原文地址:http://blog.csdn.net/qq_25806863/article/details/76423425
这两个集合主要是内部对数据的存储方式不一样,一个用的数组,一个用的链表。
关于数组合链表在之前有个简单的比较
array list
O(N)
。O(1)
。O(N)
,如果发生在最后一个元素上也就是高端,时间是O(1)
当插入和删除都只对高端操作,数组也比较合适,否则就应用 链表 linked list
使用链表是因为链表在内存中是不连续的,不用因为一个元素的插入或删除而引起其他元素的变动。
链表由一系列节点组成,节点不需要在内存中相连,因此每一个节点除了自身元素外还要包含自己的后继元的节点的链(link),称之为next链
。最后一个节点的next链引用null。
O(N)
,跟数组一样。O(N)
。要查找第x个元素,花费的时间是O(x), 效率不如数组。删除A2:
在A1后插入Ax:
让每一个节点拥有其前驱节点的引用就成了双链表。
import java.util.Arrays;
import java.util.ConcurrentModificationException;
import java.util.Iterator;
import java.util.NoSuchElementException;
import java.util.function.Consumer;
/**
* @author sll on 2017/6/30.
*/
public class MyArrayList<Element> implements Iterable<Element> {
private static final int DEFAULT_CAPACITY = 5;
private int size;
private Element[] elements;
private int modCount = 0;
public MyArrayList() {
elements = (Element[]) new Object[DEFAULT_CAPACITY];
clear();
}
public void clear() {
size = 0;
ensureCapacity(DEFAULT_CAPACITY);
}
public int size() {
return size;
}
public boolean isEmpty() {
return size == 0;
}
public void trimToSize() {
ensureCapacity(size);
}
//这里是查找,使用数组直接就查出来了,所以一次查询的时间是 O(1)
public Element get(int index) {
if (index < 0 || index >= size) {
throw new ArrayIndexOutOfBoundsException();
}
return elements[index];
}
public Element set(int index, Element newElement) {
if (index < 0 || index >= size) {
throw new ArrayIndexOutOfBoundsException();
}
Element old = elements[index];
elements[index] = newElement;
return old;
}
public boolean add(Element e) {
return add(size, e);
}
//这里的增加和删除就要移动很多项,所以一次插入或删除的速度是 O(N)
public boolean add(int index, Element e) {
if (elements.length == size) {
ensureCapacity(size * 2 + 1);
}
System.arraycopy(elements, index, elements, index + 1, size - index);
elements[index] = e;
size++;
modCount++;
return true;
}
public Element remove(int index) {
Element removeElement = elements[index];
int moveNum = size - index - 1;
if (moveNum > 0) {
System.arraycopy(elements, index + 1, elements, index, moveNum);
}
size--;
elements[size] = null;
modCount++;
return removeElement;
}
public void ensureCapacity(int newCapacity) {
if (newCapacity < size)
return;
elements = Arrays.copyOf(elements, newCapacity);
}
public Itr iterator() {
return new Itr();
}
private class Itr implements Iterator<Element> {
private int exceptMocCount = modCount;
private int current = 0;
@Override
public boolean hasNext() {
return current < size();
}
@Override
public Element next() {
if (exceptMocCount != modCount) {
throw new ConcurrentModificationException();
}
if (!hasNext()) {
throw new NoSuchElementException();
}
return elements[current++];
}
@Override
public void remove() {
if (exceptMocCount != modCount) {
throw new ConcurrentModificationException();
}
MyArrayList.this.remove(--current);
}
@Override
public void forEachRemaining(Consumer super Element> action) {
}
}
}
import java.util.ConcurrentModificationException;
import java.util.Iterator;
import java.util.NoSuchElementException;
/**
* @author sll on 2017/6/30.
*/
public class MyLinkedList implements Iterable {
private int theSize = 0;
private int modCount = 0;
private Node startNode, endNode;//额外的头节点和尾节点
public MyLinkedList() {
clear();
}
private void clear() {
startNode = new Node(null, null, null);
endNode = new Node(null, startNode, null);
startNode.next = endNode;
theSize = 0;
modCount++;
}
public int size() {
return theSize;
}
public boolean isEmpty() {
return size() == 0;
}
public boolean add(E x) {
return add(size(), x);
}
public boolean add(int index, E x) {
addBefore(getNode(index), x);
return true;
}
public E get(int index) {
return getNode(index).data;
}
public E set(int index, E x) {
Node p = getNode(index);
E old = p.data;
p.data = x;
return old;
}
public E remove(int index) {
return remove(getNode(index));
}
private void addBefore(Node p, E x) {
Node newNode = new Node<>(x, p.prev, p);
newNode.prev.next = newNode;
p.prev = newNode;
theSize++;
modCount++;
}
private Node getNode(int index) {
Node p;
if (index < 0 || index > size())
throw new IndexOutOfBoundsException();
if (index < size() / 2) {
p = startNode.next;
for (int i = 0; i < index; i++) {
p = p.next;
}
} else {
p = endNode;
for (int i = size(); i > index; i--) {
p = p.prev;
}
}
return p;
}
private E remove(Node x) {
x.prev.next = x.next;
x.next.prev = x.prev;
theSize--;
modCount++;
return x.data;
}
@Override
public Iterator iterator() {
return new MyIterator();
}
private static class Node {
public E data;
public Node prev;
public Node next;
public Node(E data, Node prev, Node next) {
this.data = data;
this.prev = prev;
this.next = next;
}
}
private class MyIterator implements Iterator {
private Node current = startNode.next;
private int expectedModCount = modCount;
private boolean okToRemove = false;
@Override
public boolean hasNext() {
return current!= endNode;
}
@Override
public E next() {
if (modCount != expectedModCount)
throw new ConcurrentModificationException();
if (!hasNext()) {
throw new NoSuchElementException();
}
E nextItem = current.data;
current = current.next;
okToRemove = true;
return nextItem;
}
@Override
public void remove() {
if (modCount != expectedModCount)
throw new ConcurrentModificationException();
if (!okToRemove) {
throw new NoSuchElementException();
}
MyLinkedList.this.remove(current.prev);
okToRemove = false;
expectedModCount++;
}
}
}
以分别用ArrayList和LinkedList用下面三个方法做一个时间对比:
//从末端添加,创建一个由N个项的List。
public static void makeList1(List list, int N) {
list.clear();
for (int i = 0; i < N; i++) {
list.add(i);
}
}
//从首端添加,创建一个由N个项的List。
public static void makeList2(List list, int N) {
list.clear();
for (int i = 0; i < N; i++) {
list.add(0, i);
}
}
//获取长度为N的List的每一个值求和。
public static void makeList3(List list) {
long total = 0;
for (int i = 0; i < list.size(); i++) {
total += list.get(i);
}
}
首先创建两个集合,N=100000,打印耗费时间(ms):
List arrayList = new ArrayList<>();
List linkedList = new LinkedList();
makeList1
makeList1(arrayList, 100000);
makeList1(linkedList, 100000);
结果可能会不一样,但也是相近的。
运行时间都是O(N)
makeList2
makeList2(arrayList, 100000);
makeList2(linkedList, 100000);
linkedList花费的时间少得多,执行时间是O(N)
而ArrayList因为每次在头部插入都需要将之前所有项向后移动一位,所以执行时间是O(N^2)
makeList3
makeList2(arrayList, 100000);
makeList2(linkedList, 100000);
makeList3(arrayList);
makeList3(linkedList);
这个每次都要查找,所以ArrayList比较快,执行时间是O(N)
而inkedList的执行时间是O(N^2)
用迭代器实现makeList3
将makeList3改成:
public static void makeList3(List list) {
long total = 0;
for (Integer i : list) {
total+=i;
}
}
两个都很相近了,都是O(N)。
参考《数据结构与算法分析java版》