List 与 Set 的 contains方法比较

Posted 2021-09-06 深蓝

一、结论

1. set.contains("obj")的效率明显高于list.contains("obj")

主要比较了 ArrayList,LinkedList 和 HashSet

2. 主要原因是 List 底层是通过遍历的方式去作比较，而 Set 是算key的hash值的形式与集合内元素比较

二、源码

1. ArrayList

ArrayList<Object> list = new ArrayList<>();
list.contains("");

// 调用indexof
    /**
     * Returns true if this list contains the specified element. 
     * More formally, returns true if and only if this list contains at least one element e such that (o==null ? e==null : o.equals(e)).
     * 参数: o – element whose presence in this list is to be tested
     * 返回: true if this list contains the specified element
     */
public boolean contains(Object o) {
    return indexOf(o) >= 0;
}

/**
 * Returns the index of the first occurrence of the specified element in this list, 
 * or -1 if this list does not contain the element. 
 * More formally, returns the lowest index i such that (o==null ? get(i)==null : o.equals(get(i))),
 * or -1 if there is no such index.
 */
public int indexOf(Object o) {
    if (o == null) {
        for (int i = 0; i < size; i++)
            if (elementData[i]==null)
                return i;
    } else {
        for (int i = 0; i < size; i++)
            if (o.equals(elementData[i]))
                return i;
    }
    return -1;
}

2. LinkedList

// 类似 ArrayList
LinkedList<Object> linkedList = new LinkedList<>();
linkedList.contains("");

public boolean contains(Object o) {
    return indexOf(o) != -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;
}

3. HashSet

HashSet<Object> set = new HashSet<>();
set.contains("");

/**
 * Returns true if this set contains the specified element. More formally, returns true if and only if this set contains an element e such that (o==null ? e==null : o.equals(e)).
 * 参数: o – element whose presence in this set is to be tested
 * 返回: true if this set contains the specified element
 */
public boolean contains(Object o) {
    return map.containsKey(o);
}

// 下面就是高效率的关键点，算hash二不是一个个比对
/**
 * Returns true if this map contains a mapping for the specified key.
 * 参数: key – The key whose presence in this map is to be tested
 * 返回: true if this map contains a mapping for the specified key.
 */
public boolean containsKey(Object key) {
    return getNode(hash(key), key) != null;
}

/**
 * 计算key.hashCode()并将哈希值的高位扩展为低位。
 */
static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

/**
 * Implements Map.get and related methods
 * 先算比对hash 再 equals
 * 参数: hash – hash for key ,key – the key
 * 返回: the node, or null if none
 */
final Node<K,V> getNode(int hash, Object key) {
    Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (first = tab[(n - 1) & hash]) != null) {
        if (first.hash == hash && // always check first node
            ((k = first.key) == key || (key != null && key.equals(k))))
            return first;
        if ((e = first.next) != null) {
            if (first instanceof TreeNode)
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);
            do {
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    return e;
            } while ((e = e.next) != null);
        }
    }
    return null;
}