
解决队列出队时间复杂度O(n) –> O(1)
int getSize(); // O(1)
boolean isEmpty(); // O(1)
void enqueue(E e); // O(1) 均摊
E dequeue(); // O(1) 均摊
E getFront(); // O(1)
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public class LoopQueue<E> implements Queue<E> { private E[] data; private int front, tail; private int size;
public LoopQueue(int capacity) { data = (E[]) new Object[capacity + 1]; front = 0; tail = 0; size = 0; }
public LoopQueue() { this(10); }
public int getCapacity() { return data.length - 1; }
@Override public boolean isEmpty() { return front == tail; }
@Override public int getSize() { return size; }
@Override public void enqueue(E e) { if ((tail + 1) % data.length == front) { resize(getCapacity() * 2); } data[tail] = e; tail = (tail + 1) % data.length; size++; }
@Override public E dequeue() { if (isEmpty()) { throw new IllegalArgumentException("队列为空"); } E ret = data[front]; data[front] = null; front = (front + 1) % data.length; size--; if (size == getCapacity() / 4 && getCapacity() / 2 != 0) { resize(getCapacity() / 2); } return ret; }
@Override public E getFront() { if (isEmpty()) { throw new IllegalArgumentException("队列为空"); } return data[front]; }
private void resize(int newCapacity) { E[] newData = (E[]) new Object[newCapacity + 1]; for (int i = 0; i < size; i++) { newData[i] = data[(i + front) % data.length]; } data = newData; front = 0; tail = size; }
@Override public String toString() { StringBuilder res = new StringBuilder(); res.append(String.format("Queue: size = %d , capacity = %dn", size, getCapacity())); res.append("front ["); for (int i = front; i != tail; i = (i + 1) % data.length) { res.append(data[i]); if ((i + 1) % data.length != tail) { res.append(", "); } } res.append("] tail"); return res.toString(); } }
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