java.util.Stack类简介
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 Test {
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");
Test.it(s);
System.out.println("------pop");
String str = s.pop();
System.out.println(str);
Test.it(s);
System.out.println("------peek");
str = s.peek();
System.out.println(str);
Test.it(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 it(Stack<String> s){
System.out.print("iterator:");
Iterator<String> it = s.iterator();
while(it.hasNext()){
System.out.print(it.next()+";");
}
System.out.print("\n");
}
}
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）
