Question

我正在使用BinaryNode实现二进制搜索树来存储数据。我在我的添加中使用CompareTo方法并包含确定项目所属的子树的方法。我在使用比较的各个地方不断收到此错误：

BST.java:50: error: cannot find symbol
            if (item.CompareTo(root.data) > 0)
                    ^
  symbol:   method CompareTo(T)
  location: variable item of type T
  where T is a type-variable:
    T extends Comparable<? super T> declared in class BST

这是我的代码，我做错了什么？

import java.util.List;
import java.util.ArrayList;
import java.util.*;
import java.io.*;

public class BST<T extends Comparable<? super T>> implements BSTInterface<T>
{
    private BinaryNode<T> root;
    private int numberOfItems;
    public List<T> preOrder = new ArrayList<T>();
    public List<T> inOrder = new ArrayList<T>();
    public List<T> postOrder = new ArrayList<T>();

    public BST()
    {
        root = null;
        numberOfItems = 0;
    }

    public BST(T rootData)
    {
        root = new BinaryNode<T>(rootData);
        numberOfItems = 1;
    }

    public BST(T rootData, BST<T> leftTree, BST<T> rightTree)
    {
        root = new BinaryNode<T>(rootData);
        numberOfItems = 1;
        root.left = leftTree.root;
        root.right = rightTree.root;
    }

    public void setTree(T rootData)
    {
        root = new BinaryNode<T>(rootData);
    }

    public boolean contains(T item)
    {
        if (root.data.equals(item))
            return true;

        else
        {
            if (item.CompareTo(root.data) > 0)
            {
                root = root.left;
                return contains(item);
            }

            else if (item.CompareTo(root.data) < 0)
            {
                root = root.right;
                return contains(item);
            }

            else
                return false;
        }
    }

    public void add(T newItem)
    {
        if (root == null)
        {
            root = new BinaryNode<T>(newItem);
            numberOfItems++;
        }

        if (newItem.equals(root.data))
            return;

        if (newItem.CompareTo(root.data) < 0)
        {
            root = root.left;
            add(newItem);
        }

        if (newItem.CompareTo(root.data) > 0)
        {
            root = root.right;
            add(newItem);
        }   
    }

Answer 1

这不是c＃：p

item.CompareTo()

change to:

item.compareTo()

compareTo来自Comparable界面。

Answer 2

如果您要实施java.lang.Comparable，则应该将该方法称为compareTo()，而不是CompareTo()。

为避免此类错误，建议您在方法前使用@Override注释。这样Eclipse就会告诉您，您没有覆盖现有方法。

Answer 3

[Comparable<T>][1]有以下签名

int     compareTo(T o)
Compares this object with the specified object for order.

但您正在使用CompareTo()请更正案例，然后再尝试

Answer 4

其compareTo()而非CompareTo() // you are using capital C which is wrong

的 int java.lang.Comparable.compareTo(? super T o)： - 的

Compares this object with the specified object for order. Returns a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object. The implementor must ensure sgn(x.compareTo(y)) == -sgn(y.compareTo(x)) for all x and y. (This implies that x.compareTo(y) must throw an exception iff y.compareTo(x) throws an exception.) The implementor must also ensure that the relation is transitive: (x.compareTo(y)>0 && y.compareTo(z)>0) implies x.compareTo(z)>0. Finally, the implementor must ensure that x.compareTo(y)==0 implies that sgn(x.compareTo(z)) == sgn(y.compareTo(z)), for all z. It is strongly recommended, but not strictly required that (x.compareTo(y)==0) == (x.equals(y)). Generally speaking, any class that implements the Comparable interface and violates this condition should clearly indicate this fact. The recommended language is "Note: this class has a natural ordering that is inconsistent with equals." In the foregoing description, the notation sgn(expression) designates the mathematical signum function, which is defined to return one of -1, 0, or 1 according to whether the value of expression is negative, zero or positive. Parameters: o - the object to be compared. Returns: a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object. Throws: ClassCastException - if the specified object's type prevents it from being compared to this object.

CompareTo方法不起作用

4 个答案: