TreeSet表现得很奇怪

Question

我遇到了TreeSet(sortedNodes)和ArrayList(nodes)的奇怪问题.在我的程序中,我有一个从Event Dispatch Thread(from ActionListener)这些行调用的方法:

        System.out.println("nodes: "+nodes.size());
        sortedNodes.addAll(nodes);
        System.out.println("sortedNodes: "+sortedNodes.size());

问题是,在某些集合上sortedNodes.size()返回的数字低于nodes.size()(在这3行上,因此内容没有变化nodes).当我打印内容时sortedNodes,除了不包含它应该包含的所有对象之外,它甚至没有排序.奇怪的是 - 如果我再次调用整个方法,它就解决了这个问题.我没有得到它 - 在相同的集合上执行相同的代码,但第一次它不起作用,第二次它没有.有任何想法吗？

编辑:如果我的问题不是很清楚,这应该有所帮助

exportTree();
exportTree();

pritns on output this:

nodes: 7
sortedNodes: 4
b2[23,57]a[46,97]b[65,77]c[43,43]

nodes: 7
sortedNodes: 7
a[46,97]b[65,77]b1[55,89]b2[23,57]b3[20,20]c[43,43]c1[99,88]

比较:

public class NodeComparator implements Comparator<Node>{
    public int compare(Node o1, Node o2) {
        return o1.getLabel().compareTo(o2.getLabel());
    }
}

节点:

public class Node {

private int order;
private String label;
private Integer[] keys;
private Node[] pointers;
private Node parent;

public Node(int order, String label, Integer[] keys, Node[] pointers) {
    this.order = order;
    this.label = label;
    this.parent = null;
    if (pointers == null) {
        this.pointers = new Node[order+1];
    } else {
        this.pointers = pointers;
    }
    if (keys == null) {
        this.keys = new Integer[order];
    } else {
        this.keys = keys;
    }
}

public Node getParent() {
    return parent;
}

public void setParent(Node parent) {
    this.parent = parent;
}

public Integer[] getKeys() {
    return keys;
}

public void setKeys(Integer[] keys) {
    this.keys = keys;
}

public String getLabel() {
    return label;
}

public void setLabel(String label) {
    this.label = label;
}

public int getOrder() {
    return order;
}

public void setOrder(int order) {
    this.order = order;
}

public Node[] getPointers() {
    return pointers;
}

public void setPointers(Node[] pointers) {
    this.pointers = pointers;
}

public Node getPointer(int i) {
    return pointers[i];
}

public void setPointer(int i, Node node) {
    pointers[i] = node;
}

public Integer getKey(int i) {
    return keys[i];
}

public void setKey(int i, Integer key) {
    keys[i] = key;
}
}

整个方法:

public void exportTree() {
    String graphInText = "";
    if (nodeShapes.isEmpty()) {
        graphInText = "empty";
    } else {
        char c = 'a';
        ArrayList<Node> nodes = new ArrayList<Node>();
        sortedNodes.clear();
        System.out.println("nodeShapes: " + nodeShapes.size());
        // populate the collection of nodes from their 2d representation(nodeShapes)
        // and label every root with different character
        for (NodeShape n : nodeShapes) {
            nodes.add(n.getNode());
            if (n.getParentLink() == null) {
                n.getNode().setLabel(c + "");
                c++;
            }
        }
        System.out.println("nodes: " + nodes.size());
        // copy all the nodes (currently every node except roots has label "0")
        sortedNodes.addAll(nodes);
        System.out.println("sortedNodes: " + sortedNodes.size());
        // check labels of every node; if it contains any of the characters
        // that were assigned to roots, use this label for every child of
        // this node and add number of the pointer at the end of the string;
        // when this is done, remove those nodes, which children have been 
        // labeled;
        // repeat whole procedure while there is no node left - and this will
        // happen, since every node is connected to a root or is a root;            
        while (!nodes.isEmpty()) {
            ArrayList<Node> nodesToBeRemoved = new ArrayList<Node>();
            for (Node n : nodes) {
                for (char c2 = 'a'; c2 <= c; c2++) {
                    if (n.getLabel().contains(c2 + "")) {
                        for (int i = 1; i <= n.getOrder(); i++) {
                            Node child = n.getPointer(i);
                            if (child != null) {
                                child.setLabel(n.getLabel() + i);
                            }
                        }
                        nodesToBeRemoved.add(n);
                    }
                }
            }
            if (!nodesToBeRemoved.isEmpty()) {
                nodes.removeAll(nodesToBeRemoved);
            }
        }
        Node[] nodesA = sortedNodes.toArray(new Node[sortedNodes.size()]);
        for (Node n : nodesA) {
            String nodeInText;
            nodeInText = n.getLabel() + "[";
            for (int i = 1; i < n.getOrder() - 1; i++) {
                nodeInText += n.getKey(i) + ",";
            }
            nodeInText += n.getKey(n.getOrder() - 1) + "]";
            graphInText += nodeInText;
        }

    }
    System.out.println(graphInText);
    label.setText(graphInText);
}

我也改变了程序,所以每次创建/删除NodeShape时,Node都被添加/删除到新集合中,我在exportTree()中使用了这个新集合而不是nodeShapes - 但它的工作方式相同,所以nodeShapes没有问题.它只是TreeSet ..当我不使用它时,一切正常(但我没有对我的节点进行排序).

Answer 1

TreeSet遵循Set语义,因此不允许重复.ArrayList允许重复,因此如果多次添加节点,它可能有比TreeSet更多的节点.

TreeSet使用Comparable语义排序.你的节点可以比较吗？你通过比较器告诉它如何排序？你应该.

TreeSet可以自然而然地为您完成基础,字符串和其他任何可比较的工作.

也许那些解释了你正在观察的一些行为.