java优先级队列如何设想工作？ [英] How java priority Queue is suppose to work?

问题描述

短篇小说，我正在执行一个图表，现​​在我在Kruskal上工作，我需要一个优先级队列。我对优先级队列的定义是，具有最小密钥的元素会先来吗？这是错的吗因为当我在队列中插入加权边缘（或数字）时，它们不会被排序。

Short story, I'm implementing a graph and now I'm working on the Kruskal, I need a priority queue. My definition of a priority queue is that the element with the smallest key would come first? Is this wrong? Because when I insert the weighted edges(or numbers) in the queue they don't end up sorted.

PriorityQueue<Integer> tja = new PriorityQueue<Integer>(); 
tja.add(55);
tja.add(99); 
tja.add(1); 
tja.add(102);
tja.add(54);
tja.add(51);
System.out.println(tja);

这将打印出来; [1，54，51，102，99，55]。这不是像我想要的那样排序！是的，我做了一个执行者，进入优先级队列，从边缘对象中提取数字，并根据该int进行比较。所以这应该是有效的，还是我完全误解了这个数据结构如何运作的整个概念？

That would print out this; [1, 54, 51, 102, 99, 55]. This is not sorted like I want them to be! And yes I made a comperator that goes into the priority queue that extracts the number from the edge object and compares based on that int. So this should work, or have I just completely misunderstood the entire concept of how this data structure works?

推荐答案

System.out.println正在调用使用迭代器的toString（）方法，不能保证尊重自然的订购。从文档：提供的迭代器方法iterator（）不能保证以任何特定顺序遍历优先级队列的元素。

System.out.println is invoking the toString() method, which is using the iterator, which is not guaranteed to respect the natural ordering. From the docs: "The Iterator provided in method iterator() is not guaranteed to traverse the elements of the priority queue in any particular order."

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