java队列实现方法(顺序队列,链式队列,循环队列)

package lang;import java.io.Serializable;import java.util.Arrays;public class ArrayQueue<T> implements Serializable{  private static final long serialVersionUID = 7333344126529379197L; private int DEFAULT_SIZE = 10; private int capacity;//保存数组的长度 private Object[] elementData;//定义一个数组用于保存顺序队列的元素 private int front = 0;//队头 private int rear = 0;//队尾 //以默认数组长度创建空顺序队列 public ArrayQueue() {  capacity = DEFAULT_SIZE;  elementData = new Object[capacity]; } //以一个初始化元素来创建顺序队列 public ArrayQueue(T element) {  this();  elementData[0] = element;  rear++; }  public ArrayQueue(int initSize) {  elementData = new Object[initSize]; }  public ArrayQueue(T element, int initSize) {  this.capacity = initSize;  elementData = new Object[capacity];  elementData[0] = element;  rear++; }  public int size() {  return rear - front; }  public void offer(T element) {  ensureCapacity(rear + 1);  elementData[rear++] = element; }  private void ensureCapacity(int minCapacity) {  //如果数组的原有长度小于目前所需的长度  int oldCapacity = elementData.length;  if (minCapacity > oldCapacity) {   int newCapacity = (oldCapacity * 3) / 2 + 1;   if (newCapacity < minCapacity)    newCapacity = minCapacity;   // minCapacity is usually close to size, so this is a win:   elementData = Arrays.copyOf(elementData, newCapacity);  } }  public T poll() {  if (isEmpty()) {   throw new IndexOutOfBoundsException("空队列异常");  }  //保留队列的front端的元素的值  T oldValue = (T) elementData[front];  //释放队列的front端的元素  elementData[front++] = null;  return oldValue; }  public T peek() {  if (isEmpty()) {   throw new IndexOutOfBoundsException("空队列异常");  }  return (T) elementData[front]; }  public boolean isEmpty() {  return rear == front; }  public void clear() {  //将底层数组所有元素赋为null  Arrays.fill(elementData, null);  front = 0;  rear = 0; } public String toString() {  if (isEmpty()) {   return "[]";  } else {   StringBuilder sb = new StringBuilder("[");   for (int i = front; i < rear; i++) {    sb.append(elementData[i].toString() + ", ");   }   int len = sb.length();   return sb.delete(len - 2, len).append("]").toString();  } }}