java.util.Stack类简介（栈）

Stack是一个后进先出（last in first out，LIFO）的堆栈，在Vector类的基础上扩展5个方法而来

Deque（双端队列）比起stack具有更好的完整性和一致性，应该被优先使用

 E push(E item) 
          把项压入堆栈顶部。 
 E pop() 
          移除堆栈顶部的对象，并作为此函数的值返回该对象。 
 E peek() 
          查看堆栈顶部的对象，但不从堆栈中移除它。 
 boolean empty() 
          测试堆栈是否为空。  
 int search(Object o) 
          返回对象在堆栈中的位置，以 1 为基数。

Stack本身通过扩展Vector而来，而Vector本身是一个可增长的对象数组（ a growable array of objects）那么这个数组的哪里作为Stack的栈顶，哪里作为Stack的栈底？

答案只能从源代码中寻找，jdk1.6

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).
     * @exception  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).
     * @exception  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;
}

通过peek()方法注释The object at the top of this stack (the last item of the Vector object，可以发现

数组（Vector）的最后一位即为Stack的栈顶

pop、peek以及search方法本身进行了同步

push方法调用了父类的addElement方法

empty方法调用了父类的size方法

Vector类为线程安全类

综上，Stack类为线程安全类(多个方法调用而产生的数据不一致问题属于原子性问题的范畴)

public class StackTest {
    public static void main(String[] args) {
        Stack<String> s = new Stack<String>();
        System.out.println("------isEmpty");  
        System.out.println(s.isEmpty());  
        System.out.println("------push");  
        s.push("1");  
        s.push("2");  
        s.push("3");  
        StackTest.t(s);  
        System.out.println("------pop");  
        String str = s.pop();  
        System.out.println(str);  
        StackTest.t(s);  
        System.out.println("------peek");  
        str = s.peek();  
        System.out.println(str);  
        StackTest.t(s);  
        System.out.println("------search");  
        int i = s.search("2");  
        System.out.println(i);  
        i = s.search("1");  
        System.out.println(i);  
        i = s.search("none");  
        System.out.println(i); 
    }
    
    public static void t(Stack<String> s){
        System.out.println("iterator");
        Iterator<String> it = s.iterator();
        while(it.hasNext()){
            System.out.print(it.next()+"    ");
        }
        System.out.println("
");
    }
}

结果：

------isEmpty
true          
------push
iterator:1;2;3;  
------pop
3        --栈顶是数组最后一个
iterator:1;2;
------peek
2        --pop取后删掉，peek只取不删
iterator:1;2;
------search    
1        --以1为基数，即栈顶为1
2        --和栈顶见的距离为2-1=1
-1        --不存在于栈中

Stack并不要求其中保存数据的唯一性，当Stack中有多个相同的item时，调用search方法，只返回与查找对象equal并且离栈顶最近的item与栈顶间距离（见源码中search方法说明）