为什么java.util里的Stack不用LinkedList实现

当人们发现可以用LinkedList取代性能不佳的Stack后，Stack就被弱化了，没人觉得没有Stack活不了（或者是多大问题）。
另一方面，JDK要向后兼容，原来已存在的类不会被移除掉，那这个不被人们重视的喜爱的Stack就依然存在。
当然，你可以实现高性能的Stack提交到官方，保持跟原来的方法和行为一致就可以了。

stack是线程安全的，而用linkedlist实现可以是非线程安全，也可以是线程安全，具体可以靠自己的实现机制。
而stack基调已经定了，Vector也是线程安全的，所以选用了Vector。
而且根据源码发现：
stack用Vector只需要增加栈元素的入栈和弹出以及搜索就可以了：

public E push(E item) {
addElement(item);

    return item;
}

/**
 * Removes the object at the top of this stack and returns that
 * object as the value of this function.
 *
 * @return  The object at the top of this stack (the last item
 *          of the <tt>Vector</tt> object).
 * @throws  EmptyStackException  if this stack is empty.
 */
public synchronized E pop() {
    E       obj;
    int     len = size();

    obj = peek();
    removeElementAt(len - 1);

    return obj;
}

/**
 * Looks at the object at the top of this stack without removing it
 * from the stack.
 *
 * @return  the object at the top of this stack (the last item
 *          of the <tt>Vector</tt> object).
 * @throws  EmptyStackException  if this stack is empty.
 */
public synchronized E peek() {
    int     len = size();

    if (len == 0)
        throw new EmptyStackException();
    return elementAt(len - 1);
}

/**
 * Tests if this stack is empty.
 *
 * @return  <code>true</code> if and only if this stack contains
 *          no items; <code>false</code> otherwise.
 */
public boolean empty() {
    return size() == 0;
}

/**
 * Returns the 1-based position where an object is on this stack.
 * If the object <tt>o</tt> occurs as an item in this stack, this
 * method returns the distance from the top of the stack of the
 * occurrence nearest the top of the stack; the topmost item on the
 * stack is considered to be at distance <tt>1</tt>. The <tt>equals</tt>
 * method is used to compare <tt>o</tt> to the
 * items in this stack.
 *
 * @param   o   the desired object.
 * @return  the 1-based position from the top of the stack where
 *          the object is located; the return value <code>-1</code>
 *          indicates that the object is not on the stack.
 */
public synchronized int search(Object o) {
    int i = lastIndexOf(o);

    if (i >= 0) {
        return size() - i;
    }
    return -1;
}

而linkedlist是具有链表形式的数据结构，不能像Vector那样直接扩展。