from the front
public Item removeFirst()
// remove and return the item from the back
public Item removeLast()
// return an iterator over items in order from front to back
public Iterator<Item> iterator()
// unit testing (required)
public static void main(String[] args)
}
public class RandomizedQueue<Item> implements Iterable<Item> {
// construct an empty randomized queue
public RandomizedQueue()
// is the randomized queue empty?
public boolean isEmpty()
// return the number of items on the randomized queue
public int size()
// add the item
public void enqueue(Item item)
// remove and return a random item
public Item dequeue()
// return a random item (but do not remove it)
public Item sample()
// return an independent iterator over items in random order
public Iterator<Item> iterator()
// unit testing (required)
public static void main(String[] args)
}
先说第一个类,因为要同时满足双向操作,所以很容易想到使用双向链表作为基础的数据结构,在每个Node
私有类中添加一个prior
指针指向节点的前驱。双向链表的节点每个占用内存48byte,正好满足题目对内存占用的要求。对于size()
方法需要返回队列中的元素数量,只要维护一个size
属性即可,每次入列出列对其进行++
和--
的操作,同时这个size
属性可以用来判断队列是否为空。
public class Deque<Item> implements Iterable<Item> {
// inner class node 16+8*2+8+8=48 bytes memory
private class Node {
Item item;
Node next;
Node prior;
}
private class ListIterator implements Iterator<Item> {
private Node current = first;
public boolean hasNext() { return current != null; }
public Item next() {
if (hasNext()) {
Item item = current.item;
current = current.next;
return item;
}
else throw new NoSuchElementException("There is no next item. ");
}
public void remove() {
throw new UnsupportedOperationException("This method is not supported. ");
}
}
private Node first;
private Node last;
private int size;
// construct an empty deque
public Deque() { }
// is the deque empty?
public boolean isEmpty() { return size == 0; }
// return the number of items on the deque
public int size() { return size; }
// add the item to the front
public void addFirst(Item item) {
if (item == null) { throw new IllegalArgumentException("A null item detected."); }
else if (isEmpty()) {
first = new Node();
first.item = item;
last = first;
size++;
}
else {
Node oldFirst = first;
first = new Node();
first.item = item;
first.next = oldFirst;
oldFirst.prior = first;
size++;
}
}
// add the item to the back
public void addLast(Item item) {
if (item == null) { throw new IllegalArgumentException("A null item detected."); }
else if (isEmpty()) {
last = new Node();
last.item = item;
first = last;
size++;
}
else {
Node oldLast = last;
last = new Node();
last.item = item;
last.prior = oldLast;
oldLast.next = last;
size++;
}
}
// remove and return the item from the front
public Item removeFirst() {
if (!isEmpty()) {
Item item = first.item;
first = first.next;
size--;
return item;
}
else {
throw new NoSuchElementException("The deque is empty.");
}
}
// remove and return the item from the back
public Item removeLast() {
if (!isEmpty()) {
Item item = last.item;
last = last.prior;
size--;
return item;
}
else {
throw new NoSuchElementException("The deque is empty.");
}
}
// return an iterator over items in order from front to back
public Iterator<Item> iterator() { return new ListIterator(); }
// unit testing (required)
public static void main(String[] args) {
// unit testing大家自己写吧
}
}
再说第二个类。随机队列的入列不需要再考虑,主要考虑当随机选取了一个节点之后,如何找到这个节点并remove掉。看了网上的一些解答,意识到这里可能确实是选择用数组作为底层的数据结构更好一些,因为数组可以在的操作时间中找到需要出列的节点。但是我既然已经用了链表那就还是用链表实现一下吧。
正常的想法是,随机生成一个索引i
,从1
开始经过i-1
次操作找到出列元素,然后对出列元素 前驱的后继、后继的前驱 进行修改(当然边界情况要考虑,只有1个元素,只有2个元素以及随机索引在头尾的情况)。这边我考虑到我们的链表已经是双向链表了,如果根据索引i
在上半部分还是下半部分(和size/2
作比较),可以选择从头部或者尾部开始向后或向前寻找,这样把最坏情况降低到,在现实中应该是会有些速度提升的吧,然而我最后的timing打分还是没全过,169/193,应该是链表的原因了,看到网上有前辈用数组可以满分,见EnPFighter. sample()
方法的实现原理和上面描述的差不多,只需要找到并返回item
就行了,更简单一些。
关于随机遍历器,我想了想只用链表解决的话,还是跟上面这段一样,要写比较多的代码并且套循环,代码不整洁,速度也不够快,所以这里还是借用了临时数组的思想。记录一个随机生成的索引target
和临时数组中未被抽到过的元素的最大所以lastIndex
,每次生成在[1, lastIndex]
之间生成随机target
,并将target
的元素和lastIndex
的元素进行调换,然后执行lastIndex--
以保证已经抽到过的不会再被抽到。
public class RandomizedQueue<Item> implements Iterable<Item> {
// inner class node 16+8*2+8+8=48 bytes memory
private class Node {
Item item;
Node next;
Node prior;
}
private class ListIterator implements Iterator<Item> {
private int lastIndex;
private Item[] temp;
public ListIterator() {
lastIndex = size - 1;
temp = (Item[]) new Object[size];
Node current = first;
for (int i = 0; i < size; i++) {
temp[i] = current.item;
current = current.next;
}
}
public boolean hasNext() { return lastIndex >= 0; }
public Item next() {
if (hasNext()) {
int target = StdRandom.uniform(lastIndex + 1);
Item item = temp[target];
temp[target] = temp[lastIndex];
temp[lastIndex] = item;
lastIndex--;
return item;
}
else throw new NoSuchElementException("There is no next item. ");
}
public void remove() {
throw new UnsupportedOperationException("This method is not supported. ");
}
}
private Node first;
private Node last;
private int size;
// construct an empty randomized queue
public RandomizedQueue() { }
// is the randomized queue empty?
public boolean isEmpty() { return size == 0; }
// return the number of items on the randomized queue
public int size() { return size; }
// add the item
public void enqueue(Item item) {
if (item == null) throw new IllegalArgumentException("The item is null.");
else {
Node oldLast = last;
last = new Node();
last.item = item;
last.next = null;
if (isEmpty()) first = last;
else {
last.prior = oldLast;
oldLast.next = last;
}
size++;
}
}
// remove and return a random item
public Item dequeue() {
if (isEmpty()) throw new NoSuchElementException("The RandomQueue is empty.");
else {
if (size == 1) {
Item item = first.item;
first = null;
last = null;
size--;
return item;
}
else {
int i = StdRandom.uniform(size) + 1;
if (i > size / 2) {
if (size == 2) {
Item item = last.item;
last = first;
last.prior = null;
first.next = null;
size--;
return item;
}
else {
if (i == size) {
Item item = last.item;
last = last.prior;
last.next = null;
size--;
return item;
}
else {
Node current = last;
for (int j = 0; j < size - i; j++) {
current = current.prior;
}
current.prior.next = current.next;
current.next.prior = current.prior;
size--;
return current.item;
}
}
}
else {
if (size == 2) {
Item item = first.item;
first = last;
last.prior = null;
first.next = null;
size--;
return item;
}
else {
if (i == 1) {
Item item = first.item;
first = first.next;
first.prior = null;
size--;
return item;
}
else {
Node current = first;
for (int j = 0; j < i - 1; j++) {
current = current.next;
}
current.prior.next = current.next;
current.next.prior = current.prior;
size--;
return current.item;
}
}
}
}
}
}
// return a random item (but do not remove it)
public Item sample() {
if (isEmpty()) throw new NoSuchElementException("The RandomQueue is empty.");
else {
int i = StdRandom.uniform(size) + 1;
if (i > size / 2) {
Node current = last;
for (int j = 0; j < size - i; j++) {
current = current.prior;
}
return current.item;
}
else {
Node current = first;
for (int j = 0; j < i - 1; j++) {
current = current.next;
}
return current.item;
}
}
}
// return an iterator over items in order from front to back
public Iterator<Item> iterator() { return new ListIterator(); }
}
最后要实现一个Permutation
类用来产生队列中随机的k
个不重复的元素。因为不重复这个条件在Iterator
的next()
方法已经实现了,所以直接调用foreach
语句就行。
public class Permutation {
public static void main(String[] args) {
RandomizedQueue<String> rq = new RandomizedQueue<String>();
int k = Integer.parseInt(args[0]);
while (!StdIn.isEmpty()) {
String item = StdIn.readString();
rq.enqueue(item);
}
for (String s: rq
) {
if (k == 0) break;
StdOut.println(s);
k--;
}
}
}
最终成绩是92分,不够满分,有部分时间和空间爆了。
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