Java list与set中contains()方法效率案例详解

// ArrayList 中的方法
public boolean contains(Object o) {
      return indexOf(o) >= 0;
}
 
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;
}

// HashSet 中的方法
public boolean add(E e) {
	 // PRESENT 是一个object对象
   return map.put(e, PRESENT)==null;
}
public boolean contains(Object o) {
      return map.containsKey(o);
}


//  HashMap 中的方法
public boolean containsKey(Object key) {
  	  return getNode(hash(key), key) != null;
}

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;
}
//  getNode 方法同样也被hashMap中的get方法所调用
public V get(Object key) {
        Node<K,V> e;
        return (e = getNode(hash(key), key)) == null ? null : e.value;
}