import java.util.ArrayDeque; import java.util.Deque;  public class IntegerStack {   private Deque<Integer> data = new ArrayDeque<Integer>();    public void push(Integer element) {     data.addFirst(element);   }    public Integer pop() {     return data.removeFirst();   }    public Integer peek() {     return data.peekFirst();   }    public String toString() {     return data.toString();   }    public static void main(String[] args) {     IntegerStack stack = new IntegerStack();     for (int i = 0; i < 5; i++) {       stack.push(i);     }     System.out.println(stack);     System.out.println("After pushing 5 elements: " + stack);     int m = stack.pop();     System.out.println("Popped element = " + m);     System.out.println("After popping 1 element : " + stack);     int n = stack.peek();     System.out.println("Peeked element = " + n);     System.out.println("After peeking 1 element : " + stack);   } }  

private transient E[] elements;  private transient int head;  private transient int tail;  /*此處存放e的位置是從elements數(shù)組最后的位置開始存儲(chǔ)的*/  public void addFirst(E e) {     if (e == null)       throw new NullPointerException();     elements[head = (head - 1) & (elements.length - 1)] = e;//首次數(shù)組容量默認(rèn)為16，head=（0-1）&(16-1)=15     if (head == tail)       doubleCapacity();   }  /*每次擴(kuò)容都按插入的先后順序重新放入一個(gè)新的數(shù)組中，最新插入的放在數(shù)組的第一個(gè)位置。*/   private void doubleCapacity() {     assert head == tail;     int p = head;     int n = elements.length;     int r = n - p; // number of elements to the right of p     int newCapacity = n << 1;     if (newCapacity < 0)       throw new IllegalStateException("Sorry, deque too big");     Object[] a = new Object[newCapacity];     System.arraycopy(elements, p, a, 0, r);     System.arraycopy(elements, 0, a, r, p);     elements = (E[])a;     head = 0;     tail = n;   }    public E removeFirst() {     E x = pollFirst();     if (x == null)       throw new NoSuchElementException();     return x;   }    public E pollFirst() {     int h = head;     E result = elements[h]; // Element is null if deque empty     if (result == null)       return null;     elements[h] = null;   // 重新設(shè)置數(shù)組中的這個(gè)位置為null，方便垃圾回收。     head = (h + 1) & (elements.length - 1);//將head的值回退，相當(dāng)于將棧的指針又向下移動(dòng)一格。例如，12--〉13     return result;   }    public E peekFirst() {     return elements[head]; // elements[head] is null if deque empty   }