java stack代替 java stack 包

前言：

记录在阅读算法 第四版 谢路云译时的疑惑和解惑思路

在1.3背包、队列、和栈 的答疑部分时的问答

文章主体：

问 Java标准库中有栈和队列吗？

答 有，也没有。Java中有一个内置的库，叫做java.util.Stack ,但你需要栈的时候请不要使用它。它新增了几个一般不属于栈的方法，例如获取第一个i元素。它还允许从栈底添加元素（而非栈顶），所以他可以当作队列来使用！ 尽管拥有这些额外的操作看起来很有用，但它们其实是累赘。我们使用某种数据结构类型不仅仅是为了获得我们能够想象的各种操作，也是为了准确地指定我们所需要的操作。这么做的主要好处在于系统能够防止我们执行一些意外的操作。java.util.Stack API 是宽接口的一个典型例子，我们通常会极力避免出现这种情况。

疑惑

看到书中推荐说不要使用它，作为Java自身的内置库我还是十分的信任，所以带着疑惑去查看了源码。

/*
 * Copyright (c) 1994, 2010, Oracle and/or its affiliates. All rights reserved.
 * ORACLE PROPRIETARY/CONFIDENTIAL. Use is subject to license terms.
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package java.util;

/**
 * The <code>Stack</code> class represents a last-in-first-out
 * (LIFO) stack of objects. It extends class <tt>Vector</tt> with five
 * operations that allow a vector to be treated as a stack. The usual
 * <tt>push</tt> and <tt>pop</tt> operations are provided, as well as a
 * method to <tt>peek</tt> at the top item on the stack, a method to test
 * for whether the stack is <tt>empty</tt>, and a method to <tt>search</tt>
 * the stack for an item and discover how far it is from the top.
 * <p>
 * When a stack is first created, it contains no items.
 *
 * <p>A more complete and consistent set of LIFO stack operations is
 * provided by the {@link Deque} interface and its implementations, which
 * should be used in preference to this class.  For example:
 * <pre>   {@code
 *   Deque<Integer> stack = new ArrayDeque<Integer>();}</pre>
 *
 * @author  Jonathan Payne
 * @since   JDK1.0
 */
public
class Stack<E> extends Vector<E> {
    /**
     * Creates an empty Stack.
     */
    public Stack() {
    }

    /**
     * Pushes an item onto the top of this stack. This has exactly
     * the same effect as:
     * <blockquote><pre>
     * addElement(item)</pre></blockquote>
     *
     * @param   item   the item to be pushed onto this stack.
     * @return  the <code>item</code> argument.
     * @see     java.util.Vector#addElement
     */
    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;
    }

    /** use serialVersionUID from JDK 1.0.2 for interoperability */
    private static final long serialVersionUID = 1224463164541339165L;
}

可以看到，源码之中只有6个方法

public Stack()
public E push(E item) 
public synchronized E pop()
public synchronized E peek() 
public boolean empty() 
public synchronized int search(Object o)

并没有暴露其他操作方法。

于是我使用直接写代码测试，一敲代码就发现，之前走入了误区，并且明白了它指的宽接口含义。

关键句在：public class Stack<E> extends Vector<E>

它继承的是Vector 一个自动增长的对象数组 。

所以也就继承了它的操作，同时它的方法也是基于vector实现的。

然后也就越发觉得书中这句话的精妙了

尽管拥有这些额外的操作看起来很有用，但它们其实是累赘。我们使用某种数据结构类型不仅仅是为了获得我们能够想象的各种操作，也是为了准确地指定我们所需要的操作。这么做的主要好处在于系统能够防止我们执行一些意外的操作。

如有不正确之处，还望指出 0 0