( c u r + s t e p ) % l e n g t h (cur + step) \% length (cur+step)%length
cur 当前指针位置
step 前进步数
length 数组长度
注意：
如果 step = 1，也就是一次走一步，可以在 >= length 时重置为 0 即可
判断空
判断满
满之后的策略可以根据业务需求决定
例如我们要实现的环形队列，满之后就拒绝入队
代码
java复制代码
public class ArrayQueue<E> implements Queue<E>, Iterable<E>{
private int head = 0;
private int tail = 0;
private final E[] array;
private final int length;
@SuppressWarnings("all")
public ArrayQueue(int capacity) {
length = capacity + 1;
array = (E[]) new Object[length];
}
@Override
public boolean offer(E value) {
if (isFull()) {
return false;
}
array[tail] = value;
tail = (tail + 1) % length;
return true;
}
@Override
public E poll() {
if (isEmpty()) {
return null;
}
E value = array[head];
head = (head + 1) % length;
return value;
}
@Override
public E peek() {
if (isEmpty()) {
return null;
}
return array[head];
}
@Override
public boolean isEmpty() {
return tail == head;
}
@Override
public boolean isFull() {
return (tail + 1) % length == head;
}
@Override
public Iterator<E> iterator() {
return new Iterator<E>() {
int p = head;
@Override
public boolean hasNext() {
return p != tail;
}
@Override
public E next() {
E value = array[p];
p = (p + 1) % array.length;
return value;
}
};
}
}
判断空、满方法 2
引入 size
java复制代码
public class ArrayQueue2<E> implements Queue<E>, Iterable<E> {
private int head = 0;
private int tail = 0;
private final E[] array;
private final int capacity;
private int size = 0;
@SuppressWarnings("all")
public ArrayQueue2(int capacity) {
this.capacity = capacity;
array = (E[]) new Object[capacity];
}
@Override
public boolean offer(E value) {
if (isFull()) {
return false;
}
array[tail] = value;
tail = (tail + 1) % capacity;
size++;
return true;
}
@Override
public E poll() {
if (isEmpty()) {
return null;
}
E value = array[head];
head = (head + 1) % capacity;
size--;
return value;
}
@Override
public E peek() {
if (isEmpty()) {
return null;
}
return array[head];
}
@Override
public boolean isEmpty() {
return size == 0;
}
@Override
public boolean isFull() {
return size == capacity;
}
@Override
public Iterator<E> iterator() {
return new Iterator<E>() {
int p = head;
@Override
public boolean hasNext() {
return p != tail;
}
@Override
public E next() {
E value = array[p];
p = (p + 1) % capacity;
return value;
}
};
}
}
判断空、满方法 3
head 和 tail 不断递增，用到索引时，再用它们进行计算，两个问题
如何保证 head 和 tail 自增超过正整数最大值的正确性
如何让取模运算性能更高
答案：让 capacity 为 2 的幂
java复制代码
public class ArrayQueue3<E> implements Queue<E>, Iterable<E> {
private int head = 0;
private int tail = 0;
private final E[] array;
private final int capacity;
@SuppressWarnings("all")
public ArrayQueue3(int capacity) {
if ((capacity & capacity - 1) != 0) {
throw new IllegalArgumentException("capacity 必须为 2 的幂");
}
this.capacity = capacity;
array = (E[]) new Object[this.capacity];
}
@Override
public boolean offer(E value) {
if (isFull()) {
return false;
}
array[tail & capacity - 1] = value;
tail++;
return true;
}
@Override
public E poll() {
if (isEmpty()) {
return null;
}
E value = array[head & capacity - 1];
head++;
return value;
}
@Override
public E peek() {
if (isEmpty()) {
return null;
}
return array[head & capacity - 1];
}
@Override
public boolean isEmpty() {
return tail - head == 0;
}
@Override
public boolean isFull() {
return tail - head == capacity;
}
@Override
public Iterator<E> iterator() {
return new Iterator<E>() {
int p = head;
@Override
public boolean hasNext() {
return p != tail;
}
@Override
public E next() {
E value = array[p & capacity - 1];
p++;
return value;
}
};
}
}
