java.util.HashMap$Node cannot be cast to java.util.HashMap$TreeNode
时间: 2023-12-17 09:02:04 浏览: 56
这个错误通常出现在将一个非空的HashMap实例转换为TreeMap实例时。因为HashMap和TreeMap底层的数据结构不同,所以在进行转换时需要进行数据的重组。如果在转换过程中,HashMap中的节点类型不符合TreeMap的要求,就会出现这个错误。
解决方法可以是在转换之前,将HashMap中的数据先转换为一个List,然后再将List转换为TreeMap。这样可以避免节点类型不匹配的问题。具体代码实现可以参考下面的示例:
```
HashMap<Integer, String> hashMap = new HashMap<>();
hashMap.put(1, "one");
hashMap.put(2, "two");
hashMap.put(3, "three");
List<Map.Entry<Integer, String>> entryList = new ArrayList<>(hashMap.entrySet());
TreeMap<Integer, String> treeMap = new TreeMap<>(Comparator.comparingInt(Map.Entry::getKey));
entryList.forEach(entry -> treeMap.put(entry.getKey(), entry.getValue()));
```
相关问题
Java将List年级班级课程转为树形结构
您可以使用递归算法将List年级、班级和课程数据转换为树形结构。以下是一个示例代码:
```java
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class TreeNode {
private String name;
private List<TreeNode> children;
public TreeNode(String name) {
this.name = name;
this.children = new ArrayList<>();
}
public String getName() {
return name;
}
public List<TreeNode> getChildren() {
return children;
}
public void addChild(TreeNode child) {
children.add(child);
}
}
public class Main {
public static void main(String[] args) {
// 生成样例数据
Map<String, List<String>> gradeData = new HashMap<>();
gradeData.put("Grade 1", List.of("Class A", "Class B"));
gradeData.put("Grade 2", List.of("Class C", "Class D"));
gradeData.put("Grade 3", List.of("Class E"));
Map<String, List<String>> classData = new HashMap<>();
classData.put("Class A", List.of("Math", "Science"));
classData.put("Class B", List.of("English"));
classData.put("Class C", List.of("History", "Geography"));
classData.put("Class D", List.of("Physics", "Chemistry"));
classData.put("Class E", List.of("Art"));
// 转换为树形结构
TreeNode root = new TreeNode("School");
for (String grade : gradeData.keySet()) {
TreeNode gradeNode = new TreeNode(grade);
root.addChild(gradeNode);
List<String> classes = gradeData.get(grade);
for (String className : classes) {
TreeNode classNode = new TreeNode(className);
gradeNode.addChild(classNode);
List<String> courses = classData.get(className);
for (String course : courses) {
TreeNode courseNode = new TreeNode(course);
classNode.addChild(courseNode);
}
}
}
// 打印树形结构
printTree(root, 0);
}
private static void printTree(TreeNode node, int depth) {
StringBuilder indent = new StringBuilder();
for (int i = 0; i < depth; i++) {
indent.append(" ");
}
System.out.println(indent + node.getName());
for (TreeNode child : node.getChildren()) {
printTree(child, depth + 1);
}
}
}
```
以上示例代码将List年级、班级和课程数据转换为一个树形结构,并打印出来。您可以根据实际情况修改数据结构和打印方式。
决策树java简单代码运行
以下是一个简单的决策树 Java 代码示例:
``` java
import java.util.Scanner;
public class DecisionTree {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
// 创建决策树
TreeNode decisionTree = createDecisionTree();
// 开始询问用户
while (true) {
TreeNode current = decisionTree;
while (current.children != null) {
System.out.println(current.question);
String answer = input.nextLine();
current = current.children.get(answer);
}
// 打印最终答案
System.out.println(current.answer);
// 询问用户是否继续
System.out.println("是否继续?(y/n)");
String response = input.nextLine();
if (!response.equalsIgnoreCase("y")) {
break;
}
}
input.close();
}
// 创建决策树
private static TreeNode createDecisionTree() {
TreeNode root = new TreeNode("动物是哺乳动物吗?");
TreeNode node1 = new TreeNode("是");
TreeNode node2 = new TreeNode("有羽毛吗?");
TreeNode node3 = new TreeNode("是");
TreeNode node4 = new TreeNode("是鸟类");
TreeNode node5 = new TreeNode("不是鸟类");
root.children.put("是", node1);
root.children.put("否", node2);
node2.children.put("是", node3);
node2.children.put("否", node5);
node3.children.put("是", node4);
node1.answer = "是哺乳动物";
node4.answer = "是有羽毛的鸟类";
node5.answer = "不是鸟类";
return root;
}
}
class TreeNode {
String question; // 问题
String answer; // 答案
Map<String, TreeNode> children; // 子节点
public TreeNode(String question) {
this.question = question;
this.children = new HashMap<>();
}
}
```
这个示例创建了一个简单的决策树,用于确定一个动物是否是哺乳动物或鸟类。在程序运行时,它会逐步询问用户一系列问题,直到最终确定动物的分类。