在 Java 中从 TreeSet 中查找并返回一个元素
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Find and return an element from a TreeSet in Java
提问by Joarder Kamal
I have the following Node Class
我有以下节点类
Class Node {
private int id;
public int getId() {
return this.id;
}
}
and then create a TreeSet with the Nodes. Next I wanted to find and return a Node object based on id matching. However, every time the findNode() function is returning the next-to-next Node not the next one. I understand it is because of calling the iterator.next() twice. How can call it only once to check with the id value as well as return the object reference. I also tried with by creating a temporary Object reference but again it was the same result.
然后用节点创建一个 TreeSet。接下来我想根据 id 匹配查找并返回一个 Node 对象。但是，每次 findNode() 函数都返回下一个节点而不是下一个。我明白这是因为两次调用 iterator.next() 。如何仅调用一次以检查 id 值并返回对象引用。我也尝试过创建一个临时的 Object 引用，但结果还是一样。
Class NodeSet {
Set<Node> set = new TreeSet<Node>();
public Node findNode(int id) {
Iterator<Node> iterator = set.iterator();
while(iterator.hasNext()) {
if(iterator.next().getId() == id)
return iterator.next();
}
return null;
}
}
回答by Debasis
Edit: The solution proposed here is logarithmic(unlike the accepted answer) and hence is much faster when the number of nodes in a treeset is large.
编辑：这里提出的解决方案是对数的（与接受的答案不同），因此当树集中的节点数量很大时，速度要快得多。
There is no getmethod in the Setinterface which the class TreeSetimplements. Keeping in mind that there exists a complete ordering between the elements of a treeset, a three line hack is as follows:
在类TreeSet实现的Set接口中没有get方法。请记住，树集的元素之间存在完整的排序，三行 hack 如下：
Object search(TreeSet treeset, Object key) {
Object ceil = treeset.ceiling(key); // least elt >= key
Object floor = treeset.floor(key); // highest elt <= key
return ceil == floor? ceil : null;
}
回答by Reddy
Class NodeSet {
Set<Node> set = new TreeSet<Node>();
public Node findNode(int id) {
Iterator<Node> iterator = set.iterator();
while(iterator.hasNext()) {
Node node = iterator.next();
if(node.getId() == id)
return node;
}
return null;
}
}
回答by Mik378
The issue happens here:
问题发生在这里：
while(iterator.hasNext()) {
if(iterator.next().getId() == id)
return iterator.next();
}
You call twice iterator.nextin the same loop explaining the "next-to-next" issue.
您iterator.next在同一个循环中调用两次来解释“next-to-next”问题。
Make a local variable to still reach the same element or better: use the for loop if you have jdk >= 5:
制作一个局部变量以仍然到达相同的元素或更好：如果您有 jdk >= 5，请使用 for 循环：
for(Node node: set) {
if(node.getId() == id)
return node;
}
As @JB Nizet suggests in his comment above, a simple Mapalready implements your code logic by essence, and thus would better fit than a TreeSetand manual check.
More precisely, a TreeMapsorted on Nodes would be relevant. (since it sounds you need the order aspect)
正如@JB Nizet 在他上面的评论中所建议的那样，一个简单Map的本质上已经实现了您的代码逻辑，因此比TreeSet手动检查更适合。
更准确地说，TreeMap对Nodes 进行排序是相关的。（因为听起来你需要订单方面）
