Prioprity源码分析--priority如何实现优先级排序
程序员文章站
2022-04-21 21:11:32
目录本文围绕的核心问题:priority是怎么实现的根据优先级排序?Demo源码分析:offer方法siftUp方法siftUpUsingComparator方法接口Comparator中的compare方法(默认比较器)我自定义的compare方法siftUpComparable方法本文围绕的核心问题:priority是怎么实现的根据优先级排序?Demopublic class Demo { public static void main(St.....
目录
接口Comparator中的compare方法(默认比较器)
本文围绕的核心问题:priority是怎么实现优先级排序?
Demo
public class Demo {
public static void main(String[] args) {
//需求:priority中age越大优先级越高
PriorityQueue<Student> priorityQueue = new PriorityQueue<>(2, new StudentComparator());
Student s1 = new Student("jack", 2200);
Student s2 = new Student("Tom", 223);
priorityQueue.offer(s1);
priorityQueue.offer(s2);
Student poll = priorityQueue.poll();
System.out.println(poll.name+":"+poll.age);
}
}
public class StudentComparator implements Comparator<Student> {
@Override
public int compare(Student current, Student middle) {
if (current.age>middle.age)
return -1;
else if (current.age<middle.age)
return 1;
else
return 0;
}
}
public class Student {
String name;
int age;
public Student(String name,int age){
this.name=name;
this.age=age;
}
}
源码分析:
- 底层的队列依靠数组实现
- 内部有比较器,可以在构造函数中指定自定义的比较器
public class PriorityQueue<E> extends AbstractQueue<E>
implements java.io.Serializable {
...
/**
* Priority queue represented as a balanced binary heap: the two
* children of queue[n] are queue[2*n+1] and queue[2*(n+1)]. The
* priority queue is ordered by comparator, or by the elements'
* natural ordering, if comparator is null: For each node n in the
* heap and each descendant d of n, n <= d. The element with the
* lowest value is in queue[0], assuming the queue is nonempty.
*/
transient Object[] queue; // non-private to simplify nested class access
/**
* The number of elements in the priority queue.
*/
private int size = 0;
/**
* The comparator, or null if priority queue uses elements'
* natural ordering.
*/
private final Comparator<? super E> comparator;
...
public PriorityQueue(int initialCapacity,
Comparator<? super E> comparator) {
// Note: This restriction of at least one is not actually needed,
// but continues for 1.5 compatibility
if (initialCapacity < 1)
throw new IllegalArgumentException();
this.queue = new Object[initialCapacity];
this.comparator = comparator;
}
}
offer方法
public boolean offer(E e) {
if (e == null)
throw new NullPointerException();
modCount++;
//i记录当前数组长度
int i = size;
//越界扩容处理
if (i >= queue.length)
grow(i + 1);
//加入新元素e,长度+1
size = i + 1;
//第一个元素直接放置
if (i == 0)
queue[0] = e;
//之后加入队列的元素进行筛选
else
siftUp(i, e);
return true;
}
siftUp方法
private void siftUp(int k, E x) {
//判断是否有比较器
if (comparator != null)
//使用自定义的比较器进行插入
siftUpUsingComparator(k, x);
else
//使用默认比较器
siftUpComparable(k, x);
}
siftUpUsingComparator方法
private void siftUpUsingComparator(int k, E x) {
//从当前位置,不断的往前遍历,通过比较器修改每个位置
while (k > 0) {
//parent是中间位置的角标
int parent = (k - 1) >>> 1;
Object e = queue[parent];
//调用比较器的compare方法,比较当前要插入的元素和中间位置的大小关系
if (comparator.compare(x, (E) e) >= 0)
//默认情况下构建的是一个完全二叉树实现的小顶堆。
//我们这里通过自定义的比较器构建一个大顶堆,在比较器中需要设置当前元素小于中间元素返回1,才能跳出循环
break;
//否则把当前元素放置到中间位置,
queue[k] = e;
//k的值变成中间位置,方便后续不断的调整优先队列中前半部分保持优先级的降序
k = parent;
}
//确定好优先级位置后放置x
queue[k] = x;
}
接口Comparator中的compare方法(默认比较器)
public interface Comparator<T> {
/**
* Compares its two arguments for order. Returns a negative integer,
* zero, or a positive integer as the first argument is less than, equal
* to, or greater than the second.<p>
* 比较两个参数进行排序,返回负数,零,正整数,
* 分别代表第一个参数小于,等于,大于第二个参数
*...
*/
int compare(T o1, T o2);
}
自定义的compare方法
目标:(让队列最后是降序排序)
// 给age降序排序
public class StudentComparator implements Comparator<Student> {
@Override
public int compare(Student current, Student middle) {
if (current.age>middle.age)
return -1;
else if (current.age<middle.age)
return 1;
else
return 0;
}
}
siftUpComparable 方法
代码类似siftUpUsingComparator,使用的是默认比较器
private void siftUpComparable(int k, E x) {
Comparable<? super E> key = (Comparable<? super E>) x;
while (k > 0) {
int parent = (k - 1) >>> 1;
Object e = queue[parent];
if (key.compareTo((E) e) >= 0)
break;
queue[k] = e;
k = parent;
}
queue[k] = key;
}
本文地址:https://blog.csdn.net/qq_38783664/article/details/110471807