利用动态数组实现循环队列

利用动态数组实现循环队列

  1. 循环队列主要是为了解决利用动态数组作为底层实现时出队操作的时间复杂度为O(n)这一问题
  2. 基于动态数组的循环队列的实现
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package StacksAndQueues.LoopQueue;

/**
* @ Description: 循环动态队列的实现
* @ Date: Created in 21:56 11/07/2018
* @ Author: Anthony_Duan
*/
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 int getSize() {
return size;
}

@Override
public boolean isEmpty() {
return front == tail;
}

@Override
public void enqueue(E e) {
if ((tail + 1) % data.length == front) {
resize(getCapacity() * 2);
}
data[tail] = e;
tail = (tail + 1) % data.length;
size++;
}

/**
* 循环队列出队的均摊时间复杂度为O(1)
*
* @return
*/
@Override
public E dequeue() {
if (isEmpty()) {
throw new IllegalArgumentException("Cannot dequeue from an empty queue.");
}

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("Queue is empty.");
}
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 = %d\n", 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();
}


public static void main(String[] args) {
LoopQueue<Integer> queue = new LoopQueue<Integer>();
for (int i = 0; i < 10; i++) {
queue.enqueue(i);
System.out.println(queue);
if (1 % 3 == 2) {
queue.dequeue();
System.out.println(queue);
}
}
}
}
  1. 循环队列复杂度分析
  • void enqueue(E e);————O(1)均摊
  • E dequeue();——————O(1)均摊
  • E getFront();——————O(1)
  • int getSize();——————O(1)
  • boolean isEmpty();———O(1)
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