list集合源码分析

List集合我们众所周知的就是有序可重复，底层是一维数组，但是我现在通过面试的情况发现自己对于集合的理解还是太浅薄了。

ArrayList 底层是一个数组，让我们了解ArrayList到底是什么。

首先ArrayList都知道查找快，增删慢，了解一下为什么？

看一下源码：

public class ArrayList<E> extends AbstractList<E>
        implements List<E>, RandomAccess, Cloneable, java.io.Serializable

可以知道ArrayList中实现了RandomAccess这个接口，追踪一下：

public interface RandomAccess {
}

what？实现是接口是空的？那怎么回事？原来是这样啊！

既然我们知道他是在集合中，那么我就顺着他的父类寻找，先看一下List，没有发现，List在向上是collection，这时我们看看他的方法：

@SuppressWarnings({"rawtypes", "unchecked"})
    public static void shuffle(List<?> list, Random rnd) {
        int size = list.size();
        if (size < SHUFFLE_THRESHOLD || list instanceof RandomAccess) {
            for (int i=size; i>1; i--)
                swap(list, i-1, rnd.nextInt(i));
        } else {
            Object arr[] = list.toArray();

            // Shuffle array
            for (int i=size; i>1; i--)
                swap(arr, i-1, rnd.nextInt(i));

            // Dump array back into list
            // instead of using a raw type here, it's possible to capture
            // the wildcard but it will require a call to a supplementary
            // private method
            ListIterator it = list.listIterator();
            for (int i=0; i<arr.length; i++) {
                it.next();
                it.set(arr[i]);
            }
        }
    }

private static void swap(Object[] arr, int i, int j) {
        Object tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;
    }

public class Collections

这样就特别清晰了！源码其实就是一个for循环！

ArrayList中初始长度为10，数组扩容1.5倍+1

private static final int DEFAULT_CAPACITY = 10;

public boolean add(E e) {
        ensureCapacityInternal(size + 1);  // Increments modCount!!
        elementData[size++] = e;
        return true;
    }

private void ensureExplicitCapacity(int minCapacity) {
        modCount++;
        // overflow-conscious code
        if (minCapacity - elementData.length > 0)
            grow(minCapacity);
    }

private void grow(int minCapacity) {
        // overflow-conscious code
        int oldCapacity = elementData.length;
        int newCapacity = oldCapacity + (oldCapacity >> 1);//右移一位除以2
        if (newCapacity - minCapacity < 0)
            newCapacity = minCapacity;
        if (newCapacity - MAX_ARRAY_SIZE > 0)
            newCapacity = hugeCapacity(minCapacity);
        // minCapacity is usually close to size, so this is a win:
        elementData = Arrays.copyOf(elementData, newCapacity);
    }

缺点是向指定的索引位置插入对象或删除对象的速度较慢.因为指向索引位置插入对象时,会将指定索引位置及之后的所有对象相应向后移动一位

Vector集合与ArrayList集合没有本质区别，因为Vector中方法和ArrayList的方法是一致的，但是每个方法上都有synchronized

关键字，所以说Vector集合是线程安全，但是也正因为如此，vector更浪费资源。

LinkList底层是一个双向链表，查询慢，插入删除快。

void linkLast(E e) {
        final Node<E> l = last;
        final Node<E> newNode = new Node<>(l, e, null);
        last = newNode;
        if (l == null)
            first = newNode;
        else
            l.next = newNode;
        size++;
        modCount++;
    }

private void linkFirst(E e) {
        final Node<E> f = first;
        final Node<E> newNode = new Node<>(null, e, f);
        first = newNode;
        if (f == null)
            last = newNode;
        else
            f.prev = newNode;
        size++;
        modCount++;
    }

void linkLast(E e) {
        final Node<E> l = last;
        final Node<E> newNode = new Node<>(l, e, null);
        last = newNode;
        if (l == null)
            first = newNode;
        else
            l.next = newNode;
        size++;
        modCount++;
    }

每次查询都要从头或尾进行遍历

public int lastIndexOf(Object o) {
        int index = size;
        if (o == null) {
            for (Node<E> x = last; x != null; x = x.prev) {
                index--;
                if (x.item == null)
                    return index;
            }
        } else {
            for (Node<E> x = last; x != null; x = x.prev) {
                index--;
                if (o.equals(x.item))
                    return index;
            }
        }
        return -1;
    }

public int indexOf(Object o) {
        int index = 0;
        if (o == null) {
            for (Node<E> x = first; x != null; x = x.next) {
                if (x.item == null)
                    return index;
                index++;
            }
        } else {
            for (Node<E> x = first; x != null; x = x.next) {
                if (o.equals(x.item))
                    return index;
                index++;
            }
        }
        return -1;
    }

就这样吧！！！！